MathsCBSE Class 10 · Pair of Linear Equations in Two Variables

Pair of Linear Equations in Two Variables: Meaning and Forms

Updated 14 October 2026 · Fact-checked

A pair of linear equations in two variables is two equations of the form ax + by + c = 0 in the same unknowns x and y. A solution is an ordered pair that satisfies both equations together. Board questions ask you to write standard forms, test points, find unknown constants and form a pair from a situation.

The solution of a pair is the common point of both lines, and it must satisfy each equation when substituted separately.

Illustration drawn for QuestionBuddy.

Understand Pair of Linear Equations: Meaning and Forms

A linear equation in two variables is an equation that can be written as $a x + b y + c = 0$ , where $a$ , $b$ , $c$ are real numbers and $a$ and $b$ are not both zero. It is called linear because each variable appears only to the power $1$ , and its graph is a straight line. For example, $2 x + 3 y = 12$ becomes $2 x + 3 y - 12 = 0$ , so $a = 2$ , $b = 3$ , $c = - 12$ .

One such equation has infinitely many solutions. For $x + y = 10$ , the pairs $(1, 9)$ , $(4, 6)$ , $(2.5, 7.5)$ all work. One condition alone cannot fix two unknowns. That is why we need a pair: two equations in the same two variables, written as $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ .

A solution of the pair is an ordered pair $(x, y)$ that makes both equations true at the same time. It works because each equation is a line, and a point satisfying both must lie on both lines, that is, at a common point. To check a point, substitute it into each equation separately. If either fails, it is not a solution of the pair.

Forms matter. You may meet $a x + b y = c$ (constant on the right) or $a x + b y + c = 0$ (all on the left). Moving a term across the equals sign changes its sign, so $3 x - 2 y = 7$ in general form is $3 x - 2 y - 7 = 0$ , with $c = - 7$ . Equations like $y = 2 x - 5$ or $\frac{x}{2} + \frac{y}{3} = 1$ are still linear; rearrange or clear fractions first. An equation with $x y$ or $x^{2}$ is not linear.

Writing a situation as a pair is the first skill of this chapter. Name the two unknowns with units, then turn each separate fact into one equation. Everything later builds on this: graphing the two lines, deciding from the coefficients whether there is one solution, none, or infinitely many, and solving by substitution or elimination.

Keep the condition in mind: the two equations must use the same variables with the same meaning. If $x$ is the price of a pen in one equation, it must be the price of a pen in the other too.

Pair of Linear Equations: Meaning and Forms: formulas to remember

General form of a linear equation in two variables: $a x + b y + c = 0, a, b, c \in R, a^{2} + b^{2} \neq = 0$; a and b are not both zero. The condition a² + b² ≠ 0 says the same thing.
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 same form before reading off coefficients.
Converting between forms: $a x + b y = k ⟺ a x + b y - k = 0 (c = - k)$; The constant changes sign when it moves across the equals sign.
Solution of a pair: $(x_{0}, y_{0}) : a_{1} x_{0} + b_{1} y_{0} + c_{1} = 0 and a_{2} x_{0} + b_{2} y_{0} + c_{2} = 0$; Both must hold. Satisfying one equation is not enough.

How to solve Pair of Linear Equations: Meaning and Forms questions

Use this routine for any question on forming equations, standard form or checking solutions.

1Read the question and decide what is unknown; name the two variables with units, for example 'let the cost of one pen be Rs x'.
2Translate each separate condition into one equation, writing it next to the sentence it came from.
3Clear fractions and brackets, then rearrange each equation into the form ax + by + c = 0 (or ax + by = c if asked).
4Read off a, b, c carefully with signs, especially c, which flips sign when moved.
5To test a point, substitute it into each equation separately and show LHS and RHS.
6To find an unknown constant, substitute the given solution into the equation containing the constant and solve.
7Check your answer against the original words or the other equation before writing the final statement.

Pair of Linear Equations: Meaning and Forms in real life

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.

