MathsCBSE Class 9 · Chapter 13

Two Variables, One Line Class 9: Linear Equations, Slope and Pairs of Equations

Q: Is solving a pair of linear equations in Class 9 now?

Yes. The 2026-27 CBSE content for this chapter includes pairs of linear equations, the graphical method, consistency and inconsistency, and the substitution and elimination methods.

Q: Do I need the cross-multiplication method?

No. The chapter teaches substitution, elimination and graphs.

Q: How is this chapter linked to Introduction to Linear Polynomials?

Chapter 2 introduced y = ax + b with slope a and y-intercept b. This chapter writes the same idea as y = mx + d, works with the general form ax + by + c = 0, and adds pairs of equations.

Q: What does 'consistent' mean?

A pair is consistent if it has at least one solution (intersecting or coincident lines) and inconsistent if it has none (parallel lines).

Q: Why is the slope of a vertical line undefined?

On a vertical line the run (change in x) is 0, and division by 0 has no meaning.

Updated 15 October 2026 · Fact-checked

Two Variables, One Line in Class 9 covers linear equations ax + by + c = 0, whose infinitely many solutions lie on a straight line. You learn slope and the form y = mx + d, then solve pairs of equations by substitution, elimination and graphs, and decide whether a pair has one, none or infinitely many solutions.

What Two Variables, One Line is about

This chapter of *Ganita Manjari* Part II is CBSE's Linear Equations in Two Variables (Unit II, Algebra). It begins at a fruit cart: mangoes at ₹60 a kg, bananas at ₹50 a kg, a bill of ₹280. With two unknown amounts, the situation becomes one equation, $60 x + 50 y = 280$ .

The chapter has two halves.

One equation, one line:

standard form $a x + b y + c = 0$ (a and b not both zero);
a solution is an ordered pair $(x, y)$ ; there are infinitely many, and when plotted they form a straight line;
slope (rise ÷ run) measures steepness; in $y = m x + d$ , m is the slope and d the y-intercept;
in real situations x and y may only take certain values (whole numbers of items), so the graph is a set of separate points on the line.

Two equations, at most one meeting point:

a pair of linear equations is solved by substitution, elimination or graphs;
two lines can intersect (one solution), be parallel (no solution) or coincide (infinitely many solutions); comparing the ratios of coefficients tells you which.

Questions range from checking a solution or finding a slope (1 to 2 marks) to word problems on tickets, fares, ages, digits and fractions (3 marks) and graphical solutions (5 marks).

Weightage: This chapter is part of Unit II, Algebra, which carries 20 of the 80 marks in the school annual exam, shared with Introduction to Polynomials, Sequences and Progressions and Exploring Algebraic Identities.Source: CBSE Mathematics syllabus, Class IX 2026-27, course structure, Unit II Algebra

Key concepts

Linear equations in two variables and standard form

NCERT section 13.1 Linear Equations in Two Variables

An equation that can be written as
$a x + b y + c = 0$
with a, b, c real numbers and a and b not both zero is a linear equation in two variables (in standard form). a and b are the coefficients of x and y; c is the constant.

Fractions are allowed, but we usually multiply through to get whole-number coefficients.
An equation in one variable such as $4 x + 7 = 0$ also counts, as $4 x + 0 \cdot y + 7 = 0$ .
If $c = 0$ , the line passes through the origin.

Worked example

Write in standard form and give a, b and c: (i) $3 y = 2 x - 5$ (ii) $\frac{x}{2} + \frac{y}{3} = 1$ (iii) $y = 4$ .

Try it first, then show the solutionSolution

(i) $2 x - 3 y - 5 = 0$ : $a = 2$ , $b = - 3$ , $c = - 5$ .
(ii) Multiply by 6: $3 x + 2 y = 6$ , so $3 x + 2 y - 6 = 0$ : $a = 3$ , $b = 2$ , $c = - 6$ .
(iii) $0 \cdot x + y - 4 = 0$ : $a = 0$ , $b = 1$ , $c = - 4$ .

Answer: (i) $2 x - 3 y - 5 = 0$ (ii) $3 x + 2 y - 6 = 0$ (iii) $0 x + y - 4 = 0$

Solutions are ordered pairs, and there are infinitely many

NCERT section 13.2 Solution of Linear Equation in Two Variables

A solution is a pair of values $(x, y)$ that makes the equation true. The order matters: the first number is x. (2, 3) solves $3 x + 2 y = 12$ because $6 + 6 = 12$ ; (1, 3) does not.

To produce solutions, choose any value for x, substitute, and solve the one-variable equation for y. Since x can be anything, there are infinitely many solutions. Choosing $x = 0$ and $y = 0$ in turn gives the two intercept points, which are often the easiest to find.

Worked example

Find three solutions of $2 x + 3 y = 12$ and check whether $(3, 2)$ is a solution.

Try it first, then show the solutionSolution

$x = 0$ : $3 y = 12$ , $y = 4$ , giving (0, 4).
$y = 0$ : $2 x = 12$ , $x = 6$ , giving (6, 0).
$x = 3$ : $3 y = 6$ , $y = 2$ , giving (3, 2).
Check (3, 2): $2 (3) + 3 (2) = 12$ . Yes.

