Two Variables, One Line: Class 9 Ganita Manjari · Chapter 13

Ganita Manjari Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 13 Oct 2026 · Updated 13 Oct 2026

85 min
2 simulations
44 practice questions
23 MCQs

What you will learn in Two Variables, One Line

Write a linear equation in two variables in the standard form ax + by + c = 0 and name its coefficients and constant term.
Test whether an ordered pair is a solution, find solutions of a linear equation, and explain why there are infinitely many.
Draw the graph of a linear equation from two of its points and read further solutions from the graph.
Find the slope of a line from two points, and read the slope and y-intercept from y = mx + d.
Form a pair of linear equations from a situation and solve it by substitution, by elimination and by drawing graphs.
Use the ratios a1/a2, b1/b2 and c1/c2 to decide whether a pair has one solution, no solution or infinitely many.

Original companion text aligned to NCERT Mathematics Class 9, Chapter 13, pp. 85–122. Not affiliated with or endorsed by NCERT.

You have already drawn rules such as $y = 2 x + 1$ as straight lines (Part 1, Chapter 2, Introduction to Linear Polynomials). Here we turn that around. We start from an equation that ties two unknown numbers together and ask which pairs of numbers fit it. The answers always line up along a straight line. Then we put two such equations side by side and hunt for the pair, if there is one, that fits both.

13.1 Linear Equations in Two Variables

At the school canteen a samosa costs ₹15 and a plate of idli costs ₹20. Ishaan orders some of each for his friends, and the bill comes to ₹110. Later he cannot remember how many of each he ordered. Can we still write down what we know?

Two amounts are unknown, so each gets its own letter. Let $x$ be the number of samosas and $y$ the number of idli plates. The samosas cost $15 x$ rupees and the plates cost $20 y$ rupees. Together they make the bill:

15 x + 20 y = 110

The left side adds up what the two kinds of food cost; the right side is what Ishaan actually paid. An equation like this has two letters, each to the first power only, and never multiplied together. It is called a linear equation in two variables (दो चरों वाला रैखिक समीकरण).

3/3Both $x = 2, y = 4$ and $x = 6, y = 1$ give a bill of ₹110.

Fig. 13.1One bill, two different orders that fit it.NCERT p. 85

Two price tags at the top: a samosa at 15 rupees each and an idli plate at 20 rupees each, with the unknown counts x samosas and y plates written below. In the second step a bill box appears in the middle reading 15x plus 20y equals 110. In the third step two trays appear at the bottom: the left tray has 2 samosas and 4 idli plates, labelled x equals 2, y equals 4 and 30 plus 80 equals 110; the right tray has 6 samosas and 1 plate, labelled x equals 6, y equals 1 and 90 plus 20 equals 110.

Look at the trays in Fig. 13.1: the bill alone does not tell us what Ishaan ordered, because two different orders fit it. Keep this in mind. Much of this chapter is about what one equation in two variables can tell us, and what it cannot.

13.1.1 Standard Form of Linear Equations in Two Variables

Any linear equation in two variables can be rearranged so that every term sits on one side and 0 sits on the other. This tidy layout makes equations easy to compare.

Definition

Linear equation in two variables

An equation that can be rearranged into $a x + b y + c = 0$ for some real numbers $a,$ $b,$ $c,$ with at least one of $a$ and $b$ non-zero.

Example $15 x + 20 y - 110 = 0,$ with $a = 15,$ $b = 20,$ $c = - 110.$

Standard form

a x + b y + c = 0

(13.1)

where

$a$: the coefficient of $x$
$b$: the coefficient of $y$
$c$: the constant term ( $a$ and $b$ not both 0)

We call $a x + b y + c = 0$ the standard form (मानक रूप). The number that multiplies $x$ is the coefficient of $x,$ and likewise for $y,$ while $c$ is the constant term. For Ishaan's bill the standard form is $15 x + 20 y - 110 = 0.$

Two remarks help in practice. First, the coefficients may be fractions or decimals. For instance, $\frac{2}{3} x - \frac{1}{4} y + \frac{5}{6} = 0$ is already in standard form. Multiplying every term by 12, the LCM of 3, 4 and 6, gives the neater $8 x - 3 y + 10 = 0,$ and the solutions do not change.