Writing situations as a pair of equations

Suppose your school canteen sells samosas and cups of tea at fixed prices. On Monday your group buys 4 samosas and 3 teas for Rs 110. On Tuesday another group buys 2 samosas and 5 teas for Rs 90.

Write the situation as a pair of linear equations in standard form, taking a samosa as Rs x and a tea as Rs y. By comparing the two bills, find the price of each item, and then find what 3 samosas and 2 teas cost.

Try it first, then show the solutionSolution

Monday: $4 x + 3 y = 110$ , so $4 x + 3 y - 110 = 0$ .
Tuesday: $2 x + 5 y = 90$ , so $2 x + 5 y - 90 = 0$ .
Double Tuesday's bill: $4 x + 10 y = 180$ .
Subtract Monday's bill: $(4 x + 10 y) - (4 x + 3 y) = 180 - 110$ , so $7 y = 70$ .
$y = 10$ .
Put in Monday's equation: $4 x + 30 = 110$ , so $4 x = 80$ , $x = 20$ .
Check in Tuesday's equation: $2 (20) + 5 (10) = 40 + 50 = 90$ , true.
3 samosas and 2 teas: $3 (20) + 2 (10) = 60 + 20 = 80$ .

Answer: Equations: 4x + 3y − 110 = 0 and 2x + 5y − 90 = 0; a samosa costs Rs 20 and a tea Rs 10; 3 samosas and 2 teas cost Rs 80.

Memory hook: Two bills, two equations: one canteen receipt is one line.

Meaning of a solution

Imagine a mobile recharge plan where you pay a fixed monthly charge plus a rate per GB of extra data. Suppose a month with 3 GB extra cost Rs 239 and a month with 5 GB extra cost Rs 299.

Taking the fixed charge as Rs f and the rate as Rs r per GB, form the pair. Compare the two bills to find f and r, then find the bill for a month with 8 GB extra.

Try it first, then show the solutionSolution

3 GB month: $f + 3 r = 239$ .
5 GB month: $f + 5 r = 299$ .
Subtract the first from the second: $2 r = 60$ , so $r = 30$ .
Put in the first: $f + 90 = 239$ , so $f = 149$ .
Check in the second: $149 + 5 (30) = 149 + 150 = 299$ , true.
8 GB bill: $149 + 8 (30) = 149 + 240 = 389$ .

Answer: f + 3r = 239 and f + 5r = 299; fixed charge Rs 149, rate Rs 30 per GB; the 8 GB bill is Rs 389.

Memory hook: Subtract the two bills: the fixed charge cancels and only the data rate is left.

Tricks to remember Pair of Linear Equations: Meaning and Forms

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

Both or nothing
A solution of a pair must pass BOTH equations. One pass and one fail means not a solution.
Use it when: Whenever you verify a point.
Cross the bridge, flip the sign
When a term crosses the equals sign, its sign flips: 2x + y = 7 becomes 2x + y − 7 = 0, so c = −7.
Use it when: Reading c in the general form.
One fact, one equation
Count the separate facts in the question. Two facts give two equations, the pair.
Use it when: Forming equations from word problems.

Pair of Linear Equations: Meaning and Forms: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Linear or not, and many solutions · 3 marks · Short answerBoard level

Which of $2 x + 3 y = 6$ , $x y + y = 4$ , $\frac{x}{2} + y = 1$ and $x^{2} - y = 3$ are linear equations in two variables? For each linear one, find three solutions.

Try it first, then show the solutionSolution

$2 x + 3 y = 6$ is linear: each variable has power 1 and they are not multiplied. $\frac{x}{2} + y = 1$ is linear too, since $\frac{x}{2}$ is just $\frac{1}{2} x$ . (1 mark)
$x y + y = 4$ has the product $x y$ (degree 2) and $x^{2} - y = 3$ has $x^{2}$ , so neither is linear. (½ mark)
For $2 x + 3 y = 6$ : $x = 0$ gives $y = 2$ ; $y = 0$ gives $x = 3$ ; $x = 6$ gives $y = - 2$ . So $(0, 2)$ , $(3, 0)$ , $(6, - 2)$ . (1 mark)
For $\frac{x}{2} + y = 1$ : $(0, 1)$ , $(2, 0)$ , $(4, - 1)$ . Any value of $x$ gives a matching $y$ , so each line has infinitely many solutions. (½ mark)