Answer: (0, 4), (6, 0) and (3, 2); (3, 2) is a solution.

The graph is a straight line

NCERT section 13.2 Solution of Linear Equation in Two Variables

Plot the solutions of a linear equation in two variables and they all lie on one straight line. Two facts go together:

every point on the line is a solution;
every solution is a point on the line.

So to draw the graph, find two solutions (a third as a check), plot them and draw the line through them. To test whether a point lies on the line, substitute its coordinates.

Special lines: $x = k$ is vertical, $y = k$ is horizontal, $x = 0$ is the y-axis and $y = 0$ is the x-axis.

Worked example

The point $(2, k)$ lies on the graph of $3 x - y = 4$ . Find k, and give the points where the line crosses the axes.

Try it first, then show the solutionSolution

Substitute: $3 (2) - k = 4$ , so $k = 2$ .
y-axis ( $x = 0$ ): $- y = 4$ , so $(0, - 4)$ .
x-axis ( $y = 0$ ): $3 x = 4$ , so $(\frac{4}{3}, 0)$ .

Answer: $k = 2$ ; the line crosses the axes at $(0, - 4)$ and $(\frac{4}{3}, 0)$ .

Slope of a line

NCERT section 13.3 Slope-intercept form of a Linear Equation in Two Variables

The slope (gradient) measures steepness:
$m = \frac{rise}{run} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
for any two points on the line; it is the same whichever two points (and whichever order) you use.

Positive slope: as x increases, y increases (the line rises to the right).
Negative slope: as x increases, y decreases.
Zero slope: a horizontal line, $y = k$ .
Undefined slope: a vertical line, $x = k$ (the run is 0).

A ramp's slope is its height gained ÷ horizontal distance.

Worked example

Find the slope of the line through $A (- 2, 5)$ and $B (4, - 1)$ , and describe the line.

Try it first, then show the solutionSolution

$m = \frac{- 1 - 5}{4 - ( - 2 )} = \frac{- 6}{6} = - 1$ .
Negative slope: y falls by 1 for every 1 step to the right.

Answer: Slope $- 1$ ; the line falls to the right.

Slope-intercept form y = mx + d

NCERT section 13.3 Slope-intercept form of a Linear Equation in Two Variables

Solving $a x + b y + c = 0$ for y (when $b \neq = 0$ ) gives
$y = m x + d$
where m is the slope and d is the y-intercept: the line cuts the y-axis at $(0, d)$ .

Why m is the slope: take two points with x-values $x_{1}$ and $x_{2}$ ; their y-values differ by $m (x_{2} - x_{1})$ , so rise ÷ run $= m$ .

The x-intercept is found by putting $y = 0$ : $x = - \frac{d}{m}$ (if $m \neq = 0$ ). Lines with equal slopes and different intercepts are parallel.

Worked example

Write $4 x + 2 y = 6$ in slope-intercept form. State the slope, the y-intercept and the x-intercept, and explain what the slope means.

Try it first, then show the solutionSolution

$2 y = - 4 x + 6$ , so $y = - 2 x + 3$ .
Slope $m = - 2$ ; y-intercept $d = 3$ , point $(0, 3)$ .
x-intercept: $0 = - 2 x + 3$ , $x = 1.5$ , point $(1.5, 0)$ .
For every increase of 1 in x, y decreases by 2.

Answer: $y = - 2 x + 3$ ; slope $- 2$ , y-intercept 3, x-intercept 1.5.

Graphs when x and y can take only certain values

The CBSE syllabus asks you to draw graphs 'when x and y assume only certain values'. In real problems the variables are often counts (numbers of pens, tickets, people) or cannot be negative (lengths, times). Then only some points of the line make sense.

The graph becomes a set of separate dots on the line, or a segment of it, not the whole line. Always state the allowed values (for example, x and y are whole numbers with $x, y \geq 0$ ).

Worked example

Notebooks cost ₹20 and pens ₹10. Riya spends exactly ₹100 on x notebooks and y pens. Write the equation and list all possible purchases. Describe the graph.

Try it first, then show the solutionSolution

$20 x + 10 y = 100$ , or $2 x + y = 10$ , with x and y whole numbers.
x can be 0 to 5: (0, 10), (1, 8), (2, 6), (3, 4), (4, 2), (5, 0).
The graph is these 6 separate points on the line $y = - 2 x + 10$ , from (0, 10) to (5, 0); points between them (like $x = 1.5$ ) make no sense here.

Answer: 6 possible purchases; the graph is 6 dots on the line $2 x + y = 10$ .

Solving a pair of equations by substitution

NCERT section 13.5 Finding Solutions to a Pair of Linear Equations

A pair of linear equations $a_{1} x + b_{1} y + c_{1} = 0$ , $a_{2} x + b_{2} y + c_{2} = 0$ arises when a situation gives two facts about two unknowns (two shopping bills, two taxi fares). A solution is an ordered pair satisfying both.

Substitution method:
1. Make one variable the subject of one equation (choose the easiest).
2. Substitute into the other equation to get an equation in one variable.
3. Solve it, then substitute back for the other variable.
4. Check in both original equations.

Worked example

Solve by substitution: $x + 2 y = 11$ and $3 x - y = 5$ .