Second, one of the letters may be missing. The equation $9 - 3 x = 0$ can be written as $- 3 x + 0 \cdot y + 9 = 0,$ so it counts as a linear equation in two variables with $b = 0.$ Similarly, $2 y = 7$ is $0 \cdot x + 2 y - 7 = 0$ with $a = 0.$ Only losing both letters is not allowed, because then the equation says nothing about $x$ or $y .$

Example 13.1NCERT Example 1 · p. 86

Putting equations in standard form

Write each equation in the form $a x + b y + c = 0$ and state $a,$ $b$ and $c .$

$3 x - 4 y = 7.5$
$\frac{y}{3} + 2 = \frac{5 x}{2}$
$6 = 0.4 y - 2.5 x$
$2 x = - 9$

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Which of these is not a linear equation in two variables?

2 x - 7 = 3 y

x = 5

x^{2} + y = 4

\frac{1}{2} x + y = 0

Show answer

Answer (C) $x^{2} + y = 4$

x^{2}

is a second power, so

x^{2} + y = 4

is not linear.

x = 5

looks like it has only one letter, but it is

x + 0 \cdot y - 5 = 0,

which is allowed.

Practice · 3 questions

Practice 13.1: questions with answers

Short answerEasyNCERT Exercise Set 13.1, Q1
Write a linear equation in two variables with $a = 0,$ $b = 5$ and $c = - \frac{3}{4} .$ Then rewrite it with whole-number coefficients.
Show solution
Answer $0 \cdot x + 5 y - \frac{3}{4} = 0,$ or $20 y - 3 = 0.$
Substituting gives $0 \cdot x + 5 y - \frac{3}{4} = 0,$ that is $5 y - \frac{3}{4} = 0.$ Multiplying every term by 4 gives $20 y - 3 = 0.$ Both describe every point with $y = 0.15,$ whatever $x$ is.
2
Short answerEasyNCERT Exercise Set 13.1, Q2
Write each equation in standard form, with a positive first coefficient. Then give its two coefficients and its constant term.
1. $y + 4 = 3 x$
  Show solution
  $3 x - y - 4 = 0 ⁣ :$ coefficient of $x$ is 3, of $y$ is $- 1,$ constant $- 4.$
2. $5 y = 2 x$
  Show solution
  $2 x - 5 y = 0 ⁣ :$ coefficients 2 and $- 5,$ constant 0.
3. $x = - 6$
  Show solution
  $x + 0 \cdot y + 6 = 0 ⁣ :$ coefficients 1 and 0, constant 6.
4. $4 y = 9$
  Show solution
  $0 \cdot x + 4 y - 9 = 0 ⁣ :$ coefficients 0 and 4, constant $- 9.$
3
Short answerEasyNCERT Exercise Set 13.1, Q3
Decide which equation describes each situation.
1. A bag of rice weighs three times as much as a bag of sugar. With the rice bag $r$ kg and the sugar bag $s$ kg, Asha writes $r = 3 s$ and Dev writes $s = 3 r .$ Who is right?
  Asha, $r = 3 s$ Dev, $s = 3 r$ Both, since they say the same thing
  Show answer
  Answer (A) Asha, $r = 3 s$
  The rice bag is the heavier one, so $r$ must be 3 times $s .$ Test it: if $s = 2,$ rice weighs 6 kg, and only $r = 3 s$ gives that.
2. In a kabaddi match two raiders together scored 23 points. One scored $x$ points and the other $y .$ Is $x + y = 23$ a correct equation? Is $x - y = 23$ ?
  Show solution
  Answer $x + y = 23$ is correct; $x - y = 23$ is not.
  'Together' means the two scores are added, so $x + y = 23$ is right. The equation $x - y = 23$ would say one raider scored 23 points more than the other, which is a different fact.

13.2 Solution of Linear Equation in Two Variables

You can already solve $2 x + 5 = 17 ⁣ :$ it has exactly one answer, $x = 6.$ With two letters the idea of an answer must change, because a single number cannot be tested in $3 x + 4 y = 24.$ We need a value for each letter.

13.2.1 Algebraic Meaning of Solution

For an equation with two letters, a solution (हल) needs a value for each letter: an $x$ -value and a $y$ -value that, used together, make both sides equal. Take $3 x + 4 y = 24.$ With $x = 4$ and $y = 3,$ the left side is $3 (4) + 4 (3) = 12 + 12 = 24.$ So $x = 4,$ $y = 3$ is a solution.

We write this solution as $(4, 3),$ with $x$ first and $y$ second. The order matters, which is why $(4, 3)$ is called an ordered pair (क्रमित युग्म). Swap the numbers and test $(3, 4) ⁣ :$ $3 (3) + 4 (4) = 9 + 16 = 25,$ not 24. So $(3, 4)$ is not a solution, even though it uses the same two numbers.