Answer: Linear: $2 x + 3 y = 6$ (e.g. $(0, 2)$ , $(3, 0)$ , $(6, - 2)$ ) and $\frac{x}{2} + y = 1$ (e.g. $(0, 1)$ , $(2, 0)$ , $(4, - 1)$ ). Not linear: $x y + y = 4$ and $x^{2} - y = 3$ .

Example 2: Standard form from brackets and decimals · 2 marks · Very short answerBoard level

Write the pair $2 (x + 3) = 3 y - 4$ and $0.5 x + 0.25 y = 2$ in the form $a x + b y + c = 0$ with integer coefficients. State $a_{1}, b_{1}, c_{1}, a_{2}, b_{2}, c_{2}$ and find $c_{1} + c_{2}$ .

Try it first, then show the solutionSolution

Expand the first: $2 x + 6 = 3 y - 4$ , so $2 x - 3 y + 10 = 0$ . (½ mark)
So $a_{1} = 2$ , $b_{1} = - 3$ , $c_{1} = 10$ . (½ mark)
Multiply the second by 4 to clear the decimals: $2 x + y = 8$ , so $2 x + y - 8 = 0$ . (½ mark)
So $a_{2} = 2$ , $b_{2} = 1$ , $c_{2} = - 8$ , and $c_{1} + c_{2} = 10 - 8 = 2$ . (½ mark)

Answer: $2 x - 3 y + 10 = 0$ and $2 x + y - 8 = 0$ ; $a_{1} = 2$ , $b_{1} = - 3$ , $c_{1} = 10$ , $a_{2} = 2$ , $b_{2} = 1$ , $c_{2} = - 8$ ; $c_{1} + c_{2} = 2$ .

Example 3: Find constants from a given solution · 3 marks · Short answerBoard level

$x = 3$ , $y = - 2$ is a solution of both $p x + 4 y = 7$ and $2 x - q y = 14$ . Find $p$ and $q$ , and hence find the value of $p^{2} + q$ .

Try it first, then show the solutionSolution

Substitute into the first: $3 p + 4 (- 2) = 7$ . (½ mark)
$3 p - 8 = 7$ , so $3 p = 15$ , $p = 5$ . (½ mark)
Substitute into the second: $2 (3) - q (- 2) = 14$ . (½ mark)
$6 + 2 q = 14$ , so $q = 4$ . (½ mark)
Check: $15 - 8 = 7$ and $6 + 8 = 14$ , both true. (½ mark)
$p^{2} + q = 25 + 4 = 29$ . (½ mark)

Answer: p = 5, q = 4, and p² + q = 29.

Example 4: Testing several points · 3 marks · Short answerHarder board level

Which of the points $(1, 2)$ , $(4, - 1)$ and $(- 2, 5)$ lie on the line $x + y = 3$ ? Which one is the solution of the pair $x + y = 3$ and $3 x + 2 y = 10$ ? Explain why the other points are not solutions of the pair. Also find the point on $x + y = 3$ at which $3 x + 2 y$ is 13, that is, 3 more than 10.

Try it first, then show the solutionSolution

$(1, 2)$ : $1 + 2 = 3$ ; $(4, - 1)$ : $4 - 1 = 3$ ; $(- 2, 5)$ : $- 2 + 5 = 3$ . All lie on $x + y = 3$ . (½ mark)
In $3 x + 2 y$ : $(1, 2)$ gives $7$ , $(4, - 1)$ gives $10$ and $(- 2, 5)$ gives $4$ . Only $(4, - 1)$ satisfies both, so it is the solution; the others lie only on the first line, so they are not common points. (1 mark)
On $x + y = 3$ , $y = 3 - x$ , so $3 x + 2 (3 - x) = x + 6$ . Setting $x + 6 = 13$ gives $x = 7$ , $y = - 4$ . (1 mark)

Answer: All three lie on $x + y = 3$ ; only $(4, - 1)$ is the solution of the pair. The point $(7, - 4)$ lies on $x + y = 3$ and gives $3 x + 2 y = 21 - 8 = 13$ .