Try it first, then show the solutionSolution

From the second: $y = 3 x - 5$ .
Substitute: $x + 2 (3 x - 5) = 11$ , so $7 x - 10 = 11$ and $x = 3$ .
$y = 3 (3) - 5 = 4$ .
Check: $3 + 8 = 11$ and $9 - 4 = 5$ .

Answer: $x = 3$ , $y = 4$

Solving a pair of equations by elimination

NCERT section 13.5 Finding Solutions to a Pair of Linear Equations

Elimination method:
1. Multiply one or both equations so that one variable has the same (or opposite) coefficient in both.
2. Subtract (or add) the equations to remove that variable.
3. Solve for the remaining variable, then substitute back.

If both variables disappear:

and you get a true statement like $0 = 0$ , the equations are the same line: infinitely many solutions;
and you get a false statement like $0 = 5$ , the lines are parallel: no solution.

Worked example

Solve by elimination: $3 x + 4 y = 10$ and $2 x - 2 y = 2$ .

Try it first, then show the solutionSolution

Multiply the second equation by 2: $4 x - 4 y = 4$ .
Add to the first: $7 x = 14$ , so $x = 2$ .
Then $2 (2) - 2 y = 2$ , so $y = 1$ .
Check: $6 + 4 = 10$ .

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

Graphical method and the nature of solutions

NCERT section 13.6 Graphical Method for Solving A Pair of Linear Equations in Two Variables

Draw both lines; a solution of the pair is a point on both.

Lines	Number of solutions	Called	Ratio test
intersect	exactly one	consistent	$\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$
parallel	none	inconsistent	$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$
coincide	infinitely many	consistent (dependent)	$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$

The ratio test is the same as comparing slopes: equal slopes mean parallel or coincident lines.

Worked example

Without drawing, decide whether each pair has one, no or infinitely many solutions: (i) $x + 2 y = 4$ , $2 x + 4 y = 12$ (ii) $2 x + 3 y = 9$ , $4 x + 6 y = 18$ (iii) $x + y = 5$ , $2 x - y = 4$ .

Try it first, then show the solutionSolution

(i) $\frac{1}{2} = \frac{2}{4} \neq = \frac{4}{12}$ : parallel lines, no solution.
(ii) $\frac{2}{4} = \frac{3}{6} = \frac{9}{18}$ : the same line, infinitely many solutions.
(iii) $\frac{1}{2} \neq = \frac{1}{- 1}$ : the lines intersect; solving gives $x = 3$ , $y = 2$ .

Answer: (i) none (ii) infinitely many (iii) exactly one, (3, 2)

Maths in real life: Two Variables, One Line

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.

Linear equation in two variables

Suppose at a fruit cart apples cost ₹120 per kg and bananas ₹40 per dozen, and you spend exactly ₹320.

Write the equation and find two possible purchases.

Try it first, then show the solutionSolution

$120 x + 40 y = 320$ , or $3 x + y = 8$ .
x = 2, y = 2: 2 kg apples, 2 dozen bananas.
x = 1, y = 5: 1 kg apples, 5 dozen bananas.

Answer: $3 x + y = 8$ ; for example (2, 2) or (1, 5).

Memory hook: One bill, two unknowns: many possible baskets.

Slope as a rate

Suppose an auto fare is $y = 15 x + 25$ rupees for x km.

What do 15 and 25 mean, and what is the fare for 8 km?

Try it first, then show the solutionSolution

Slope 15: each extra km adds ₹15.
Intercept 25: the fixed starting charge.
$15 \times 8 + 25 = 145$ .

Answer: ₹15 per km plus ₹25 fixed; ₹145 for 8 km.

Memory hook: Slope is the rate, intercept is the start.

Slope of a ramp

Suppose your school builds a ramp that rises 0.6 m over a horizontal run of 9 m.

What is its slope, and is it gentler than $\frac{1}{12}$ ?

Try it first, then show the solutionSolution

Slope $= \frac{0.6}{9} = \frac{1}{15}$ .
$\frac{1}{15} < \frac{1}{12}$ , so it is gentler.

Answer: $\frac{1}{15}$ ; yes, gentler than $\frac{1}{12}$ .

Memory hook: Rise over run: smaller is gentler.

Pair of equations

Suppose at an exhibition 2 adult and 3 child tickets cost ₹550, and 3 adult and 2 child tickets cost ₹600.

Find the price of each ticket.

Try it first, then show the solutionSolution

$2 a + 3 c = 550$ , $3 a + 2 c = 600$ .
Adding: $a + c = 230$ ; subtracting: $a - c = 50$ .
$a = 140$ , $c = 90$ .

Answer: Adult ₹140, child ₹90.

Memory hook: Two facts pin down two unknowns.

Parallel lines: no solution

Suppose two buses leave 10 km apart on the same straight road at the same moment, both travelling at 40 km/h in the same direction.

Will the rear bus ever catch the front one?

Try it first, then show the solutionSolution

Distances: $y = 40 t$ and $y = 40 t + 10$ .
Equal slopes, different intercepts: parallel lines that never meet.

Answer: No; the gap stays 10 km.

Memory hook: Same speed, different start: never meet.

