MathsCBSE Class 10 · Chapter 3

Pair of Linear Equations in Two Variables Class 10 Notes

Updated 13 October 2026 · Fact-checked

A pair of linear equations in two variables is two equations of the form ax + by + c = 0 solved together. Each is a straight line, so the pair has one solution, none or infinitely many. The board asks you to test consistency using coefficient ratios, solve by graph, substitution or elimination, and model word problems.

What Pair of Linear Equations in Two Variables is about

This chapter is about two straight-line equations in the same two unknowns, $x$ and $y$ , solved together. A solution is a point $(x, y)$ that lies on both lines. You learn to find it by drawing the lines (graphical method) and by algebra (substitution and elimination). You also learn to tell, just from the coefficients, whether the pair has one solution, none, or infinitely many.

It builds on Class 9 linear equations in two variables and their graphs. It links to Polynomials (a linear polynomial gives a straight-line graph), to Coordinate Geometry (plotting points, intercepts, areas of triangles on a grid), and to Quadratic Equations, where you again turn word problems into equations.

Board questions take a few clear shapes: MCQs and assertion-reason items on the ratio conditions, find- $k$ questions for a given number of solutions, solve-and-hence questions using substitution or elimination, graph questions that also ask for the vertices or area of a triangle, and word problems on ages, digits, fractions, costs and fares. These often appear as case-based questions.

Weightage: This chapter is part of Unit II, Algebra, which carries 20 of the 80 theory marks together with Polynomials, Quadratic Equations and Arithmetic Progressions.Source: CBSE Mathematics syllabus 2026-27, Unit II Algebra

In real board papers: Standard, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.6 marks per paper; Basic, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.1 marks per paper. See the questions

Key concepts

Pair of linear equations and what a solution means

NCERT section 3.1 Introduction

A linear equation in two variables has the form $a x + b y + c = 0$ , where $a$ , $b$ , $c$ are real numbers and $a$ , $b$ are not both zero. On its own it has infinitely many solutions: every point on its straight-line graph satisfies it.

A pair of linear equations is two such equations in the same two variables:

$a_{1} x + b_{1} y + c_{1} = 0$
$a_{2} x + b_{2} y + c_{2} = 0$

A solution of the pair is an ordered pair $(x, y)$ that satisfies both equations at the same time. Geometrically, it is a point common to both lines.

To test a point, substitute it into each equation separately. If even one equation fails, the point is not a solution of the pair. Before comparing coefficients, always write both equations in the same form. This avoids sign errors in later work.

Worked example

(i) Verify that $(2, - 1)$ is a solution of $3 x - 4 y = 10$ .
(ii) Find $k$ so that $(2, - 1)$ is also a solution of $k x + 3 y = 7$ .
(iii) Explain why $(2, - 1)$ is then the only solution of the pair formed.

Try it first, then show the solutionSolution

(i) LHS $= 3 (2) - 4 (- 1) = 6 + 4 = 10 =$ RHS, so $(2, - 1)$ satisfies $3 x - 4 y = 10$ .
(ii) Substitute $x = 2$ , $y = - 1$ in $k x + 3 y = 7$ : $2 k - 3 = 7$ .
$2 k = 10$ , so $k = 5$ .
(iii) The pair is $3 x - 4 y = 10$ and $5 x + 3 y = 7$ .
$\frac{a _{1}}{a _{2}} = \frac{3}{5}$ and $\frac{b _{1}}{b _{2}} = \frac{- 4}{3}$ , and these are unequal.
So the lines intersect in exactly one point. That point is $(2, - 1)$ , since it lies on both lines.

Answer: $k = 5$ . The pair $3 x - 4 y = 10$ , $5 x + 3 y = 7$ has the unique solution $x = 2$ , $y = - 1$ .

Graphical method of solution

NCERT section 3.2 Graphical Method of Solution of a Pair of Linear Equations

Each linear equation is a straight line, so you can solve a pair by drawing both lines on the same axes. The point where they cross lies on both lines, so its coordinates are the solution.

Method:

Make a table of at least two points for each line. A third point is your check: if it is not in line, you have made a mistake.
Choosing $x = 0$ and $y = 0$ gives the intercepts, which are quick to plot.
Use a clear scale, label each line with its equation, and mark the intersection point.
Write the answer as $x = \dots$ , $y = \dots$ .

Boards often extend this. They ask for the vertices of the triangle formed by the two lines and an axis, or its area. The vertices are the intersection point and the points where each line meets that axis.

Worked example

Solve graphically: $x + y = 5$ and $2 x - y = 4$ . Also find the vertices and area of the triangle formed by these lines and the $x$ -axis.

Try it first, then show the solutionSolution

For $x + y = 5$ : points $(0, 5)$ , $(5, 0)$ , $(2, 3)$ .
For $2 x - y = 4$ , that is $y = 2 x - 4$ : points $(0, - 4)$ , $(2, 0)$ , $(3, 2)$ .
Plot both sets of points on the same axes and join each set with a straight line.
The lines cross at $(3, 2)$ . Check: $3 + 2 = 5$ and $2 (3) - 2 = 4$ .
Line $x + y = 5$ meets the $x$ -axis at $(5, 0)$ . Line $2 x - y = 4$ meets it at $(2, 0)$ .
The vertices are $(3, 2)$ , $(2, 0)$ and $(5, 0)$ .
Base on the $x$ -axis $= 5 - 2 = 3$ units. Height = $y$ -coordinate of $(3, 2)$ $= 2$ units.
Area $= \frac{1}{2} \times 3 \times 2 = 3$ square units.

Answer: $x = 3$ , $y = 2$ . The vertices are $(3, 2)$ , $(2, 0)$ , $(5, 0)$ and the area is $3$ square units.

Consistent, inconsistent and dependent pairs

NCERT section 3.2 Graphical Method of Solution of a Pair of Linear Equations

Two lines in a plane can behave in only three ways, so a pair of linear equations has only three possible outcomes.

Intersecting lines: exactly one common point, so a unique solution. The pair is consistent.
Coincident lines: one line lies on top of the other, so every point is common and there are infinitely many solutions. The pair is consistent and also called dependent, because one equation is a multiple of the other.
Parallel lines: they never meet, so there is no solution. The pair is inconsistent.

So consistent means at least one solution, and inconsistent means none. A dependent pair is always consistent; do not mix these words up. In a graph question, state both the shape (parallel, coincident or intersecting) and the word (consistent or inconsistent).

Worked example

Draw the graphs of $2 x + 3 y = 6$ and $4 x + 6 y = 18$ . State whether the pair is consistent or inconsistent, and confirm your answer using coefficient ratios.

Try it first, then show the solutionSolution

For $2 x + 3 y = 6$ : points $(0, 2)$ , $(3, 0)$ , $(- 3, 4)$ .
$4 x + 6 y = 18$ simplifies to $2 x + 3 y = 9$ : points $(0, 3)$ , $(3, 1)$ , $(- 3, 5)$ .
Plot and join each set. Each line drops 2 units for every 3 units to the right, so the two lines are parallel and never meet.
No point lies on both lines, so the pair has no solution.
Check: $\frac{a _{1}}{a _{2}} = \frac{2}{4} = \frac{1}{2}$ , $\frac{b _{1}}{b _{2}} = \frac{3}{6} = \frac{1}{2}$ , $\frac{c _{1}}{c _{2}} = \frac{6}{18} = \frac{1}{3}$ .
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , which confirms parallel lines.

Answer: The lines are parallel, so the pair is inconsistent and has no solution.

Algebraic conditions for the number of solutions

NCERT section 3.2 Graphical Method of Solution of a Pair of Linear Equations

You can predict the outcome without drawing anything. For $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ :

$\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ : intersecting lines, unique solution.
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ : coincident lines, infinitely many solutions.
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ : parallel lines, no solution.

Why it works. Suppose $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = t$ . Then the $x$ and $y$ part of the first equation is exactly $t$ times that of the second. If the constants are also in ratio $t$ , the first equation is just $t$ times the second, so it is the same line. If the constants are not in ratio $t$ , the same expression would have to equal two different numbers at once, which is impossible.

In find- $k$ questions for infinitely many or no solutions, always test the constant ratio as well.

Worked example

Find the value of $k$ for which the pair $x + (k + 1) y = 5$ and $(k + 1) x + 9 y = 8 k - 1$ has infinitely many solutions.

Try it first, then show the solutionSolution

Both equations are in the form $a x + b y = c$ , so compare: $a_{1} = 1$ , $b_{1} = k + 1$ , $c_{1} = 5$ ; $a_{2} = k + 1$ , $b_{2} = 9$ , $c_{2} = 8 k - 1$ .
For infinitely many solutions: $\frac{1}{k + 1} = \frac{k + 1}{9} = \frac{5}{8 k - 1}$ .
From the first two ratios: $(k + 1)^{2} = 9$ , so $k + 1 = \pm 3$ and $k = 2$ or $k = - 4$ .
Test $k = 2$ : the ratios are $\frac{1}{3}$ , $\frac{3}{9} = \frac{1}{3}$ and $\frac{5}{15} = \frac{1}{3}$ . All are equal.
Test $k = - 4$ : $\frac{1}{- 3} = - \frac{1}{3}$ and $\frac{- 3}{9} = - \frac{1}{3}$ , but $\frac{5}{- 33} \neq = - \frac{1}{3}$ . This value gives parallel lines (no solution), so reject it.

Answer: $k = 2$

Substitution method

NCERT section 3.3.1 Substitution Method

In substitution, you use one equation to write one variable in terms of the other, then put that expression into the second equation. This leaves one equation in one variable, which you can solve directly.

Why it works. At the solution, both equations are true for the same $x$ and $y$ . So whatever $y$ equals in the first equation, it must equal the same thing in the second.

Steps:

Pick the variable with coefficient $1$ or $- 1$ if you can. This avoids fractions.
Express it, for example $y = 3 x - 7$ , and label it as equation (3).
Substitute into the other equation, not the one you just used.
Solve for the first variable, then back-substitute into (3).
Check in the equation you did not use.