How many solutions are there? Pick any value of $x$ and put it in; you are left with a one-variable equation for $y .$ With $x = 0$ we get $4 y = 24,$ so $(0, 6)$ is a solution. With $x = 1$ we get $4 y = 21,$ so $(1, \frac{21}{4})$ is one too. With $x = - 4$ we get $4 y = 36,$ so $(- 4, 9)$ works. Every value of $x$ gives exactly one $y,$ and there are endless values to choose. So the equation has infinitely many solutions.

13.2.2 Geometric Visualisation of Solutions

Now plot some solutions of $3 x + 4 y = 24$ as points on a grid. First make $y$ the subject. From $4 y = - 3 x + 24$ we get

y = - \frac{3}{4} x + 6

This has the form $y = m x + d,$ which you met in Part 1 as $y = a x + b .$ You saw there that such a rule always gives a straight line. So every solution of $3 x + 4 y = 24$ is a point on one line, and every point on that line is a solution. The line is the graph (आलेख) of the equation: a picture of all its infinitely many solutions at once.

4/4 $(3, 4)$ gives $3 (3) + 4 (4) = 25,$ so it is not on the line.

Fig. 13.2Every solution of

3 x + 4 y = 24

lies on one straight line.NCERT Fig. 13.1 · p. 89

A grid from minus 5 to 9 across and minus 2 to 10 up, with working on the right. Step 1 plots (0, 6) and (8, 0), from x equals 0 and x equals 8. Step 2 adds (4, 3) and (minus 4, 9). Step 3 draws one straight line through all four points, noting that they are all on one line. Step 4 marks the point (3, 4) with a dashed ring just off the line and notes that it gives 25, not 24, so it is not a solution.

SimulationIs the point on the line?

A coordinate grid from minus 6 to 10 on both axes. Three fixed points, C (minus 4, 9), A (0, 6) and B (8, 0), lie on the line 3x plus 4y equals 24 and are joined by measured segments, so the line is drawn from C to B. A fourth point P starts at (6, 6). You can drag P; it snaps to half units and shows its coordinates, so you can test whether it satisfies the equation.

Fig. 13.3The segments from

C

through

A

B

are part of the line

3 x + 4 y = 24.

Drag

P

and test its coordinates.

Try this

Drag $P$ onto the line at a point with whole-number coordinates, other than $A,$ $B$ and $C .$ Check that $3 x + 4 y = 24$ there.
Put $P$ at $(6, 1.5),$ then at $(6, 2) .$ Which one is on the line? Test both in the equation.
Put $P$ at $(3, 4) .$ It looks close to the line: what does $3 x + 4 y$ give there?

Since two points fix a line, any two solutions are enough to draw it. The quickest two to find are usually the axis crossings: every point of the $y$ -axis has $x = 0,$ and every point of the $x$ -axis has $y = 0.$

Example 13.2NCERT Example 2 · p. 89

A graph from two easy points

Find two solutions of $x + 2 y = 5$ and draw its graph. Use the graph to find four more solutions.

Try it yourself first. Then reveal the solution one step at a time.

4/4Grid corners on the line are more solutions.

Fig. 13.4Two axis crossings fix the line

x + 2 y = 5;

more solutions can then be read off.NCERT Fig. 13.2 · p. 90

A grid from minus 3 to 8 across and minus 2 to 4 up. Step 1 rings the point (0, 2.5) on the y-axis, found by putting x equals 0. Step 2 rings (5, 0) on the x-axis, found by putting y equals 0. Step 3 joins the two points with a straight line. Step 4 marks four more points where the line crosses grid corners: (minus 1, 3), (1, 2), (3, 1) and (7, minus 1).

Example 13.3NCERT Example 3 · p. 90

Two lines, two cases

Find two solutions of each equation and draw its graph: (i) $5 x + 2 y = 10$ (ii) $3 x - 4 y = 0.$

Try it yourself first. Then reveal the solution one step at a time.

Example 13.4NCERT Example 4 · p. 92

Finding missing coefficients

The pair $(2, - 5)$ is a solution of both $p x - 2 y = 16$ and $3 x + q y = - 14.$ Find $p$ and $q .$

Try it yourself first. Then reveal the solution one step at a time.

13.3 Slope-Intercept Form of a Linear Equation in Two Variables

13.3.1 Understanding Slope

A wheelchair ramp at a railway station climbs gently, while a playground slide climbs sharply. Both go up; what differs is how much each rises for every step forward. Lines on a graph differ in the same way, and mathematics captures this with one number, the slope (ढाल) of the line, also called its gradient.