Topics in this chapter

Two Variables, One Line formula sheet

Standard form: $a x + b y + c = 0, a, b not both 0$
Slope from two points: $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$; Rise ÷ run; undefined for vertical lines.
Slope-intercept form: $y = m x + d$; m slope, d y-intercept: crosses the y-axis at $(0, d)$ .
From standard form: $y = - \frac{a}{b} x - \frac{c}{b} (b \neq = 0)$; So the slope is $- \frac{a}{b}$ .
x-intercept: $x = - \frac{d}{m} for y = m x + d$; Put $y = 0$ .
Special lines: $x = k (vertical), y = k (horizontal, slope 0)$
Through the origin: $c = 0 \Rightarrow (0, 0) lies on a x + b y + c = 0$
Parallel lines: $m_{1} = m_{2}, d_{1} \neq = d_{2}$
Unique solution: $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$
No solution: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$
Infinitely many solutions: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$
Two-digit number: $10 x + y (tens digit x, units digit y); reversed: 10 y + x$

How the school exam asks Two Variables, One Line

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

Question type	How it is asked	Method
Standard form and checking solutions	Write an equation in standard form; check whether a point is a solution; find k if a point lies on a line.	Move every term to one side; clear fractions. Substitute coordinates and check whether both sides match; for k, solve the resulting equation.
Slope and intercepts	Find the slope from two points or from an equation; find intercepts; interpret the slope in context (cost per km, ramp steepness).	Use rise ÷ run, or rearrange to $y = m x + d$ . Put $x = 0$ or $y = 0$ for the intercepts. Explain m as 'change in y for each 1 increase in x'.
Forming a pair of equations	Tickets, bills, taxi fares, ages, digits, fractions, coins.	Define both variables with units. Write one equation per fact. For digits use $10 x + y$ ; for ages shift both ages by the same years.
Solving algebraically	Solve by substitution or elimination (sometimes the method is specified).	Substitution when a variable has coefficient 1; otherwise elimination. Always check the answer in both equations and in the story.
Solving graphically	Draw both lines, read the solution, and sometimes find the area of a triangle formed with an axis.	Make a table of at least two (better three) points for each line, use a clear scale, read the intersection, and check it algebraically.
Nature of solutions	Decide whether a pair has one, none or infinitely many solutions; find k for coincident or parallel lines.	Compare $\frac{a _{1}}{a _{2}}$ , $\frac{b _{1}}{b _{2}}$ , $\frac{c _{1}}{c _{2}}$ ; set the needed ratios equal and solve for k.

Tricks to remember Two Variables, One Line

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

Rise over run
Slope is how far you climb (rise) for each step forward (run).
Use it when: Getting the slope formula the right way up.
m moves, d decides where
In $y = m x + d$ , m sets the tilt and d sets where the line crosses the y-axis.
Use it when: Reading slope and intercept quickly.
Cross, never, always
Intersecting lines cross once (one solution); parallel lines never meet (none); coincident lines always meet (infinitely many).
Use it when: The nature-of-solutions table.
Same, same, different = parallel
Ratios a and b the same but c different: parallel lines, no solution.
Use it when: The ratio test.
Check in both
A solution of a pair must work in both equations; test it before you write it down.
Use it when: Catching arithmetic slips.

Two Variables, One Line: exam-level solved questions

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

Example 1: Spot the solution · 1 mark · MCQExam level

Which ordered pair is a solution of $2 x + y = 7$ ?
(a) (3, 2)
(b) (2, 3)
(c) (1, 4)
(d) (0, 6)

Try it first, then show the solutionSolution

(2, 3): $4 + 3 = 7$ .
The others give 8, 6 and 6.

Answer: (b)

Example 2: Slope from an equation · 1 mark · MCQExam level

The slope of the line $y = - 4 x + 9$ is
(a) 9
(b) −4
(c) 4
(d) $\frac{9}{4}$

Try it first, then show the solutionSolution

In $y = m x + d$ , m is the slope.

Answer: (b) −4

Example 3: Parallel lines (assertion-reason) · 1 mark · Assertion-reasonExam level

Assertion (A): The pair $x + 2 y = 4$ and $2 x + 4 y = 12$ has no solution.
Reason (R): Lines with equal slopes and different y-intercepts are parallel.
(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

$y = - \frac{1}{2} x + 2$ and $y = - \frac{1}{2} x + 3$ : equal slopes, different intercepts.
So the lines are parallel and never meet.

Answer: (a)

Example 4: Finding unknown coefficients · 2 marks · Very short answerExam level

$(- 2, 3)$ satisfies both $a x + 4 y = 18$ and $5 x + b y = 2$ . Find a and b.

Try it first, then show the solutionSolution

$- 2 a + 12 = 18$ gives $a = - 3$ .
$- 10 + 3 b = 2$ gives $b = 4$ .

Answer: $a = - 3$ , $b = 4$

Example 5: Ticket prices · 3 marks · Short answerHarder exam level

3 adult and 4 child tickets cost ₹1000; 2 adult and 5 child tickets cost ₹900. Find the price of each ticket.

Try it first, then show the solutionSolution

$3 a + 4 c = 1000$ and $2 a + 5 c = 900$ .
Multiply by 2 and 3: $6 a + 8 c = 2000$ and $6 a + 15 c = 2700$ .
Subtract: $7 c = 700$ , so $c = 100$ ; then $3 a = 600$ , $a = 200$ .
Check: $400 + 500 = 900$ .