Example 5: Forming the pair from a situation and verifying · 5 marks · Long answerHarder board level

A school hall has some rows of chairs with the same number of chairs in each row. If 3 chairs are added to each row and 2 rows are removed, the total number of chairs is unchanged. If 3 chairs are removed from each row and 4 rows are added, the total is again unchanged. Taking $x$ rows and $y$ chairs per row, form a pair of linear equations in standard form. Verify that 8 rows of 9 chairs satisfies the pair, and find the total number of chairs.

Try it first, then show the solutionSolution

Total chairs $= x y$ . (½ mark)
First condition: $(x - 2) (y + 3) = x y$ . (½ mark)
Expand: $x y + 3 x - 2 y - 6 = x y$ , so $3 x - 2 y - 6 = 0$ . (1 mark)
Second condition: $(x + 4) (y - 3) = x y$ . (½ mark)
Expand: $x y - 3 x + 4 y - 12 = x y$ , so $- 3 x + 4 y - 12 = 0$ , i.e. $3 x - 4 y + 12 = 0$ . (1 mark)
Check $(8, 9)$ : $3 (8) - 2 (9) - 6 = 24 - 18 - 6 = 0$ and $3 (8) - 4 (9) + 12 = 24 - 36 + 12 = 0$ , both true. (1 mark)
Total chairs $= 8 \times 9 = 72$ ; as a check, $(8 - 2) (9 + 3) = 72$ and $(8 + 4) (9 - 3) = 72$ . (½ mark)

Answer: 3x − 2y − 6 = 0 and 3x − 4y + 12 = 0; 8 rows of 9 chairs works, total 72 chairs.

Example 6: Case study: metro smart cards · 4 marks · Case-basedHOTS / competency

Suppose two friends top up metro smart cards. A ride on a short route costs Rs x and a ride on a long route costs Rs y. Aman takes 6 short and 4 long rides in a week and spends Rs 240. Sara takes 3 short and 5 long rides and spends Rs 237.
(i) Write the pair of equations in the form $a x + b y + c = 0$ . (1 mark)
(ii) Aman thinks the fares are $x = 20$ , $y = 30$ . Is he right? Justify. (1 mark)
(iii) Find the fares by elimination. Next week Sara has Rs 500 on her card and must take 4 short rides. How many long rides can she afford? (2 marks)

Try it first, then show the solutionSolution

(i) $6 x + 4 y - 240 = 0$ and $3 x + 5 y - 237 = 0$ . (1 mark)
(ii) $6 (20) + 4 (30) = 240$ , true; $3 (20) + 5 (30) = 210 \neq = 237$ . The guess satisfies only Aman's equation, so he is wrong. (1 mark)
(iii) Double Sara's equation: $6 x + 10 y = 474$ . Subtract Aman's: $6 y = 234$ , so $y = 39$ . Then $6 x = 240 - 156 = 84$ , so $x = 14$ . Check: $42 + 195 = 237$ . (1 mark)
4 short rides cost $4 \times 14 = 56$ , leaving $500 - 56 = 444$ . $444 \div 39 = 11.38 \dots$ , so she can afford 11 long rides ( $11 \times 39 = 429$ ), with Rs 15 left. (1 mark)

Answer: (i) $6 x + 4 y - 240 = 0$ , $3 x + 5 y - 237 = 0$ . (ii) No: $(20, 30)$ fails Sara's equation. (iii) Short ride Rs 14, long ride Rs 39; Sara can afford 11 long rides.