If the variable disappears and you get a true statement like $0 = 0$ , there are infinitely many solutions. A false statement like $0 = 5$ means no solution.

Worked example

Solve $3 x - y = 7$ and $2 x + 5 y = - 1$ by substitution. Hence find $m$ if the solution satisfies $y = m x + 3$ .

Try it first, then show the solutionSolution

From $3 x - y = 7$ : $y = 3 x - 7$ ...(3)
Substitute (3) in $2 x + 5 y = - 1$ : $2 x + 5 (3 x - 7) = - 1$ .
$2 x + 15 x - 35 = - 1$ , so $17 x = 34$ and $x = 2$ .
From (3): $y = 3 (2) - 7 = - 1$ .
Check in $2 x + 5 y = - 1$ : $2 (2) + 5 (- 1) = 4 - 5 = - 1$ . Correct.
Substitute in $y = m x + 3$ : $- 1 = 2 m + 3$ , so $2 m = - 4$ and $m = - 2$ .

Answer: $x = 2$ , $y = - 1$ and $m = - 2$

Elimination method

NCERT section 3.3.2 Elimination Method

In elimination, you make the coefficients of one variable equal in size, then add or subtract the equations so that variable cancels.

Why it works. If two equations are true, then their sum and their difference are also true. Multiplying an equation by a non-zero number does not change its line, so the solution is unchanged.

Steps:

Multiply one or both equations so that one variable has the same coefficient, ignoring sign. Using the LCM keeps the numbers small.
Subtract if the signs are the same. Add if the signs are opposite.
Solve for the remaining variable, then substitute back.

A useful shortcut: when the coefficients are swapped, as in $a x + b y = c$ and $b x + a y = d$ , add and subtract the equations directly. You get simple values for $x + y$ and $x - y$ .

If both variables cancel, read the result. $0 = 0$ means infinitely many solutions. $0 =$ a non-zero number means no solution.

Worked example

Solve: $47 x + 31 y = 63$ and $31 x + 47 y = 15$ .

Try it first, then show the solutionSolution

The coefficients are swapped, so add the equations: $78 x + 78 y = 78$ .
Divide by $78$ : $x + y = 1$ ...(3)
Subtract the second equation from the first: $16 x - 16 y = 48$ .
Divide by $16$ : $x - y = 3$ ...(4)
Add (3) and (4): $2 x = 4$ , so $x = 2$ .
From (3): $y = 1 - 2 = - 1$ .
Check: $47 (2) + 31 (- 1) = 94 - 31 = 63$ and $31 (2) + 47 (- 1) = 62 - 47 = 15$ . Both are correct.

Answer: $x = 2$ , $y = - 1$

Situational (word) problems

NCERT section 3.3 Algebraic Methods of Solving a Pair of Linear Equations

Many board questions describe a real situation and leave you to build the equations. Work in four stages.

Define the unknowns clearly, for example: let the numerator be $x$ and the denominator be $y$ .
Translate each condition into an equation, one sentence at a time.
Solve by substitution or elimination.
Check your answer against the original words, not just your equations, and state it with units.

Standard set-ups:

Ages: ' $t$ years ago' means subtract $t$ from every person's age.
Two-digit number with tens digit $x$ and units digit $y$ : the number is $10 x + y$ and the reversed number is $10 y + x$ .
Fractions: write the fraction as $\frac{x}{y}$ .
Costs and fares: fixed charge plus rate times quantity.
Geometry: for example, the angles of a triangle add up to $18 0^{\circ}$ .

Most lost marks come from the translation step, so write each equation next to the sentence it comes from.

Worked example

A fraction becomes $\frac{1}{2}$ if $1$ is added to its numerator and $4$ is added to its denominator. It becomes $\frac{2}{5}$ if $3$ is subtracted from both its numerator and denominator. Find the fraction.

Try it first, then show the solutionSolution

Let the fraction be $\frac{x}{y}$ .
First condition: $\frac{x + 1}{y + 4} = \frac{1}{2}$ , so $2 x + 2 = y + 4$ , which gives $2 x - y = 2$ ...(1)
Second condition: $\frac{x - 3}{y - 3} = \frac{2}{5}$ , so $5 x - 15 = 2 y - 6$ , which gives $5 x - 2 y = 9$ ...(2)
From (1): $y = 2 x - 2$ .
Substitute in (2): $5 x - 2 (2 x - 2) = 9$ , so $x + 4 = 9$ and $x = 5$ .
$y = 2 (5) - 2 = 8$ .
Check: $\frac{5 + 1}{8 + 4} = \frac{6}{12} = \frac{1}{2}$ and $\frac{5 - 3}{8 - 3} = \frac{2}{5}$ . Both conditions hold.

Answer: The fraction is $\frac{5}{8}$ .

Maths in real life: Pair of Linear Equations in Two Variables

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Graphical method of solution

Imagine two water tanks on your school roof. Tank A holds 600 litres and is being drained at 40 litres per minute for cleaning. At the same moment, Tank B, which holds up to 1000 litres, starts with 100 litres and is filled at 60 litres per minute.

Let $x$ be the time in minutes and $y$ the litres of water in a tank. Write an equation for each tank. Solve the pair graphically to find when both tanks hold the same amount of water and what that amount is. Then use your equations to say which tank has more water after 8 minutes, and by how much.

Try it first, then show the solutionSolution

Tank A: $y = 600 - 40 x$ , which is $40 x + y = 600$ .
Tank B: $y = 100 + 60 x$ , which is $60 x - y = - 100$ .
Table for Tank A: $x = 0, y = 600$ ; $x = 5, y = 400$ ; $x = 15, y = 0$ .
Table for Tank B: $x = 0, y = 100$ ; $x = 5, y = 400$ ; $x = 10, y = 700$ .
Plot both sets of points on the same axes and join each set with a straight line. Label each line with its equation.
The two lines meet at $(5, 400)$ .
Check: Tank A gives $600 - 40 (5) = 400$ and Tank B gives $100 + 60 (5) = 400$ . Both are correct.
After 8 minutes, Tank A has $600 - 40 (8) = 280$ litres.
After 8 minutes, Tank B has $100 + 60 (8) = 580$ litres.
Difference: $580 - 280 = 300$ litres.

Answer: Both tanks hold 400 litres after 5 minutes. After 8 minutes, Tank B has 300 litres more than Tank A (580 litres against 280 litres).

Memory hook: Where the draining line meets the filling line, the two tanks are level: the crossing point is the solution.

Consistent, inconsistent and dependent pairs

Suppose you ask a fruit seller about his prices. He says, "3 bananas and 2 apples cost Rs 50." When you ask again, he says, "6 bananas and 4 apples cost Rs 100."

Let a banana cost Rs $x$ and an apple cost Rs $y$ . Form the pair of equations. State whether their graphs intersect, are parallel or are coincident, and whether the pair is consistent or inconsistent. Can you find the actual prices from these two statements? Support your answer with two different price pairs that fit both statements.

Try it first, then show the solutionSolution

The first statement gives $3 x + 2 y = 50$ .
The second statement gives $6 x + 4 y = 100$ .
Compare the coefficients: $\frac{a _{1}}{a _{2}} = \frac{3}{6} = \frac{1}{2}$ , $\frac{b _{1}}{b _{2}} = \frac{2}{4} = \frac{1}{2}$ and $\frac{c _{1}}{c _{2}} = \frac{50}{100} = \frac{1}{2}$ .
All three ratios are equal. The second equation is just 2 times the first, so the lines are coincident.
The pair is consistent and dependent, with infinitely many solutions.
Try $x = 10$ , $y = 10$ : $3 (10) + 2 (10) = 50$ . This fits both statements.
Try $x = 6$ , $y = 16$ : $3 (6) + 2 (16) = 18 + 32 = 50$ . This also fits both statements.
Two different price pairs fit, so the actual prices cannot be found. You need a statement that is not a multiple of the first.

Answer: The lines are coincident, so the pair is consistent and dependent with infinitely many solutions. The prices cannot be fixed: for example, Rs 10 per banana and Rs 10 per apple fits, and so does Rs 6 per banana and Rs 16 per apple.

Memory hook: A doubled statement gives no new information: same line, infinitely many answers.

Algebraic conditions for the number of solutions

Imagine you and a classmate buy pens and notebooks from the same stationery shop at fixed prices. You say, "2 pens and 3 notebooks cost me Rs 120." Your classmate says, "4 pens and $k$ notebooks cost me Rs 250."

Let a pen cost Rs $x$ and a notebook cost Rs $y$ . Find the value of $k$ for which the two statements cannot both be true. Is there any value of $k$ for which the statements give infinitely many possible prices? Justify your answer.

Try it first, then show the solutionSolution

Your statement gives $2 x + 3 y = 120$ .
Your classmate's statement gives $4 x + k y = 250$ .
The statements cannot both be true when the pair has no solution. This needs $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ .
Here $\frac{a _{1}}{a _{2}} = \frac{2}{4} = \frac{1}{2}$ .
Set $\frac{3}{k} = \frac{1}{2}$ , which gives $k = 6$ .
Check the constants: $\frac{c _{1}}{c _{2}} = \frac{120}{250} = \frac{12}{25}$ , and $\frac{12}{25} \neq = \frac{1}{2}$ . So the pair has no solution when $k = 6$ .
Infinitely many prices would need $\frac{2}{4} = \frac{3}{k} = \frac{120}{250}$ .
But $\frac{2}{4} = \frac{1}{2}$ while $\frac{120}{250} = \frac{12}{25}$ . These are never equal, so no value of $k$ gives infinitely many solutions.
For every $k \neq = 6$ , the ratios $\frac{a _{1}}{a _{2}}$ and $\frac{b _{1}}{b _{2}}$ differ, so there is exactly one pair of prices.

Answer: When $k = 6$ , the pair has no solution, so the statements contradict each other. No value of $k$ gives infinitely many solutions. For every other value of $k$ , there is exactly one pair of prices.