Answer: Adult ₹200, child ₹100

Example 6: Taxi fares · 3 marks · Short answerHarder exam level

A taxi charges a fixed amount plus a rate per km. An 8 km ride costs ₹180 and a 14 km ride costs ₹288. Find the fixed charge, the rate, and the fare for 20 km.

Try it first, then show the solutionSolution

$f + 8 r = 180$ and $f + 14 r = 288$ .
Subtract: $6 r = 108$ , so $r = 18$ ; $f = 180 - 144 = 36$ .
20 km: $36 + 20 \times 18 = 396$ .

Answer: Fixed ₹36, ₹18 per km; ₹396 for 20 km.

Example 7: A two-digit number · 3 marks · Short answerHarder exam level

The digits of a two-digit number differ by 3. The number plus the number formed by reversing its digits is 121. Find the number (with the larger digit in the tens place).

Try it first, then show the solutionSolution

Number $= 10 x + y$ , reversed $= 10 y + x$ : $11 (x + y) = 121$ , so $x + y = 11$ .
$x - y = 3$ .
Adding: $2 x = 14$ , so $x = 7$ , $y = 4$ .

Answer: 74 (check: $74 + 47 = 121$ )

Example 8: A fraction from two conditions · 3 marks · Short answerHarder exam level

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

Try it first, then show the solutionSolution

$3 (n + 1) = 2 (d + 1)$ : $3 n - 2 d = - 1$ .
$2 (n - 1) = d - 1$ : $d = 2 n - 1$ .
$3 n - 2 (2 n - 1) = - 1$ gives $n = 3$ , $d = 5$ .

Answer: $\frac{3}{5}$

Example 9: Graphical solution and a triangle · 5 marks · Long answerHarder exam level

Solve $x + y = 5$ and $2 x - y = 4$ graphically. Find the coordinates of the vertices of the triangle formed by the two lines and the y-axis, and its area.

Try it first, then show the solutionSolution

$x + y = 5$ : (0, 5), (5, 0), (3, 2).
$2 x - y = 4$ : (0, −4), (2, 0), (3, 2).
The lines meet at (3, 2), so $x = 3$ , $y = 2$ .
They cut the y-axis at (0, 5) and (0, −4): vertices (0, 5), (0, −4), (3, 2).
Base on the y-axis $= 5 - (- 4) = 9$ ; height $=$ distance of (3, 2) from the y-axis $= 3$ ; area $= \frac{1}{2} \times 9 \times 3 = 13.5$ square units.

Answer: Solution (3, 2); triangle (0, 5), (0, −4), (3, 2) with area 13.5 square units.

Example 10: Choosing a mobile plan · 3 marks · Short answerHOTS / competency

Plan A: ₹99 a month plus ₹0.50 per minute. Plan B: ₹149 a month plus ₹0.25 per minute. For how many minutes do they cost the same? Which plan is cheaper for 300 minutes?

Try it first, then show the solutionSolution

A: $y = 0.5 m + 99$ ; B: $y = 0.25 m + 149$ .
Equal when $0.25 m = 50$ , so $m = 200$ minutes (₹199 each).
At 300 minutes: A $= 249$ , B $= 224$ : B is cheaper, since its slope is smaller.

Answer: Same at 200 minutes; Plan B is cheaper at 300 minutes.

Example 11: Two robots · 4 marks · Case-basedHOTS / competency

Robot 1 starts at the origin and repeatedly moves 2 units right then 3 units up. Robot 2 starts at (6, 0) and repeatedly moves 1 unit right then 3 units up. Find the lines along which their corner points lie, and decide whether their paths meet at a corner point. If they move at the same time, do they arrive there together?

Try it first, then show the solutionSolution

Robot 1's corner points (2, 3), (4, 6), … lie on $y = \frac{3}{2} x$ .
Robot 2's corner points (7, 3), (8, 6), … lie on $y = 3 (x - 6) = 3 x - 18$ .
$\frac{3}{2} x = 3 x - 18$ gives $x = 12$ , $y = 18$ .
Robot 1 reaches (12, 18) after 6 moves; Robot 2 also after 6 moves ( $6 + 6 = 12$ , $3 \times 6 = 18$ ).
So both reach (12, 18) after the same number of moves and would meet there.

Answer: They meet at (12, 18), both after 6 moves.

Case-study practice (Section E style)

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

The school fair food stall

A class runs a food stall selling samosas and glasses of juice at fixed prices. On Day 1 it sells 30 samosas and 20 juices and collects ₹1100. On Day 2 it sells 25 samosas and 30 juices and collects ₹1250.

(i) · 1 mark
Taking the price of a samosa as ₹s and a juice as ₹j, write the two equations.
Answer
Answer: $30 s + 20 j = 1100$ and $25 s + 30 j = 1250$
One equation per day: number sold × price, added, equals the takings.
(ii) · 1 mark
Show that $s = 20$ , $j = 25$ satisfies the Day 1 equation.
Answer
Answer: Yes
$30 \times 20 + 20 \times 25 = 600 + 500 = 1100$ .
(iii) · 2 marks
Solve the pair by elimination and find the cost of 10 samosas and 10 juices.
Answer
Answer: ₹20 and ₹25; ₹450
Simplify: $3 s + 2 j = 110$ and $5 s + 6 j = 250$ . Multiply the first by 3: $9 s + 6 j = 330$ ; subtract: $4 s = 80$ , $s = 20$ , then $j = 25$ (1 mark). Cost $= 10 \times 20 + 10 \times 25 = 450$ (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.