Example 7: An unknown solution and an unknown constant · 3 marks · Short answerHarder board level

The point $(k + 1, 2 k - 3)$ is a solution of both $3 x + 2 y = 11$ and $5 x - y = m$ . Find $k$ , the solution and $m$ . Then find the point where the line $5 x - y = m$ crosses the x-axis.

Try it first, then show the solutionSolution

Substitute in the first: $3 (k + 1) + 2 (2 k - 3) = 11$ , so $7 k - 3 = 11$ and $k = 2$ . (1 mark)
The solution is $(3, 1)$ . Substitute in the second: $m = 5 (3) - 1 = 14$ . (1 mark)
On the x-axis $y = 0$ , so $5 x = 14$ and $x = \frac{14}{5}$ . The line crosses at $(\frac{14}{5}, 0)$ . Check: $9 + 2 = 11$ . (1 mark)

Answer: $k = 2$ , solution $(3, 1)$ , $m = 14$ ; the line $5 x - y = 14$ crosses the x-axis at $(\frac{14}{5}, 0)$ .

Example 8: Forming and solving a pair from ages · 5 marks · Long 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. Form a pair of linear equations in standard form, solve it by elimination, and find their present ages.

Try it first, then show the solutionSolution

Let the father's present age be $x$ years and the son's be $y$ years. (½ mark)
Five years ago: $x - 5 = 7 (y - 5)$ , so $x - 7 y + 30 = 0$ . (1 mark)
Five years hence: $x + 5 = 3 (y + 5)$ , so $x - 3 y - 10 = 0$ . (1 mark)
Subtract the first from the second: $(x - 3 y - 10) - (x - 7 y + 30) = 0$ , so $4 y - 40 = 0$ , $y = 10$ . (1 mark)
Substitute in $x - 3 y - 10 = 0$ : $x = 30 + 10 = 40$ . (½ mark)
Check: $40 - 5 = 35 = 7 \times 5$ and $40 + 5 = 45 = 3 \times 15$ , both true. (½ mark)
So the father is 40 years old and the son is 10 years old. (½ mark)

Answer: Equations x − 7y + 30 = 0 and x − 3y − 10 = 0; father 40 years, son 10 years.

Where students lose marks in Pair of Linear Equations: Meaning and Forms

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

Mistake: Reading the wrong sign for c
Students copy the constant from ax + by = k without changing its sign.
Wrong working
3x − 4y = 9, so c = 9
Right working
3x − 4y − 9 = 0, so c = −9
Do this: Always rewrite with 0 on the right before naming a, b, c.
Usually costs: ½ to 1 mark, and later ratio work goes wrong.
Mistake: Checking a point in only one equation
A point on one line is assumed to solve the pair.
Wrong working
(1, 2): 1 + 2 = 3 true, so it solves x + y = 3 and 3x + 2y = 10
Right working
3(1) + 2(2) = 7 ≠ 10, so (1, 2) is not a solution of the pair
Do this: Write LHS and RHS for both equations every time.
Usually costs: Full marks for the part.
Mistake: Not clearing fractions before naming coefficients
Coefficients are read off while fractions remain, then mixed with integer forms.
Wrong working
x/2 + y/3 = 1, so a = 1, b = 1, c = −1
Right working
Multiply by 6: 3x + 2y − 6 = 0, so a = 3, b = 2, c = −6 (or a = 1/2, b = 1/3, c = −1)
Do this: Clear fractions by the LCM first and keep the whole equation scaled.
Usually costs: ½ to 1 mark.
Mistake: Mixing up the meaning of variables between conditions
x is used for one quantity in the first sentence and another in the second.
Wrong working
Let x = cost of pen. 'A notebook and 2 pens cost 50': x + 2y = 50
Right working
Let x = pen, y = notebook: 2x + y = 50
Do this: Write 'let x = ..., y = ...' with units at the top and reread it for each equation.
Usually costs: Most of the marks in a word problem.