Memory hook: Equal x- and y-ratios but a different money ratio: parallel lines, so someone's bill is wrong.

Substitution method

Suppose that in a school cricket match, a batter scores 64 runs and every run comes from boundaries. She hits 13 boundaries in all. Some are fours and the rest are sixes.

Form a pair of linear equations and solve it by substitution to find how many fours and how many sixes she hit. Then find what percentage of her runs came from sixes, correct to one decimal place.

Try it first, then show the solutionSolution

Let the number of fours be $x$ and the number of sixes be $y$ .
Number of boundaries: $x + y = 13$ ... (1)
Runs scored: $4 x + 6 y = 64$ ... (2)
From (1), $y = 13 - x$ ... (3)
Substitute (3) into (2): $4 x + 6 (13 - x) = 64$ .
Expand: $4 x + 78 - 6 x = 64$ , so $- 2 x = - 14$ and $x = 7$ .
From (3), $y = 13 - 7 = 6$ .
Check in (2): $4 (7) + 6 (6) = 28 + 36 = 64$ . This is correct.
Runs from sixes: $6 \times 6 = 36$ .
Percentage: $\frac{36}{64} \times 100 = 56.25%$ , which is $56.3%$ correct to one decimal place.

Answer: She hit 7 fours and 6 sixes. Sixes gave 36 of her 64 runs, which is about 56.3% of her score.

Memory hook: Count first, runs second: write sixes as 13 minus fours, then substitute.

Elimination method

Imagine two sections of your class visit a science museum on a school trip. Section A takes 5 teachers and 40 students and pays Rs 1400 in total. Section B takes 4 teachers and 30 students and pays Rs 1080. Every teacher pays the same fare, and every student pays the same fare.

Find the ticket price for a teacher and for a student by the elimination method. Then find what Section C must pay for 6 teachers and 45 students.

Try it first, then show the solutionSolution

Let a teacher's ticket cost Rs $a$ and a student's ticket cost Rs $s$ .
Section A: $5 a + 40 s = 1400$ ... (1)
Section B: $4 a + 30 s = 1080$ ... (2)
Multiply (1) by 4: $20 a + 160 s = 5600$ ... (3)
Multiply (2) by 5: $20 a + 150 s = 5400$ ... (4)
Subtract (4) from (3): $10 s = 200$ , so $s = 20$ .
Substitute into (1): $5 a + 800 = 1400$ , so $5 a = 600$ and $a = 120$ .
Check in (2): $4 (120) + 30 (20) = 480 + 600 = 1080$ . This is correct.
Section C pays $6 (120) + 45 (20) = 720 + 900 = 1620$ .

Answer: A teacher's ticket costs Rs 120 and a student's ticket costs Rs 20. Section C must pay Rs 1620.

Memory hook: Make the teachers match, then subtract: the teachers cancel and the student fare is left.

Situational (word) problems

Suppose you hear a mother and her daughter chatting at a family function. The mother says, "Five years ago, I was 7 times as old as you were. Five years from now, I will be 3 times as old as you will be."

Form a pair of linear equations and find their present ages. Then find how many years from now the mother's age will be exactly twice the daughter's age.

Try it first, then show the solutionSolution

Let the mother's present age be $m$ years and the daughter's be $d$ years.
Five years ago: $m - 5 = 7 (d - 5)$ , so $m = 7 d - 30$ ... (1)
Five years later: $m + 5 = 3 (d + 5)$ , so $m = 3 d + 10$ ... (2)
From (1) and (2): $7 d - 30 = 3 d + 10$ , so $4 d = 40$ and $d = 10$ .
From (2): $m = 3 (10) + 10 = 40$ .
Check with the words: five years ago the ages were 35 and 5, and $35 = 7 \times 5$ . Five years later they will be 45 and 15, and $45 = 3 \times 15$ . Both conditions hold.
Let the mother's age be twice the daughter's after $t$ years: $40 + t = 2 (10 + t)$ .
Then $40 + t = 20 + 2 t$ , so $t = 20$ .
Check: after 20 years the ages will be 60 and 30, and $60 = 2 \times 30$ .

Answer: The mother is 40 years old and the daughter is 10 years old. The mother's age will be twice the daughter's age 20 years from now.

Memory hook: Years ago or later, shift both ages by the same amount before you compare.

Topics in this chapter

Pair of Linear Equations in Two Variables formula sheet

General form of a pair: $a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0$; Write both equations in this form, or both as ax + by = c, before comparing coefficients.
Unique solution (intersecting lines, consistent): $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$; Use for 'one solution' questions. The constants do not matter here.
Infinitely many solutions (coincident lines, consistent and dependent): $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$; All three ratios must be equal. In find-k questions, test every value of k you get against the constant ratio.
No solution (parallel lines, inconsistent): $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$; The first two ratios are equal and the third is different.
Reading the result of elimination or substitution: $0 = 0 \Rightarrow infinitely many; 0 = k (k \neq = 0) \Rightarrow no solution$; Use when both variables cancel out while you are solving.
Swapped-coefficient shortcut: $a x + b y = c, b x + a y = d \Rightarrow x + y = \frac{c + d}{a + b}, x - y = \frac{c - d}{a - b}$; Add and subtract the equations directly when the coefficients are swapped (needs a ≠ b and a + b ≠ 0).
Intercepts of a line: $a x + b y = c : meets x -axis at (\frac{c}{a}, 0), y -axis at (0, \frac{c}{b})$; Quick points for plotting, and the base vertices of a triangle formed with an axis.
Area of triangle formed with an axis: $Area = \frac{1}{2} \times base on axis \times distance of intersection point from that axis$; With the x-axis, the height is the y-coordinate of the intersection point, ignoring sign.
Two-digit number: $number = 10 x + y, reversed = 10 y + x$; Here x is the tens digit and y is the units digit. Digit sum is x + y.
Age shift: $t years ago: (x - t), (y - t); t years later: (x + t), (y + t)$; Shift every person's age, not just one.

How the board asks Pair of Linear Equations in Two Variables

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Nature of the pair from coefficients (MCQ or assertion-reason)	You are given two equations and asked whether the lines intersect, are parallel or coincide, or whether the pair is consistent. It is sometimes framed as an assertion about the number of solutions.	Write both equations in the same form. Compute all three ratios, then match the pattern: ratios differ, so one solution; first two equal and third different, so none; all equal, so infinitely many.
Find k for a given number of solutions	Find k (or a and b) so that the pair has a unique solution, no solution or infinitely many solutions. The unknown is often inside expressions like k + 1 or 2a − b.	Set up the correct ratio condition. Solve the equation from the first two ratios, then test each value in the constant ratio and reject values that break the condition. For two unknowns, form two equations from the equal ratios and solve them as a pair.
Solve and hence find	Solve a pair by substitution or elimination, then use the solution to find m in y = mx + c, or to evaluate an expression.	Use the method asked for. If none is named, choose the faster one. Check the solution in the unused equation, then substitute into the extra condition.
Graphical solution with a triangle	Draw both lines, find the solution from the graph, and find the vertices or area of the triangle formed with the x-axis or y-axis.	Make a table of three points per line, label the lines, and mark the intersection. Find where each line meets the required axis, then use half times base times height.
Word problems on ages, digits, fractions, costs and fares	A short paragraph with two conditions, usually set as a 3-mark or 5-mark question.	Define the variables, write one equation per condition, solve, check against the words, and state the final answer in a sentence with units.
Geometry-based situations	Angles of a triangle given in terms of x and y, or the length and breadth of a rectangle that change under stated conditions.	Use the fact behind the figure, such as the angle sum of 180 degrees or area as length times breadth. Write two linear equations and solve them.
Case-study (Section E)	A real context, such as ticket sales, a shop bill or a fare chart, with parts of 1, 1 and 2 marks. The 2-mark part often adds a new condition.	Fix the variables once and reuse them in every part. Solve the core pair carefully, because the later parts depend on it.

Tricks to remember Pair of Linear Equations in Two Variables

Short memory aids. Say each one aloud once, then use it on a question the same day.

Differ, Two, All
Ratios differ: one answer. Two ratios equal: no answer. All three equal: all points are answers.
Use it when: Instant recall of the three conditions in MCQs and assertion-reason questions.
Same sign subtract, opposite sign add
After matching coefficients, look at the signs of the matched terms. If they are the same, subtract; if they are opposite, add.
Use it when: Avoids the most common elimination slip.
Swapped coefficients: add then subtract
If the x-coefficient of one equation is the y-coefficient of the other, add the equations to get x + y and subtract them to get x − y.
Use it when: Turns big-number pairs into two-line solutions.
Third point check
Always plot three points per line. If the third point is not in line with the other two, recheck your table.
Use it when: Catches calculation errors before the graph is ruined.
Square roots give two k values
When k + 1 = ±3 or similar appears, test both values against the constant ratio.
Use it when: Stops you keeping a value that actually gives parallel lines.
Check in the unused equation
Substitute your answer into the equation you did not use for back-substitution.
Use it when: A ten-second check that catches sign errors.

Pair of Linear Equations in Two Variables: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Value of an expression from the solution · 1 mark · MCQBoard level

If $x = a$ , $y = b$ is the solution of the pair $x + y = 5$ and $x - y = 1$ , then the value of $a^{2} - b^{2}$ is
(a) $1$
(b) $5$
(c) $13$
(d) $25$

Try it first, then show the solutionSolution

Since $(a, b)$ is the solution, it satisfies both equations: $a + b = 5$ and $a - b = 1$ .
$a^{2} - b^{2} = (a + b) (a - b) = 5 \times 1 = 5$ .
Check by solving: adding the equations gives $2 a = 6$ , so $a = 3$ and $b = 2$ . Then $a^{2} - b^{2} = 9 - 4 = 5$ .
Option (c), $13$ , is the value of $a^{2} + b^{2}$ : a trap for anyone who misreads the sign.