Solve by substitution: $x + 2 y = 11$ and $3 x - y = 5$ . (2 marks)

Step	Marks
$y = 3 x - 5$ , so $x + 2 (3 x - 5) = 11$ and $7 x = 21$	1
$x = 3$ , $y = 4$	1

2 kg of apples and 1 kg of grapes cost ₹300; 1 kg of apples and 2 kg of grapes cost ₹270. Find the price per kg of each. (3 marks)

Step	Marks
$2 a + g = 300$ and $a + 2 g = 270$	1
Adding: $3 a + 3 g = 570$ , so $a + g = 190$ ; subtracting: $a - g = 30$	1
$a = 110$ , $g = 80$ : apples ₹110/kg, grapes ₹80/kg	1

Solve graphically: $x + 2 y = 8$ and $2 x - y = 1$ . State the nature of the solution. (5 marks)

Step	Marks
Table for $x + 2 y = 8$ : (0, 4), (8, 0), (2, 3)	1
Table for $2 x - y = 1$ : (0, −1), (1, 1), (2, 3)	1
Both lines plotted on one graph with a clear scale	1
They meet at (2, 3): $x = 2$ , $y = 3$ ; check $2 + 6 = 8$ and $4 - 3 = 1$	1
The lines intersect at one point, so the pair is consistent with a unique solution	1

Where students lose marks in Two Variables, One Line

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

Mistake: Writing the ordered pair the wrong way round
The value found first is written first.
Wrong working
$y = 4$ found first, so the solution is (4, 3).
Right working
The solution is $(x, y) = (3, 4)$ .
Do this: Always write (x-value, y-value).
Usually costs: The final answer mark.
Mistake: Slope as run over rise
The formula is turned upside down.
Wrong working
Through (1, 2) and (4, 11): $m = \frac{4 - 1}{11 - 2} = \frac{1}{3}$
Right working
$m = \frac{11 - 2}{4 - 1} = 3$
Do this: y-difference on top, x-difference below.
Usually costs: 1 mark.
Mistake: Reading the slope before rearranging
The coefficient of x in standard form is taken as the slope.
Wrong working
$4 x + 2 y = 6$ : slope 4
Right working
$y = - 2 x + 3$ : slope $- 2$
Do this: Make y the subject first.
Usually costs: 1 to 2 marks.
Mistake: Sign errors when subtracting equations
Only the first term's sign is changed.
Wrong working
$(6 a + 15 c) - (6 a + 8 c) = 2700 - 2000$ written as $7 c = 4700$
Right working
$7 c = 700$ , so $c = 100$
Do this: Put the subtracted equation in brackets and change every sign.
Usually costs: The rest of the solution.
Mistake: Drawing a full line when only whole-number values make sense
Every graph is drawn as a continuous line.
Wrong working
Number of pens and notebooks shown as a solid line including negative and fractional values.
Right working
Show only the points with whole-number, non-negative coordinates on the line.
Do this: Ask what values x and y can take in the story.
Usually costs: 1 mark in graph questions.

NCERT exercise guide: Two Variables, One Line

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 Set 13.1

What it tests: Writing equations in standard form, identifying a, b and c, and forming equations from situations.

How to approach it: Bring all terms to one side, clear fractions or decimals, and read off the coefficients with their signs.

Like the standard-form questions in Exercise Set 13.1

A shop sells pencils at ₹5 and erasers at ₹3. Asha spends ₹47. Write this as a linear equation in standard form and give a, b, c.

Show the solutionSolution

Let x pencils and y erasers be bought: $5 x + 3 y = 47$ .
Standard form: $5 x + 3 y - 47 = 0$ .
$a = 5$ , $b = 3$ , $c = - 47$ .

Answer: $5 x + 3 y - 47 = 0$ ; $a = 5$ , $b = 3$ , $c = - 47$ .

Exercise Set 13.2

What it tests: Finding solutions, checking given pairs, drawing graphs, and finding unknown coefficients when a point lies on a line.

How to approach it: Substitute chosen values for one variable; plot at least two points; substitute a given point to find an unknown.

Like the unknown-coefficient questions in Exercise Set 13.2

$(- 2, 3)$ is a solution of both $a x + 4 y = 18$ and $5 x + b y = 2$ . Find a and b.

Show the solutionSolution

First: $- 2 a + 12 = 18$ , so $a = - 3$ .
Second: $- 10 + 3 b = 2$ , so $b = 4$ .

Answer: $a = - 3$ , $b = 4$

Exercise Set 13.3

What it tests: Slopes of lines from graphs and coordinates, the slope of a ramp, and slope-intercept form.

How to approach it: Use rise ÷ run from any two points; for a ramp, height ÷ horizontal length.