Exam tips

Begin every word problem with a clear 'let' statement; examiners award a mark for correctly defined variables and equations.
Show the substitution of a point into each equation on separate lines, with LHS = RHS stated.
When a case-study part gives a suggested answer, test it in both equations before trusting it; it may be deliberately wrong.
Before any ratio or elimination work, write both equations in the same form so the signs of c match.
Simplify products like (x + 2)(y − 3) = xy fully; the xy terms cancel and leave a linear equation.

Practice questions on Pair of Linear Equations: Meaning and Forms

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

Q1 · 3 marks · Short answerHarder board level
The point $(2, - 1)$ is a solution of both $a x + 3 y = 5$ and $4 x - b y = 3 a$ . Find $a$ and $b$ , write both equations in standard form, and find the solution of the first equation that has $x = - 4$ .
Hint
Find $a$ from the first equation before you touch the second, because the second contains $a$ too.
Answer
First: $2 a - 3 = 5$ , so $a = 4$ . Second: $8 + b = 3 (4) = 12$ , so $b = 4$ . Standard forms: $4 x + 3 y - 5 = 0$ and $4 x - 4 y - 12 = 0$ (or $x - y - 3 = 0$ ). With $x = - 4$ in the first: $- 16 + 3 y = 5$ , so $y = 7$ ; the solution is $(- 4, 7)$ .
Q2 · 3 marks · Short answerBoard level
If $x = 2$ , $y = 1$ is a solution of $a x + b y = 8$ and $2 a x - b y = 10$ , find $a$ and $b$ .
Hint
Substituting gives a new pair in a and b.
Answer
2a + b = 8 and 4a − b = 10. Adding: 6a = 18, a = 3, b = 2. Check: 6 + 2 = 8, 12 − 2 = 10.
Q3 · 3 marks · Short answerHarder board level
The sum of the digits of a two-digit number is 9. The number formed by reversing the digits is 27 more than the original number. Taking the tens digit as $x$ and the units digit as $y$ , form a pair of linear equations in standard form, solve it by elimination, and check that the number is also 4 times the sum of its digits.
Hint
The number is $10 x + y$ and the reversed number is $10 y + x$ .
Answer
$x + y - 9 = 0$ and $(10 y + x) - (10 x + y) = 27$ , i.e. $9 y - 9 x = 27$ , so $x - y + 3 = 0$ . Adding: $2 x - 6 = 0$ , so $x = 3$ and $y = 6$ . The number is 36. Check: $63 - 36 = 27$ , and $4 \times 9 = 36$ .
Q4 · 4 marks · Long answerHarder board level
The sum of two numbers is 3 times their difference, and the larger exceeds twice the smaller by 2. Taking the larger number as $x$ and the smaller as $y$ , form the pair in standard form and check whether $x = 8$ , $y = 4$ is its solution. Find the product of the numbers if so.
Hint
x + y = 3(x − y) simplifies to a linear equation.
Answer
x + y = 3x − 3y gives 2x − 4y = 0, i.e. x − 2y = 0. Second: x − 2y − 2 = 0. Check (8, 4): 8 − 8 = 0 true, 8 − 8 − 2 = −2 ≠ 0 false. Not a solution. In fact x − 2y = 0 and x − 2y = 2 cannot both hold, so no such numbers exist and no product can be found.
Q5 · 3 marks · Short answerHOTS / competency
Kabir says: " $x y = 6$ and $x + y = 5$ is a pair of linear equations in two variables, with the single solution $(2, 3)$ ." Find both errors in his statement, find every solution of his pair, and explain why a pair of linear equations can never have exactly two solutions.
Hint
Put $y = 5 - x$ into $x y = 6$ . Two lines can meet at one point, at no point, or at every point.
Answer
Error 1: $x y = 6$ has the product $x y$ (degree 2), so the pair is not linear. Error 2: the solution is not single. With $y = 5 - x$ : $x (5 - x) = 6$ , so $x^{2} - 5 x + 6 = 0$ and $x = 2$ or $3$ . The solutions are $(2, 3)$ and $(3, 2)$ . Two different straight lines meet in at most one point, and if they share two points they are the same line and share all their points. So a linear pair has 0, 1 or infinitely many solutions, never exactly two.