Answer: (b) $5$

Example 2: Assertion-reason on an inconsistent pair · 1 mark · Assertion-reasonBoard level

Assertion (A): The pair $2 x - 3 y = 8$ and $4 x - 6 y = 9$ has no solution.
Reason (R): If $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , the lines are parallel and the pair is inconsistent.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

Standard form: $2 x - 3 y - 8 = 0$ and $4 x - 6 y - 9 = 0$ .
$\frac{a _{1}}{a _{2}} = \frac{2}{4} = \frac{1}{2}$ , $\frac{b _{1}}{b _{2}} = \frac{- 3}{- 6} = \frac{1}{2}$ , $\frac{c _{1}}{c _{2}} = \frac{- 8}{- 9} = \frac{8}{9}$ .
Since $\frac{1}{2} = \frac{1}{2} \neq = \frac{8}{9}$ , the lines are parallel and never meet. So A is true.
R is the standard condition for parallel lines, so R is true, and it is exactly the rule used to prove A.

Answer: (a) Both A and R are true and R is the correct explanation of A.

Example 3: Two unknowns for infinitely many solutions · 3 marks · Short answerHarder board level

Find the values of $a$ and $b$ for which the pair $3 x + 2 y = 8$ and $(2 a - b) x + (a + b) y = 24$ has infinitely many solutions.

Try it first, then show the solutionSolution

For infinitely many solutions, $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ , so $\frac{3}{2 a - b} = \frac{2}{a + b} = \frac{8}{24}$ . (1 mark)
Since $\frac{8}{24} = \frac{1}{3}$ , we get $2 a - b = 9$ and $a + b = 6$ . (1 mark)
Adding: $3 a = 15$ , so $a = 5$ ; then $b = 6 - 5 = 1$ . (½ mark)
Check: the second equation becomes $9 x + 6 y = 24$ , which is $3$ times $3 x + 2 y = 8$ , so the two lines coincide. (½ mark)

Answer: $a = 5$ and $b = 1$ .

Example 4: Graphical solution with a triangle on the y-axis · 5 marks · Long answerHarder board level

Solve graphically: $2 x - y = 2$ and $x + y = 7$ . Find the coordinates of the vertices of the triangle formed by these two lines and the $y$ -axis, and find its area.

Try it first, then show the solutionSolution

Table for $2 x - y = 2$ , that is $y = 2 x - 2$ : points $(0, - 2)$ , $(1, 0)$ , $(3, 4)$ . (1 mark)
Table for $x + y = 7$ , that is $y = 7 - x$ : points $(0, 7)$ , $(7, 0)$ , $(3, 4)$ . (1 mark)
Plot both lines on the same axes. They meet at $(3, 4)$ , so $x = 3$ , $y = 4$ . Check: $6 - 4 = 2$ and $3 + 4 = 7$ . (1 mark)
The first line cuts the $y$ -axis at $(0, - 2)$ and the second at $(0, 7)$ . So the vertices are $(3, 4)$ , $(0, - 2)$ and $(0, 7)$ . (1 mark)
Base along the $y$ -axis $= 7 - (- 2) = 9$ units. Height = distance of $(3, 4)$ from the $y$ -axis $= 3$ units. Area $= \frac{1}{2} \times 9 \times 3 = 13.5$ square units. (1 mark)

Answer: Solution $x = 3$ , $y = 4$ ; vertices $(3, 4)$ , $(0, - 2)$ , $(0, 7)$ ; area $13.5$ square units.

Example 5: Two-digit number problem · 3 marks · Short answerBoard level

A two-digit number is 4 times the sum of its digits. If 18 is added to the number, its digits are reversed. Find the number.

Try it first, then show the solutionSolution

Let the tens digit be $x$ and the units digit be $y$ . The number is $10 x + y$ .
First condition: $10 x + y = 4 (x + y)$ , so $6 x = 3 y$ , that is $y = 2 x$ ... (1). (1 mark)
Second condition: $10 x + y + 18 = 10 y + x$ , so $9 y - 9 x = 18$ , that is $y - x = 2$ ... (2). (1 mark)
Substitute (1) in (2): $2 x - x = 2$ , so $x = 2$ and $y = 4$ . The number is $24$ . (½ mark)
Check: $4 \times (2 + 4) = 24$ , and $24 + 18 = 42$ , which is $24$ reversed. (½ mark)

Answer: The number is $24$ .

Example 6: Investment with swapped interest rates · 5 marks · Long answerHarder board level

Meena invested part of her savings in Scheme A at 8% per annum simple interest and the rest in Scheme B at 10% per annum simple interest. Her total interest for one year was Rs 880. Had she invested the same amounts the other way round (the Scheme A amount at 10% and the Scheme B amount at 8%), her interest for one year would have been Rs 920. Find the amount invested in each scheme and her total investment.

Try it first, then show the solutionSolution

Let the amount in Scheme A be Rs $x$ and in Scheme B be Rs $y$ . (½ mark)
Interest for one year: $\frac{8 x}{100} + \frac{10 y}{100} = 880$ , so $8 x + 10 y = 88000$ ... (1). (1 mark)
Rates swapped: $\frac{10 x}{100} + \frac{8 y}{100} = 920$ , so $10 x + 8 y = 92000$ ... (2). (1 mark)
The coefficients swap, so add (1) and (2): $18 x + 18 y = 180000$ , giving $x + y = 10000$ ... (3). (½ mark)
Subtract (1) from (2): $2 x - 2 y = 4000$ , giving $x - y = 2000$ ... (4). (½ mark)
Add (3) and (4): $2 x = 12000$ , so $x = 6000$ ; then $y = 10000 - 6000 = 4000$ . (1 mark)
Check: 8% of Rs 6000 plus 10% of Rs 4000 is Rs 480 + Rs 400 = Rs 880; 10% of Rs 6000 plus 8% of Rs 4000 is Rs 600 + Rs 320 = Rs 920. (½ mark)

Answer: Rs 6000 in Scheme A and Rs 4000 in Scheme B; total investment Rs 10000.

Example 7: Find the error: equal ratios are not enough · 3 marks · Short answerHOTS / competency

Riya was asked how many solutions the pair $3 x + 6 y = 9$ and $x + 2 y = 4$ has. She wrote: 'Since $\frac{3}{1} = \frac{6}{2}$ , the lines coincide, so there are infinitely many solutions.'
(i) What is wrong with her reasoning?
(ii) Use the substitution method to show what really happens, and state the correct conclusion with its graphical meaning.

Try it first, then show the solutionSolution

(i) Equal ratios $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}}$ only tell you the lines are parallel or coincident. You must also compare $\frac{c _{1}}{c _{2}}$ . Here $\frac{c _{1}}{c _{2}} = \frac{- 9}{- 4} = \frac{9}{4}$ , but $\frac{a _{1}}{a _{2}} = 3$ . Since $3 \neq = \frac{9}{4}$ , the lines do not coincide. (1 mark)
(ii) From $x + 2 y = 4$ : $x = 4 - 2 y$ . (½ mark)
Substitute in $3 x + 6 y = 9$ : $3 (4 - 2 y) + 6 y = 9$ , so $12 - 6 y + 6 y = 9$ , that is $12 = 9$ . (½ mark)
This statement is false whatever the value of $y$ , so no pair $(x, y)$ satisfies both equations. (½ mark)
The pair is inconsistent. Dividing the first equation by $3$ gives $x + 2 y = 3$ , so the lines $x + 2 y = 3$ and $x + 2 y = 4$ are parallel and distinct. (½ mark)

Answer: Riya ignored the constant terms. The pair has no solution: the lines are parallel and never meet.

Example 8: Auto-rickshaw or cab app? · 4 marks · Case-basedHOTS / competency

In a town, an auto-rickshaw fare is a fixed charge plus a fixed rate per km. A 6 km ride costs Rs 108 and a 10 km ride costs Rs 164. A cab app charges Rs 60 fixed plus Rs 12 per km.
(i) Taking the auto's fixed charge as Rs $f$ and its rate as Rs $r$ per km, write the pair of equations. (1 mark)
(ii) Find $f$ and $r$ . (2 marks)
(iii) For what distance do the auto and the cab cost the same? Which is cheaper for a 25 km trip? (1 mark)

Try it first, then show the solutionSolution

(i) $f + 6 r = 108$ and $f + 10 r = 164$ . (1 mark)
(ii) Subtract the first equation from the second: $4 r = 56$ , so $r = 14$ . (1 mark)
Then $f = 108 - 6 \times 14 = 108 - 84 = 24$ . Check: $24 + 10 \times 14 = 164$ . (1 mark)
(iii) For $d$ km, the auto costs $24 + 14 d$ rupees and the cab $60 + 12 d$ rupees. They are equal when $24 + 14 d = 60 + 12 d$ , so $2 d = 36$ and $d = 18$ km (both cost Rs 276). For 25 km: auto $24 + 350 = 374$ , cab $60 + 300 = 360$ , so the cab is cheaper by Rs 14. (1 mark)

Answer: (i) $f + 6 r = 108$ , $f + 10 r = 164$ . (ii) Fixed charge Rs 24, rate Rs 14 per km. (iii) Same cost at 18 km; for 25 km the cab is cheaper by Rs 14.

Example 9: Rectangle with two changes in area · 5 marks · Long answerHarder board level

If the length of a rectangular garden is increased by 4 m and its breadth is decreased by 2 m, its area decreases by $16 m^{2}$ . If instead its length is decreased by 3 m and its breadth is increased by 3 m, its area increases by $21 m^{2}$ . Find the length and breadth of the garden. Hence find the cost of fencing the garden at Rs 45 per metre.

Try it first, then show the solutionSolution

Let the length be $l$ m and the breadth be $b$ m, so the area is $l b m^{2}$ . (½ mark)
First condition: $(l + 4) (b - 2) = l b - 16$ , so $l b - 2 l + 4 b - 8 = l b - 16$ , giving $- 2 l + 4 b = - 8$ , that is $l - 2 b = 4$ ... (1). (1 mark)
Second condition: $(l - 3) (b + 3) = l b + 21$ , so $l b + 3 l - 3 b - 9 = l b + 21$ , giving $3 l - 3 b = 30$ , that is $l - b = 10$ ... (2). (1 mark)
Subtract (1) from (2): $(l - b) - (l - 2 b) = 10 - 4$ , so $b = 6$ . From (2): $l = 16$ . (1 mark)
Check: original area $= 96$ ; $(20) (4) = 80 = 96 - 16$ and $(13) (9) = 117 = 96 + 21$ . Both conditions hold. (½ mark)
Perimeter $= 2 (16 + 6) = 44$ m, so the cost of fencing $= 44 \times 45 =$ Rs 1980. (1 mark)

Answer: Length 16 m, breadth 6 m; fencing costs Rs 1980.