Like the ramp question in Exercise Set 13.3

A wheelchair ramp rises 0.5 m over a horizontal distance of 6 m. Find its slope. Guidelines suggest a slope no steeper than $\frac{1}{12}$ . Does it meet them?

Show the solutionSolution

Slope $= \frac{0.5}{6} = \frac{1}{12} \approx 0.083$ .
It equals the limit $\frac{1}{12}$ , so it just meets the guideline.

Answer: $\frac{1}{12}$ ; it just meets the guideline.

Exercise Set 13.4

What it tests: Forming pairs of linear equations from movie tickets and snacks, taxi fares, and tickets sold at a sports meet.

How to approach it: Name the two unknowns with units, then write one equation for each piece of information.

Like the taxi-fare situation in Exercise Set 13.4

A taxi charges a fixed amount ₹x plus ₹y per km. A 5 km trip costs ₹130 and a 9 km trip costs ₹210. Form the pair of equations.

Show the solutionSolution

5 km: $x + 5 y = 130$ .
9 km: $x + 9 y = 210$ .

Answer: $x + 5 y = 130$ and $x + 9 y = 210$ (which give $y = 20$ , $x = 30$ ).

Exercise Set 13.5

What it tests: Word problems solved by forming and solving a pair: integers, bats and balls, taxi fares, fractions, ages, digits and notes.

How to approach it: Form the pair, solve by elimination or substitution, and check against every condition in the story.

Like the fraction problems in Exercise Set 13.5

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

Show the solutionSolution

Let the fraction be $\frac{n}{d}$ : $3 (n + 1) = 2 (d + 1)$ gives $3 n - 2 d = - 1$ .
$2 (n - 1) = d - 1$ gives $d = 2 n - 1$ .
Substitute: $3 n - 4 n + 2 = - 1$ , so $n = 3$ and $d = 5$ .
Check: $\frac{4}{6} = \frac{2}{3}$ and $\frac{2}{4} = \frac{1}{2}$ .

Answer: $\frac{3}{5}$

End-of-chapter Exercises

What it tests: Reading slope and intercept from graphs, x-intercepts, comparing mobile plans, conditions for unique or infinitely many solutions, coincident lines, and robots moving along lines; starred questions are enrichment.

How to approach it: Turn each situation into $y = m x + d$ or a pair of equations; use ratios for the nature of solutions.

Like the coincident-lines question at the end of the chapter

Find k so that $x + 2 y = 3$ and $2 x + k y = 6$ represent the same line.

Show the solutionSolution

Coincident lines need $\frac{1}{2} = \frac{2}{k} = \frac{3}{6}$ .
$\frac{2}{k} = \frac{1}{2}$ gives $k = 4$ , and $\frac{3}{6} = \frac{1}{2}$ also holds.

Answer: $k = 4$

Practice questions on Two Variables, One Line

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 · MCQExam level
The standard form of $y = 3 x - 2$ is
(a) $3 x + y - 2 = 0$
(b) $3 x - y - 2 = 0$
(c) $3 x - y + 2 = 0$
(d) $x - 3 y - 2 = 0$
Hint
Move y to the right-hand side.
Answer
(b)
Q2 · 1 mark · MCQExam level
Which of these lines passes through the origin?
(a) $2 x + 3 y = 6$
(b) $x - y = 1$
(c) $2 x + 3 y = 0$
(d) $y = 4$
Hint
The constant must be 0.
Answer
(c)
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): The pair $2 x + 3 y = 9$ and $4 x + 6 y = 18$ has infinitely many solutions.
Reason (R): The second equation is 2 times the first.
(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
Same line?
Answer
(a)
Q4 · 2 marks · Very short answerExam level
If $x = 2$ , $y = - 1$ is a solution of $3 x + k y = 8$ , find k.
Hint
Substitute.
Answer
$6 - k = 8$ , so $k = - 2$ .
Q5 · 2 marks · Very short answerHarder exam level
Find the slope of the line through (1, 2) and (4, 11), and its equation in the form $y = m x + d$ .
Hint
Then use one point to find d.
Answer
$m = \frac{9}{3} = 3$ ; $2 = 3 + d$ gives $d = - 1$ : $y = 3 x - 1$ .
Q6 · 2 marks · Very short answerHarder exam level
Give a formula for the x-intercept of $y = m x + d$ and use it for $y = 2 x - 6$ .
Hint
Put y = 0.
Answer
$x = - \frac{d}{m}$ (for $m \neq = 0$ ); for $y = 2 x - 6$ , $x = 3$ .
Q7 · 3 marks · Short answerHarder exam level
4 pens and 3 notebooks cost ₹170; 2 pens and 5 notebooks cost ₹190. Find the cost of a pen and of a notebook.
Hint
Double the second equation.
Answer
$4 p + 10 n = 380$ ; subtracting $4 p + 3 n = 170$ gives $7 n = 210$ , $n = 30$ , $p = 20$ : pen ₹20, notebook ₹30.
Q8 · 3 marks · Short answerHarder exam level
Four years ago a mother was 5 times as old as her daughter. Eight years from now she will be twice as old. Find their present ages.
Hint
$M - 4 = 5 (D - 4)$ and $M + 8 = 2 (D + 8)$ .
Answer
$M = 5 D - 16$ ; $5 D - 8 = 2 D + 16$ gives $D = 8$ , $M = 24$ . Check: 20 and 4; 32 and 16.
Q9 · 3 marks · Short answerHarder exam level
Find k for which $x + 2 y = 3$ and $2 x + k y = 7$ have no solution. What happens if instead the second equation is $2 x + 4 y = 6$ ?
Hint
Compare ratios.
Answer
$\frac{1}{2} = \frac{2}{k}$ gives $k = 4$ ; then $\frac{3}{7} \neq = \frac{1}{2}$ , so the lines are parallel: no solution. With $2 x + 4 y = 6$ , all ratios equal $\frac{1}{2}$ : the same line, infinitely many solutions.
Q10 · 5 marks · Long answerHarder exam level
Solve graphically: $x - y = 1$ and $x + y = 7$ . Find the area of the triangle formed by the two lines and the x-axis.
Hint
x-intercepts are 1 and 7.
Answer
Lines meet at (4, 3). They cut the x-axis at (1, 0) and (7, 0); base 6, height 3, area $= \frac{1}{2} \times 6 \times 3 = 9$ square units.
Q11 · 3 marks · Short answerHOTS / competency
Pens cost ₹15 and pencils ₹5. In how many ways can you spend exactly ₹60 on pens and pencils (whole numbers of each, either can be zero)? Describe the graph of the possibilities.
Hint
$3 x + y = 12$ with whole numbers.
Answer
(0, 12), (1, 9), (2, 6), (3, 3), (4, 0): 5 ways. The graph is 5 separate points on the line $y = 12 - 3 x$ , not the whole line.
Q12 · 3 marks · Short answerHOTS / competency
For which values of p does the pair $3 x + p y = 5$ and $6 x + 8 y = 9$ have a unique solution? What happens when p takes the other value?
Hint
Unique when $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ .
Answer
Unique unless $\frac{3}{6} = \frac{p}{8}$ , i.e. $p \neq = 4$ . When $p = 4$ : $\frac{3}{6} = \frac{4}{8} \neq = \frac{5}{9}$ , so the lines are parallel and there is no solution.