Q6 · 1 mark · MCQBoard level
If $x = a$ , $y = b$ is the solution of $x + y = 5$ and $2 x - y = 7$ , then $a^{2} + b^{2}$ equals
(a) $17$
(b) $25$
(c) $13$
(d) $15$
Hint
Add the equations to remove $y$ .
Answer
(a) $17$ . Adding: $3 x = 12$ , so $a = 4$ and $b = 1$ . Then $a^{2} + b^{2} = 16 + 1 = 17$ . (25 is the trap of squaring $a + b$ .)
Q7 · 1 mark · MCQBoard level
If $(3, - 1)$ satisfies both $k x + 2 y = 7$ and $3 x - m y = 11$ , then $k + m$ equals
(a) $5$
(b) $1$
(c) $3$
(d) $- 1$
Hint
Find each constant from its own equation.
Answer
(a) $5$ . $3 k - 2 = 7$ gives $k = 3$ ; $9 + m = 11$ gives $m = 2$ ; so $k + m = 5$ .
Q8 · 1 mark · Assertion-reasonBoard level
Assertion (A): $(2, 1)$ is a solution of $3 x - y = 5$ , so it is a solution of the pair $3 x - y = 5$ and $x + 2 y = 3$ .
Reason (R): A solution of a pair of linear equations must satisfy both equations.
(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
Test the point in the second equation as well.
Answer
(d). $3 (2) - 1 = 5$ is true, but $2 + 2 (1) = 4 \neq = 3$ , so $(2, 1)$ is not a solution of the pair and A is false. R is the correct rule, and it is exactly why A fails.
Q9 · 4 marks · Case-basedHOTS / competency
Suppose your school fair sells adult and child tickets at fixed prices, Rs x for an adult and Rs y for a child. One family pays Rs 260 for 2 adults and 3 children. Another family pays Rs 290 for 3 adults and 2 children.
(i) Write the pair of linear equations. (1 mark)
(ii) A friend says an adult ticket is Rs 40 and a child ticket is Rs 60. Test this and say whether you accept or reject it. (1 mark)
(iii) Find the correct ticket prices by elimination. (2 marks)
Hint
Adding and subtracting the two equations gives x + y and x − y.
Answer
(i) 2x + 3y = 260 and 3x + 2y = 290. (ii) 2(40) + 3(60) = 260 is true, but 3(40) + 2(60) = 240 ≠ 290, so reject it. (iii) Adding: 5x + 5y = 550, so x + y = 110. Subtracting the first from the second: x − y = 30. So 2x = 140, x = 70, y = 40. Check: 140 + 120 = 260 and 210 + 80 = 290. Adult ticket Rs 70, child ticket Rs 40.

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.General form of a linear equation in two variables?Show answer
ax + by + c = 0, with a and b not both zero.
2.How many solutions does a single linear equation in two variables have?Show answer
Infinitely many: every point on its line.
3.What is a solution of a pair?Show answer
An ordered pair (x, y) satisfying both equations.
4.Write 4x = 3y + 7 in general form.Show answer
4x − 3y − 7 = 0.
5.Is 2x + 3xy = 5 linear?Show answer
No, the xy term has degree 2.
6.Geometric meaning of a solution of a pair?Show answer
A point common to both lines.

Pair of Linear Equations: Meaning and Forms: frequently asked questions

Why do we need two equations instead of one?

One equation in two unknowns has infinitely many solutions. A second independent condition is needed to pin down one pair of values, which is the common point of the two lines.

Does the form ax + by = c or ax + by + c = 0 matter?

Both are correct, but use the same form for both equations. The sign of c is different in the two forms, so mixing them causes errors in later work.

Can a pair of linear equations have no solution?

Yes. If the lines are parallel there is no common point, so no solution. You study this in the topic on consistent and inconsistent pairs.

Is the cross-multiplication method in the syllabus?

No. Cross-multiplication and equations reducible to linear form are not in the 2026-27 syllabus. Use substitution and elimination.