Example 10: Two cars: same direction and towards each other · 5 marks · Long answerHarder board level

Two towns A and B are 180 km apart on a straight highway. At the same moment, one car leaves A and another leaves B. If both cars travel in the direction from A towards B, the car from A catches up with the car from B after 9 hours. If the cars travel towards each other, they meet after 2 hours. Find the speed of each car. Hence find how far from A the cars meet when they travel towards each other.

Try it first, then show the solutionSolution

Let the speed of the car from A be $x$ km/h and of the car from B be $y$ km/h. Since the car from A catches up, $x > y$ . (½ mark)
Same direction: in 9 hours the car from A covers 180 km more than the car from B, so $9 x - 9 y = 180$ , that is $x - y = 20$ ... (1). (1 mark)
Towards each other: together they cover 180 km in 2 hours, so $2 x + 2 y = 180$ , that is $x + y = 90$ ... (2). (1 mark)
Add (1) and (2): $2 x = 110$ , so $x = 55$ ; then $y = 90 - 55 = 35$ . (1 mark)
Check: $9 (55 - 35) = 180$ and $2 (55 + 35) = 180$ . (½ mark)
Towards each other, in 2 hours the car from A travels $55 \times 2 = 110$ km, so they meet 110 km from A. (The car from B travels $35 \times 2 = 70$ km, and $110 + 70 = 180$ .) (1 mark)

Answer: Car from A: 55 km/h, car from B: 35 km/h; they meet 110 km from A.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Alphonso mango boxes at a Ratnagiri stall

Sakhubai runs a roadside stall near Ratnagiri, selling Alphonso mangoes in two box sizes. Every large box sells at the same price, and so does every small box. On Saturday she sold 8 large boxes and 12 small boxes and collected Rs 9600. On Sunday she sold 10 large boxes and 5 small boxes and collected Rs 9000. Let the price of a large box be Rs $x$ and the price of a small box be Rs $y$ .

(i) · 1 mark
Write the two equations described above in their simplest form.
Answer
Answer: $2 x + 3 y = 2400$ and $2 x + y = 1800$
Saturday: $8 x + 12 y = 9600$ ; divide by $4$ to get $2 x + 3 y = 2400$ . Sunday: $10 x + 5 y = 9000$ ; divide by $5$ to get $2 x + y = 1800$ .
(ii) · 1 mark
Without solving, show that this pair of equations has exactly one solution.
Answer
Answer: $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , so the lines intersect: the pair is consistent with a unique solution.
$\frac{a _{1}}{a _{2}} = \frac{2}{2} = 1$ and $\frac{b _{1}}{b _{2}} = \frac{3}{1} = 3$ . Since $1 \neq = 3$ , the lines intersect at exactly one point.
(iii) · 2 marks
Find the price of each box. Hence find how much a customer pays for 3 large boxes and 4 small boxes.
Answer
Answer: Large box Rs 750, small box Rs 300; 3 large and 4 small cost Rs 3450.
Subtract $2 x + y = 1800$ from $2 x + 3 y = 2400$ : $2 y = 600$ , so $y = 300$ (½ mark). Then $2 x = 1800 - 300 = 1500$ , so $x = 750$ (½ mark). Check: $8 \times 750 + 12 \times 300 = 6000 + 3600 = 9600$ . Cost: $3 \times 750 + 4 \times 300 = 2250 + 1200 = 3450$ , so the customer pays Rs 3450 (1 mark).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Find the value of $k$ for which the pair $2 x + k y = 5$ and $4 x + 6 y = 9$ has no solution. (2 marks)

Step	Marks
Writes the condition $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , that is $\frac{2}{4} = \frac{k}{6} \neq = \frac{5}{9}$	1
Solves $\frac{k}{6} = \frac{1}{2}$ to get $k = 3$	0.5
Verifies $\frac{1}{2} \neq = \frac{5}{9}$ and concludes $k = 3$	0.5

Solve $5 x - 2 y = 4$ and $3 x + 4 y = 18$ by elimination. Hence find $p$ if $p x - y = 5$ . (3 marks)

Step	Marks
Multiplies the first equation by 2 to get $10 x - 4 y = 8$ , and adds it to the second: $13 x = 26$	1
Gets $x = 2$ and substitutes to get $y = \frac{10 - 4}{2} = 3$	1
Substitutes in $p x - y = 5$ : $2 p - 3 = 5$ , so $p = 4$	1

At a school fair, 3 adult tickets and 5 child tickets cost Rs 1200, while 2 adult tickets and 7 child tickets cost Rs 1240. (i) Find the price of each ticket. (ii) On one day, 50 tickets were sold for Rs 7600 in total. How many adult tickets were sold? (5 marks)

Step	Marks
Lets the adult ticket cost Rs $A$ and the child ticket cost Rs $C$ , and forms $3 A + 5 C = 1200$ and $2 A + 7 C = 1240$	1
Eliminates $A$ : $6 A + 10 C = 2400$ and $6 A + 21 C = 3720$ , so $11 C = 1320$ and $C = 120$	1
Finds $3 A = 1200 - 600$ , so $A = 200$ . Adult ticket Rs 200, child ticket Rs 120	1
Lets $a$ adult and $c$ child tickets be sold, and forms $a + c = 50$ and $200 a + 120 c = 7600$	1
Solves: $200 a + 120 (50 - a) = 7600$ , so $80 a = 1600$ , giving $a = 20$ and $c = 30$ . Concludes that 20 adult tickets were sold	1

Where students lose marks in Pair of Linear Equations in Two Variables

The slips examiners see most often, and what to write instead.

Mistake: Not testing the constant ratio in a find-k question
Students solve the equation from the first two ratios and stop, forgetting that 'infinitely many' needs all three ratios equal.
Wrong working
$(k + 1)^{2} = 9$ , so $k = 2$ or $k = - 4$ .
Right working
$(k + 1)^{2} = 9$ , so $k = 2$ or $k = - 4$ . For $k = - 4$ : $\frac{5}{- 33} \neq = - \frac{1}{3}$ , so reject it. Hence $k = 2$ .
Do this: After solving, write a separate line testing each value in the third ratio.
Usually costs: 1 mark in a 2- or 3-mark question, and the final answer mark.
Mistake: Mixing two forms when taking the constant ratio
One equation is written as ax + by = c and the other as ax + by + c = 0, so the sign of one constant flips.
Wrong working
$2 x + 3 y = 5$ and $4 x + 6 y - 10 = 0$ : $\frac{c _{1}}{c _{2}} = \frac{5}{- 10} = - \frac{1}{2}$ , so the lines are parallel.
Right working
Rewrite as $4 x + 6 y = 10$ . Then $\frac{2}{4} = \frac{3}{6} = \frac{5}{10} = \frac{1}{2}$ , so the lines are coincident and there are infinitely many solutions.
Do this: Put both equations in the same form before writing any ratio.
Usually costs: The whole 1-mark MCQ, or 1 to 2 marks in a longer question.
Mistake: Sign errors when subtracting equations
Only the first term is subtracted, or a negative right-hand side is not handled correctly.
Wrong working
$(3 x + 2 y) - (3 x - 4 y) = 12 - 6$ , so $- 2 y = 6$ .
Right working
$(3 x + 2 y) - (3 x - 4 y) = 12 - (- 6)$ , so $6 y = 18$ and $y = 3$ , then $x = 2$ .
Do this: Put the whole subtracted equation in brackets, including the right-hand side, then expand.
Usually costs: Usually all marks after that step, which is 1 to 2 marks.
Mistake: Wrong age equation
The time shift is applied to one person only.
Wrong working
Five years ago the mother was three times as old as her son: $M - 5 = 3 S$ .
Right working
$M - 5 = 3 (S - 5)$ , which gives $M - 3 S = - 10$ .
Do this: Write each age at that time first, for example 'son then $= S - 5$ ', then form the equation.
Usually costs: 1 mark for forming the equation, and the answer mark.
Mistake: Confusing dependent with inconsistent
Students think that infinitely many solutions means the pair has no proper answer.
Wrong working
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ , so the pair is inconsistent.
Right working
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ , so the lines are coincident. The pair is consistent (dependent) with infinitely many solutions.
Do this: Remember: inconsistent only ever means no solution.
Usually costs: 1 mark (often the full MCQ or assertion-reason mark).
Mistake: An incomplete graph
Students rush the graph, plot only two points, leave the lines unlabelled, or forget to state the solution.
Wrong working
Plots only $(0, 5)$ and $(5, 0)$ , draws both lines without labels, and writes 'solution where lines cross'.
Right working
Tables with three points per line, both lines labelled, intersection marked $(3, 2)$ , and the line 'Hence $x = 3$ , $y = 2$ '.
Do this: Always make a table, label each line and write the final answer in words.
Usually costs: 1 to 2 marks in a 3- or 5-mark graph question.

NCERT exercise guide: Pair of Linear Equations in Two Variables

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 3.1

What it tests: Turning a situation into a pair of linear equations, drawing both lines on the same graph, and comparing $\frac{a _{1}}{a _{2}}$ , $\frac{b _{1}}{b _{2}}$ and $\frac{c _{1}}{c _{2}}$ to decide whether the lines intersect, are parallel or coincide. You also decide whether a pair is consistent or inconsistent, solve consistent pairs graphically, and find the vertices of a triangle formed by the lines and an axis.