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 is the standard form of a linear equation in two variables?Show answer
$a x + b y + c = 0$ , with a and b not both zero.
2.How many solutions does one linear equation in two variables have?Show answer
Infinitely many.
3.What does the graph of such an equation look like?Show answer
A straight line; every point on it is a solution.
4.Write the slope formula.Show answer
$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
5.In $y = m x + d$ , what are m and d?Show answer
m is the slope; d is the y-intercept.
6.What is the slope of a horizontal line and of a vertical line?Show answer
0 for a horizontal line; undefined for a vertical line.
7.What do parallel lines tell you about a pair of equations?Show answer
There is no solution (inconsistent).
8.When does a pair have infinitely many solutions?Show answer
When the two equations represent the same line: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ .
9.Name the two algebraic methods for solving a pair.Show answer
Substitution and elimination.
10.If c = 0 in $a x + b y + c = 0$ , which point is always on the line?Show answer
The origin (0, 0).

Best order to study it

Learn the standard form and practise rewriting equations (Exercise Set 13.1).
Find solutions, plot them and draw lines; check points on lines (Exercise Set 13.2).
Learn slope from two points and from $y = m x + d$ , with real meanings like ramps and rates (Exercise Set 13.3).
Practise forming pairs of equations from situations (Exercise Set 13.4).
Solve pairs by substitution and elimination, including word problems (Exercise Set 13.5).
Finish with graphical solutions and the ratio test for the nature of solutions, then the end-of-chapter exercises.

Quick revision checklist

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

Can you write $\frac{x}{3} - \frac{y}{2} = 1$ in standard form?
Can you find three solutions of $x - 2 y = 4$ ?
Can you find k if (k, 3) lies on $2 x + y = 9$ ?
Can you find the slope through (−1, 4) and (3, −4)?
Can you rewrite $3 x - 6 y = 12$ as $y = m x + d$ ?
Can you list all whole-number solutions of $x + 2 y = 6$ ?
Can you solve a pair by substitution and by elimination?
Can you solve a pair by drawing graphs and reading the meeting point?
Can you decide the nature of solutions using ratios?
Can you form equations for a two-digit number problem?

Two Variables, One Line: frequently asked questions

Is solving a pair of linear equations in Class 9 now?

Yes. The 2026-27 CBSE content for this chapter includes pairs of linear equations, the graphical method, consistency and inconsistency, and the substitution and elimination methods.

Do I need the cross-multiplication method?

No. The chapter teaches substitution, elimination and graphs.

How is this chapter linked to Introduction to Linear Polynomials?

Chapter 2 introduced y = ax + b with slope a and y-intercept b. This chapter writes the same idea as y = mx + d, works with the general form ax + by + c = 0, and adds pairs of equations.

What does 'consistent' mean?

A pair is consistent if it has at least one solution (intersecting or coincident lines) and inconsistent if it has none (parallel lines).

Why is the slope of a vertical line undefined?

On a vertical line the run (change in x) is 0, and division by 0 has no meaning.

Based on: CBSE Mathematics syllabus, Class IX 2026-27; NCERT Ganita Manjari, Part II, Mathematics Textbook for Grade 9 (First Edition 2026)