How to approach it: Compare the three ratios before you draw anything, so you know whether to expect one meeting point, none, or a whole common line. For each line, make a table of at least three points with whole-number coordinates; the axis intercepts are usually the easiest. Plot both lines on the same axes, read off the meeting point and verify it in both equations. If a triangle is asked, list its vertices first, then use area $= \frac{1}{2} \times$ base $\times$ height.

Like the questions in Exercise 3.1 that ask you to check consistency and then solve graphically

Check whether the pair $x - y = 1$ and $2 x + y = 8$ is consistent. If it is, solve it graphically.

Show the solutionSolution

Standard form: $x - y - 1 = 0$ and $2 x + y - 8 = 0$ . So $a_{1} = 1$ , $b_{1} = - 1$ , $c_{1} = - 1$ and $a_{2} = 2$ , $b_{2} = 1$ , $c_{2} = - 8$ .
$\frac{a _{1}}{a _{2}} = \frac{1}{2}$ and $\frac{b _{1}}{b _{2}} = \frac{- 1}{1} = - 1$ . Since $\frac{1}{2} \neq = - 1$ , the lines intersect at exactly one point. The pair is consistent with a unique solution.
Table for $x - y = 1$ , that is $y = x - 1$ : points $(0, - 1)$ , $(1, 0)$ , $(3, 2)$ .
Table for $2 x + y = 8$ , that is $y = 8 - 2 x$ : points $(0, 8)$ , $(4, 0)$ , $(3, 2)$ .
Plot both sets of points on the same axes and join each set with a straight line. The two lines cross at $(3, 2)$ .
Verify: $3 - 2 = 1$ is true and $2 (3) + 2 = 8$ is true.

Answer: The pair is consistent (unique solution): $x = 3$ , $y = 2$ .

Exercise 3.2

What it tests: Solving pairs by the substitution method, using the solution to find an unknown constant (the 'hence find' type), and forming and solving word problems on numbers, costs, fees, ages and dimensions.

How to approach it: Choose the equation in which one variable has coefficient $1$ or $- 1$ and make that variable the subject; this avoids fractions. Substitute into the other equation, solve the one-variable equation, then substitute back to get the second variable. In word problems, start with 'Let ... be ...' with units, form both equations, and check your answer against the original wording, not just your own equations.

Like the word problems on fixed charge plus a rate in Exercise 3.2

A gym charges a one-time joining fee plus a fixed fee every month. Asha paid Rs 3400 in total for 4 months, and Ravi paid Rs 5500 in total for 7 months. Using the substitution method, find the joining fee and the monthly fee. Hence find the total cost of a 12-month membership.

Show the solutionSolution

Let the joining fee be Rs $F$ and the monthly fee be Rs $m$ .
Asha: $F + 4 m = 3400$ ... (1). Ravi: $F + 7 m = 5500$ ... (2).
From (1): $F = 3400 - 4 m$ .
Substitute in (2): $3400 - 4 m + 7 m = 5500$ , so $3 m = 2100$ and $m = 700$ .
Then $F = 3400 - 4 \times 700 = 3400 - 2800 = 600$ .
Check in (2): $600 + 7 \times 700 = 600 + 4900 = 5500$ , which is correct.
12 months: $600 + 12 \times 700 = 600 + 8400 = 9000$ .

Answer: Joining fee Rs 600, monthly fee Rs 700; a 12-month membership costs Rs 9000.

Exercise 3.3

What it tests: Solving pairs by the elimination method, including pairs where substitution would give awkward fractions, and word problems on coins, ages, digits, fractions and speeds. You also learn to read special outcomes such as $0 = 0$ or $0 = 5$ .

How to approach it: Multiply one or both equations so that one variable gets the same (or opposite) coefficient, then subtract (or add) to remove it. When the coefficients of $x$ and $y$ simply swap between the two equations, add and subtract the equations directly to get $x + y$ and $x - y$ . If elimination leaves a true statement like $0 = 0$ , the pair has infinitely many solutions; a false statement like $0 = 5$ means there is no solution.

Like the coin and money word problems in Exercise 3.3

A purse contains only Rs 5 coins and Rs 2 coins. There are 25 coins in all, and their total value is Rs 89. Use the elimination method to find how many coins of each kind there are.

Show the solutionSolution

Let the number of Rs 5 coins be $x$ and the number of Rs 2 coins be $y$ .
Number of coins: $x + y = 25$ ... (1).
Total value: $5 x + 2 y = 89$ ... (2).
Multiply (1) by $2$ : $2 x + 2 y = 50$ ... (3).
Subtract (3) from (2): $3 x = 39$ , so $x = 13$ .
From (1): $y = 25 - 13 = 12$ .
Check: $13 + 12 = 25$ coins and $5 \times 13 + 2 \times 12 = 65 + 24 = 89$ , which matches.

Answer: 13 coins of Rs 5 and 12 coins of Rs 2.

Pair of Linear Equations in Two Variables: previous year questions

Standard: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.6 marks per paper; 60 questions in all, counting OR alternatives. In 4 of 4 sample papers.
Basic: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.1 marks per paper; 64 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Pair of Linear Equations in Two Variables

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026 Basic, set 430/1/1, Q20 · 1 mark · Assertion-reasonCompetency-based

Assertion (A): The value of $p$ for which the system of equations $4 x + p y + 8 = 0$ and $2 x + 2 y + 2 = 0$ is consistent is $4$ .

Reason (R): The system of equations $a_{1} x + b_{1} y = c_{1}$ and $a_{2} x + b_{2} y = c_{2}$ is consistent with infinitely many solutions, if $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ .

Select the correct answer from the codes (A), (B), (C) and (D) as given in Questions 19 and 20.

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

Try it first, then show our solutionSolution

For $p = 4$ : $\frac{a _{1}}{a _{2}} = 2$ , $\frac{b _{1}}{b _{2}} = \frac{4}{2} = 2$ , $\frac{c _{1}}{c _{2}} = \frac{8}{2} = 4$ . Since $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , the system is inconsistent, so A is false.
½ mark
R is the standard condition for infinitely many solutions, so R is true.
½ mark

Answer: Option (D)

Where the marks are

Shows the system is inconsistent at $p = 4$ , so A is false (½ mark)
States R is true (½ mark)

See it in the full 2026 board paper, set 430/1/1

2026 Basic, set 430/1/1, Q21 · 2 marks · Very short answer

Solve the following system of equations for x and y : $\frac{x}{2} + \frac{2 y}{3} = - 1$ and $x - \frac{y}{3} = 3$

Try it first, then show our solutionSolution

Clear the fractions. Multiply the first equation by 6: $3 x + 4 y = - 6$ ...(1). Multiply the second equation by 3: $3 x - y = 9$ ...(2).
Subtract (2) from (1): $5 y = - 15 \Rightarrow y = - 3$ .
1 mark
Substitute in (2): $3 x - (- 3) = 9 \Rightarrow 3 x = 6 \Rightarrow x = 2$ . Hence $x = 2, y = - 3$ .
1 mark

Answer: x = 2, y = −3

Where the marks are

Obtains one correct value, e.g. $y = - 3$ (1 mark)
Obtains the other value, $x = 2$ (1 mark)

See it in the full 2026 board paper, set 430/1/1

2026 Basic, set 430/1/1, Q28(a) · 3 marks · Short answer

Solve the following system of equations graphically : $x + 3 y = 6$ ; $2 x - 3 y = 12$

Try it first, then show our solutionSolution

For $x + 3 y = 6$ , i.e. $y = \frac{6 - x}{3}$ : table $x$ : 0, 3, 6 and $y$ : 2, 1, 0. Plot (0, 2), (3, 1), (6, 0) and join them to get the first line.
1 mark
For $2 x - 3 y = 12$ , i.e. $y = \frac{2 x - 12}{3}$ : table $x$ : 0, 3, 6 and $y$ : −4, −2, 0. Plot (0, −4), (3, −2), (6, 0) and join them to get the second line.
1 mark
The two lines intersect at (6, 0), which lies on the x-axis. Hence the solution is $x = 6, y = 0$ .
1 mark

Answer: x = 6, y = 0, i.e. the point (6, 0)

Where the marks are

Correct table and graph of $x + 3 y = 6$ (1 mark)
Correct table and graph of $2 x - 3 y = 12$ (1 mark)
Reads the intersection point (6, 0) as the solution (1 mark)

See it in the full 2026 board paper, set 430/1/1

Every Pair of Linear Equations in Two Variables question, by year

Standard (71 questions, sample papers included)

For 2027

Sample paper 2026-27: Q3 (1 mark, solved), Q36 (4 marks, solved)

2026

Set 30/1/1: Q5 (1 mark, solved), Q32 (5 marks, solved)

Set 30/2/1: Q6 (1 mark, solved), Q32 (5 marks, solved)

Set 30/3/1: Q8 (1 mark, solved), Q34(a) (5 marks, solved), Q34(b) (5 marks, solved)

Set 30/4/1: Q8 (1 mark, solved), Q26(a) (3 marks, solved), Q26(b) (3 marks, solved)

Set 30/5/1: Q19 (1 mark, solved), Q30 (3 marks, solved), Q32 (5 marks, solved)

Sample paper 2025-26: Q3 (1 mark, solved), Q31(A) (3 marks, solved), Q31(B) (3 marks, solved)

2025

Set 30/1/1: Q2 (1 mark, solved), Q32 (5 marks, solved)

Set 30/2/1: Q14 (1 mark, solved), Q20 (1 mark, solved), Q36 (4 marks, solved)

Set 30/3/1: Q3 (1 mark, solved)

Set 30/4/1: Q6 (1 mark, solved), Q21 (2 marks, solved), Q27 (3 marks, solved)

Set 30/5/1: Q21(a) (2 marks, solved), Q21(b) (2 marks, solved), Q28 (3 marks, solved)

Set 30/6/1: Q17 (1 mark, solved), Q21(a) (2 marks, solved), Q21(b) (2 marks, solved), Q28 (3 marks, solved)

Sample paper 2024-25: Q2 (1 mark, solved), Q32(A) (5 marks, solved), Q32(B) (5 marks, solved)

2024

Set 30/1/1: Q18 (1 mark, solved), Q21 (2 marks, solved), Q29 (3 marks, solved)

Set 30/2/1: Q1 (1 mark, solved), Q31 (3 marks, solved), Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Set 30/3/1: Q1 (1 mark, solved), Q22(a) (2 marks, solved), Q22(b) (2 marks, solved), Q28 (3 marks, solved)

Set 30/4/1: Q1 (1 mark, solved), Q11 (1 mark, solved), Q28 (3 marks, solved)

Set 30/5/1: Q4 (1 mark, solved), Q36 (4 marks, solved)

Sample paper 2023-24: Q3 (1 mark, solved), Q28(a) (3 marks, solved), Q28(b) (3 marks, solved)

2023

Set 30/1/1: Q2 (1 mark, solved), Q21(a) (2 marks, solved), Q21(b) (2 marks, solved), Q26 (3 marks, solved)

Set 30/2/1: Q6 (1 mark, solved), Q8 (1 mark, solved), Q30(a) (3 marks, solved), Q30(b) (3 marks, solved)

Set 30/4/1: Q7 (1 mark, solved), Q36 (4 marks, solved)

Set 30/5/1: Q2 (1 mark, solved), Q29 (3 marks, solved)

Set 30/6/1: Q4 (1 mark, solved), Q13 (1 mark, solved), Q36 (4 marks, solved)

Basic (76 questions, sample papers included)

For 2027

Sample paper 2026-27: Q3 (1 mark, solved), Q4 (1 mark, solved), Q31 (3 marks, solved)

2026

Set 430/1/1: Q20 (1 mark, solved), Q21 (2 marks, solved), Q28(a) (3 marks, solved), Q28(b) (3 marks, solved)

Set 430/2/1: Q3 (1 mark, solved), Q21(a) (2 marks, solved), Q21(b) (2 marks, solved), Q28(a) (3 marks, solved), Q28(b) (3 marks, solved)

Set 430/3/1: Q4 (1 mark, solved), Q21 (2 marks, solved), Q28(a) (3 marks, solved), Q28(b) (3 marks, solved)

Set 430/4/1: Q35(a) (5 marks, solved), Q35(b) (5 marks, solved)

Set 430/5/1: Q36 (4 marks, solved)

Sample paper 2025-26: Q2 (1 mark, solved), Q29(A) (3 marks, solved), Q29(B) (3 marks, solved)

2025

Set 430/1/1: Q20 (1 mark, solved), Q21 (2 marks, solved), Q28(a) (3 marks, solved), Q28(b) (3 marks, solved)

Set 430/2/1: Q3 (1 mark, solved), Q21(a) (2 marks, solved), Q21(b) (2 marks, solved), Q28(a) (3 marks, solved), Q28(b) (3 marks, solved)

Set 430/3/1: Q4 (1 mark, solved), Q21 (2 marks, solved), Q28(a) (3 marks, solved), Q28(b) (3 marks, solved)

Set 430/4/1: Q14 (1 mark, solved), Q36 (4 marks, solved)

Set 430/5/1: Q32 (5 marks, solved)

Set 430/6/1: Q5 (1 mark, solved), Q32(a) (5 marks, solved)

Sample paper 2024-25: Q2 (1 mark, solved), Q31 (3 marks, solved)

2024

Set 430/1/1: Q17 (1 mark, solved), Q22(A) (2 marks, solved), Q22(B) (2 marks, solved), Q27 (3 marks, solved)

Set 430/2/1: Q2 (1 mark, solved), Q15 (1 mark, solved), Q22(a) (2 marks, solved), Q22(b) (2 marks, solved)

Set 430/3/1: Q5 (1 mark, solved), Q35(a) (5 marks, solved), Q35(b) (5 marks, solved)

Set 430/4/1: Q19 (1 mark, solved), Q38 (4 marks, solved)

Set 430/5/1: Q3 (1 mark, solved), Q33 (5 marks, solved)

Sample paper 2023-24: Q4 (1 mark, solved), Q21 (2 marks, solved), Q28(a) (3 marks, solved), Q28(b) (3 marks, solved)

2023

Set 430/1/1: Q20 (1 mark, solved), Q24(a) (2 marks, solved), Q24(b) (2 marks, solved)

Set 430/2/1: Q13 (1 mark, solved), Q16 (1 mark, solved)

Set 430/4/1: Q7 (1 mark, solved), Q9 (1 mark, solved), Q28(A) (3 marks, solved), Q28(B) (3 marks, solved)

Set 430/5/1: Q8 (1 mark, solved), Q26(a) (3 marks, solved), Q26(b) (3 marks, solved)

Set 430/6/1: Q8 (1 mark, solved), Q13 (1 mark, solved), Q37 (4 marks, solved)

Practice questions on Pair of Linear Equations in Two Variables

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The pair of equations $x + 2 y + 5 = 0$ and $- 3 x - 6 y + 1 = 0$ has
(a) a unique solution
(b) exactly two solutions
(c) infinitely many solutions
(d) no solution
Hint
Compare all three ratios, keeping the signs.
Answer
(d) no solution.
$\frac{a _{1}}{a _{2}} = - \frac{1}{3}$ , $\frac{b _{1}}{b _{2}} = \frac{2}{- 6} = - \frac{1}{3}$ , $\frac{c _{1}}{c _{2}} = \frac{5}{1} = 5$ . The first two are equal and the third is different, so the lines are parallel.
Q2 · 1 mark · Assertion-reasonBoard level
Assertion (A): The lines $x + y = 4$ and $2 x + 2 y = 8$ are coincident.
Reason (R): A pair of coincident lines has infinitely many solutions.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Ask yourself: does R explain why the lines coincide, or what happens because they coincide?
Answer
(b).
A is true because $\frac{1}{2} = \frac{1}{2} = \frac{4}{8}$ . R is also true, but it is a consequence of coinciding, not the reason the lines coincide.
Q3 · 2 marks · Short answerBoard level
Solve $4 x - y = 5$ and $3 x + 2 y = 12$ by the substitution method. Hence find $p$ if the solution also satisfies $p x + 2 y = 14$ .
Hint
Make $y$ the subject of the first equation.
Answer
$y = 4 x - 5$ ; then $3 x + 2 (4 x - 5) = 12$ gives $11 x = 22$ , so $x = 2$ , $y = 3$ .
Then $2 p + 6 = 14$ , so $p = 4$ .
Q4 · 3 marks · Short answerBoard level
Ten years ago, a mother was 4 times as old as her daughter. Ten years from now, the mother will be twice as old as her daughter. Find their present ages.
Hint
Subtract 10 from both ages for 'ten years ago' and add 10 to both for 'ten years from now'.
Answer
Let the ages be $M$ and $D$ years. $M - 10 = 4 (D - 10)$ gives $M - 4 D = - 30$ ; $M + 10 = 2 (D + 10)$ gives $M - 2 D = 10$ .
Subtracting: $2 D = 40$ , so $D = 20$ and $M = 50$ . Check: $40 = 4 \times 10$ and $60 = 2 \times 30$ .
Mother 50 years, daughter 20 years.
Q5 · 3 marks · Short answerBoard level
The sum of the numerator and the denominator of a fraction is 18. If the denominator is increased by 2, the fraction becomes $\frac{1}{3}$ . Find the fraction.
Hint
Cross-multiply the second condition to clear the fraction before solving.
Answer
Let the fraction be $\frac{x}{y}$ . Then $x + y = 18$ and $\frac{x}{y + 2} = \frac{1}{3}$ , which gives $3 x - y = 2$ .
Adding: $4 x = 20$ , so $x = 5$ and $y = 13$ . Check: $\frac{5}{15} = \frac{1}{3}$ .
The fraction is $\frac{5}{13}$ .
Q6 · 3 marks · Short answerHarder board level
In $△ A B C$ , $A B = A C$ . The angles are $∠ A = (2 x)^{\circ}$ , $∠ B = (x + y)^{\circ}$ and $∠ C = (3 y - 20)^{\circ}$ . Find $x$ and $y$ , and hence all three angles of the triangle.
Hint
Angles opposite equal sides are equal; that gives one equation, and the angle sum gives the other.
Answer
$∠ B = ∠ C$ : $x + y = 3 y - 20$ , so $x = 2 y - 20$ . Angle sum: $2 x + (x + y) + (3 y - 20) = 180$ , so $3 x + 4 y = 200$ .
Substituting: $6 y - 60 + 4 y = 200$ , so $y = 26$ and $x = 32$ .
$∠ A = 6 4^{\circ}$ , $∠ B = ∠ C = 5 8^{\circ}$ .
Q7 · 3 marks · Short answerHarder board level
For what value(s) of $p$ does the pair $p x + 3 y = p - 3$ and $12 x + p y = p$ have (i) no solution, (ii) infinitely many solutions? For which values of $p$ is the solution unique?
Hint
Set $\frac{p}{12} = \frac{3}{p}$ first, then test each value of $p$ against $\frac{c _{1}}{c _{2}}$ .
Answer
$\frac{p}{12} = \frac{3}{p}$ gives $p^{2} = 36$ , so $p = 6$ or $p = - 6$ .
$p = 6$ : ratios $\frac{1}{2}, \frac{1}{2}, \frac{3}{6} = \frac{1}{2}$ , so infinitely many solutions. $p = - 6$ : ratios $- \frac{1}{2}, - \frac{1}{2}, \frac{- 9}{- 6} = \frac{3}{2}$ , so no solution.
(i) $p = - 6$ ; (ii) $p = 6$ ; the solution is unique for every $p \neq = \pm 6$ .
Q8 · 5 marks · Long answerHarder board level
Solve graphically: $x + 2 y = 8$ and $2 x - y = 6$ . Find the coordinates of the vertices of the triangle formed by these lines and the $x$ -axis, and find its area.
Hint
Use the $x$ -intercepts of both lines in your tables; they are two of the vertices.
Answer
Points for $x + 2 y = 8$ : $(0, 4)$ , $(8, 0)$ , $(4, 2)$ . Points for $2 x - y = 6$ : $(0, - 6)$ , $(3, 0)$ , $(4, 2)$ . The lines meet at $(4, 2)$ , so $x = 4$ , $y = 2$ .
Vertices: $(4, 2)$ , $(8, 0)$ , $(3, 0)$ . Base $= 8 - 3 = 5$ units, height $= 2$ units.
Area $= \frac{1}{2} \times 5 \times 2 = 5$ square units.
Q9 · 4 marks · Short answerHarder board level
Aman says that at a sports shop, 2 bats and 3 balls cost Rs 1150. Bina, describing the same shop's prices, says that 4 bats and 6 balls cost Rs 2400.
(i) Show that both statements cannot be true together, and explain what this means on a graph.
(ii) If Bina's amount had been Rs 2300 instead, could you find the price of one bat? Explain.
(iii) The shopkeeper confirms Aman's statement and adds that 1 bat and 2 balls cost Rs 650. Hence find the price of one bat and one ball.
Hint
What would 4 bats and 6 balls cost if Aman were right? In (iii), make the bat price the subject of the new equation.
Answer
(i) $2 x + 3 y = 1150$ and $4 x + 6 y = 2400$ : $\frac{2}{4} = \frac{3}{6} = \frac{1}{2}$ but $\frac{1150}{2400} = \frac{23}{48} \neq = \frac{1}{2}$ . The lines are parallel, so there is no solution: both statements cannot be true.
(ii) With Rs 2300, the second equation is just double the first, so the lines coincide. There are infinitely many solutions, so the price of a bat cannot be found without another independent fact.
(iii) $x + 2 y = 650$ gives $x = 650 - 2 y$ . Then $2 (650 - 2 y) + 3 y = 1150$ , so $1300 - y = 1150$ and $y = 150$ , $x = 350$ . Check: $700 + 450 = 1150$ and $350 + 300 = 650$ .
A bat costs Rs 350 and a ball costs Rs 150.
Q10 · 4 marks · Case-basedHOTS / competency
Rohan's mobile data Plan A charges a fixed monthly rental plus a fixed rate per GB. His bills were:
- June: 8 GB used, bill Rs 380
- July: 14 GB used, bill Rs 530
(i) Taking the rental as Rs $f$ and the rate as Rs $r$ per GB, write the pair of equations. (1 mark)
(ii) Find $f$ and $r$ . (1 mark)
(iii) Plan B charges Rs 300 rental plus Rs 15 per GB. For what monthly usage do Plans A and B cost the same, and which plan is cheaper above that usage? Plan C charges Rs 240 rental plus Rs 25 per GB. Using the pair of equations for Plans A and C, explain why they never cost the same. (2 marks)
Hint
For $g$ GB, write each plan's cost as rental plus rate times $g$ . For Plans A and C, compare the ratios of the coefficients.
Answer
(i) $f + 8 r = 380$ and $f + 14 r = 530$ .
(ii) Subtracting: $6 r = 150$ , so $r = 25$ ; $f = 380 - 200 = 180$ . Rental Rs 180, rate Rs 25 per GB.
(iii) For $g$ GB: Plan A costs $180 + 25 g$ and Plan B costs $300 + 15 g$ rupees. Equal when $10 g = 120$ , so $g = 12$ GB (both Rs 480). Above 12 GB, Plan B is cheaper because its rate per GB is lower (for 20 GB: A Rs 680, B Rs 600).
For Plans A and C, with cost Rs $c$ : $25 g - c = - 180$ and $25 g - c = - 240$ . Here $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = 1$ but $\frac{c _{1}}{c _{2}} = \frac{180}{240} = \frac{3}{4} \neq = 1$ , so the lines are parallel and the pair has no solution. Plan C is always Rs 60 dearer than Plan A.
Q11 · 3 marks · Short answerHarder board level
Five years ago, a father was 7 times as old as his son. Five years from now, the father will be 3 times as old as his son. Find their present ages. Hence find after how many years the father will be exactly twice as old as his son.
Hint
Shift both ages in each condition. For the last part, let the number of years be $t$ and add $t$ to both present ages.
Answer
Let the ages be $F$ and $S$ years. $F - 5 = 7 (S - 5)$ gives $F - 7 S = - 30$ ; $F + 5 = 3 (S + 5)$ gives $F - 3 S = 10$ .
Subtracting: $4 S = 40$ , so $S = 10$ and $F = 40$ . Check: $35 = 7 \times 5$ and $45 = 3 \times 15$ .
After $t$ years: $40 + t = 2 (10 + t)$ , so $t = 20$ . Check: $60 = 2 \times 30$ .
Father 40 years, son 10 years; the father will be twice as old after 20 years.
Q12 · 3 marks · Short answerHarder board level
Find the values of $p$ and $q$ for which the pair $x + 2 y = 3$ and $(p - q) x + (p + q) y = 12$ has infinitely many solutions. Hence solve the pair $p x - q y = 10$ and $q x + p y = 10$ .
Hint
All three ratios must equal $\frac{3}{12}$ . Use the values of $p$ and $q$ to write the second pair with numbers.
Answer
$\frac{1}{p - q} = \frac{2}{p + q} = \frac{3}{12} = \frac{1}{4}$ , so $p - q = 4$ and $p + q = 8$ . Adding: $p = 6$ , $q = 2$ .
The second pair is $6 x - 2 y = 10$ and $2 x + 6 y = 10$ . From the first, $y = 3 x - 5$ ; then $2 x + 6 (3 x - 5) = 10$ , so $20 x = 40$ , giving $x = 2$ and $y = 1$ .
Check: $12 - 2 = 10$ and $4 + 6 = 10$ .
$p = 6$ , $q = 2$ ; solution $x = 2$ , $y = 1$ .
Q13 · 3 marks · Short answerHOTS / competency
For a show, 300 tickets were sold. Each ticket was either regular at Rs 150 or premium at Rs 250, and any seat could be sold as either type.
(i) On Saturday, all 300 tickets brought in Rs 55000. Find the number of tickets of each type.
(ii) On Sunday, an employee reported that all 300 tickets brought in Rs 80000. Solve the pair for Sunday and explain why the report must be wrong. What is the greatest amount 300 tickets could bring in?
Hint
In (ii) the algebra gives an answer. Ask whether it makes sense as a number of tickets.
Answer
(i) Let $x$ regular and $y$ premium tickets be sold. $x + y = 300$ and $150 x + 250 y = 55000$ , that is $3 x + 5 y = 1100$ . Subtract $3 x + 3 y = 900$ : $2 y = 200$ , so $y = 100$ and $x = 200$ . Check: $30000 + 25000 = 55000$ .
(ii) $3 x + 5 y = 1600$ with $3 x + 3 y = 900$ gives $2 y = 700$ , so $y = 350$ and $x = - 50$ . A number of tickets cannot be negative, and 350 premium tickets is more than 300, so the report is wrong. The greatest possible amount is when all 300 are premium: $300 \times 250 =$ Rs 75000, which is less than Rs 80000.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What does a solution of a pair of linear equations represent on a graph?Show answer
The point where the two lines meet, which lies on both lines.
2.Condition for a unique solution?Show answer
$\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , which means the lines intersect.
3.Condition for no solution?Show answer
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , which means the lines are parallel.
4.Condition for infinitely many solutions?Show answer
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ , which means the lines are coincident.
5.Is a dependent pair consistent?Show answer
Yes. It has infinitely many solutions, so it is consistent.
6.During elimination you get $0 = 7$ . What does it mean?Show answer
There is no solution. The lines are parallel and the pair is inconsistent.
7.A two-digit number has tens digit $x$ and units digit $y$ . What is the reversed number?Show answer
$10 y + x$
8.Where does $3 x + 4 y = 12$ meet the $y$ -axis?Show answer
At $(0, 3)$ .

Best order to study it

Revise what a linear equation in two variables is and how to plot a straight line from a table of values.
Learn what a solution of a pair means, then practise the graphical method, including triangles formed with an axis.
Learn the three cases (intersecting, coincident, parallel) and the ratio conditions, then do find-k questions.
Master the substitution method, then the elimination method, including the swapped-coefficient shortcut and reading 0 = 0 or 0 = k.
Practise word problems on ages, digits, fractions, costs and geometry, and finish with case-based sets.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you write the three ratio conditions and name the type of lines for each?
Can you say which pairs are consistent, which are inconsistent and which are dependent?
Can you find k for infinitely many solutions and reject the wrong root?
Can you solve a pair by substitution and check it in the unused equation?
Can you solve a pair by elimination without sign slips on a negative right-hand side?
Can you solve a swapped-coefficient pair by adding and subtracting?
Can you draw both lines from tables and find the area of the triangle they make with the x-axis?
Can you write the equations for an age problem with '5 years ago' correctly?
Can you write a two-digit number and its reverse in terms of its digits?

Pair of Linear Equations in Two Variables: frequently asked questions

Is the cross-multiplication method in the 2026-27 syllabus?

No. Cross-multiplication and equations reducible to a pair of linear equations are not in the 2026-27 syllabus. You only need the graphical, substitution and elimination methods.

Which method should I use if the question does not name one?

Use substitution when one variable has coefficient 1 or minus 1. Use elimination when the coefficients are larger or are swapped. If the question names a method, you must use that method to get full marks.

Do I need graph paper for the graphical method in the board exam?

Graph paper is provided when a graph question is set. Draw neatly with a sensible scale, show a table of values for each line, label both lines and clearly state the solution read from the graph.

What is the difference between consistent and dependent?

A consistent pair has at least one solution. A dependent pair is a special consistent pair whose lines coincide, so it has infinitely many solutions. Only parallel lines, with no solution, are inconsistent.

How can I check my answer quickly in the exam?

Substitute your values of x and y into both original equations, especially the one you did not use at the end. For word problems, also check that your answer fits the conditions given in the words of the question.

Based on: CBSE Mathematics syllabus 2026-27; NCERT Mathematics, Textbook for Class X (Reprint 2026-27)