Read a graph from left to right, the way you read a sentence. If the line climbs as you go, its slope is positive. If it drops, its slope is negative. The steeper the climb or the drop, the bigger the size of the slope.

To measure the slope, pick two points on the line, $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2}),$ and walk from $A$ to $B$ in two moves: first across, then up or down. The across move, $x_{2} - x_{1},$ is called the run, and the up-or-down move, $y_{2} - y_{1},$ is called the rise. Both carry signs: a move to the left, or downwards, counts as negative. The slope is the rise divided by the run.

Slope = rise ÷ run

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

(13.2)

where

$y_{2} - y_{1}$: the rise from the first point to the second
$x_{2} - x_{1}$: the run, which must not be 0

4/4From $A$ to $C (9, 8) ⁣ :$ rise 6 and run 8, and $\frac{6}{8} = \frac{3}{4}$ again.

Fig. 13.5The rise-to-run ratio is the same for any two points on a line.NCERT Fig. 13.7 · p. 96

A first-quadrant grid from 0 to 10 across and 0 to 9 up, with a rising line through A (1, 2) and B (5, 5). Step 2 draws an arrow 4 units to the right from A, labelled run equals 4. Step 3 draws an arrow 3 units up to B, labelled rise equals 3, and the side panel gives rise over run equals 3 over 4. Step 4 marks C (9, 8) and draws a larger dashed triangle above the line from A to C, with rise 6 and run 8, and the panel shows 6 over 8 equals 3 over 4.

The slope came out the same for $A$ to $B$ and for $A$ to $C .$ That is no accident. On a straight line every rise-and-run triangle has the same shape, only scaled up or down, so the ratio never changes. It does not matter which two points you choose.

SimulationRise and run between two points

A coordinate grid from minus 2 to 10 on both axes with two draggable points, A at (1, 2) and B at (5, 5), joined by a measured segment. The right-angled legs of the segment are drawn and labelled delta x and delta y, the run and the rise, as sizes without signs. Points snap to whole units and show their coordinates as you drag them.

Fig. 13.6The legs

Δ x

(run) and

Δ y

(rise) are drawn between

A

and

B;

they show sizes, so decide the signs yourself.

Try this

Read $Δ x$ and $Δ y$ for $A (1, 2)$ and $B (5, 5) .$ What is the slope $Δ y \div Δ x$ ?
Drag $B$ to $(9, 8),$ further along the same line. The run and rise change: does the slope?
Drag $B$ to $(5, - 1) .$ Is the rise now up or down, and what is the sign of the slope? Then put $B$ straight above $A,$ at $(1, 7) ⁣ :$ why can the slope not be worked out?

Example 13.5NCERT Example 5 · p. 97

Slope from three points

The points $P (- 2, - 1),$ $Q (0, 3)$ and $R (3, 9)$ lie on one line. Find its slope, and check it with a second pair of points.

Try it yourself first. Then reveal the solution one step at a time.

Example 13.6NCERT Example 6 · p. 97

A falling line

Find the slope of the line through $A (- 3, 4)$ and $B (3, 1) .$ Does it matter which point you take first?

Try it yourself first. Then reveal the solution one step at a time.

13.3.2 Slope-Intercept Form

In Part 1 you met the rule $y = a x + b,$ where $a$ set the steepness and $b$ was the height at which the line crossed the $y$ -axis. This chapter, like your textbook, writes the same rule with the letters $m$ and $d .$

Slope-intercept form

y = m x + d

(13.3)

where

$m$: the slope of the line
$d$: the $y$ -intercept, so $(0, d)$ is on the graph

Here $m$ is the slope and $d$ is the $y$ -intercept (अंतःखंड). Since both can be read straight off the equation, $y = m x + d$ is called the slope-intercept form. But why should $m$ be exactly the rise-over-run slope? Here is the reason.

DerivationWhy

m

is the slope

Let $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ be any two points on $y = m x + d,$ with $x_{1} \neq x_{2} .$ Both points satisfy the equation, so $y_{1} = m x_{1} + d$ and $y_{2} = m x_{2} + d .$

Subtract the first from the second. The two $d$ 's cancel, leaving $y_{2} - y_{1} = m x_{2} - m x_{1},$ and taking out the common factor $m$ gives $y_{2} - y_{1} = m (x_{2} - x_{1}) .$ In words:

rise = m \times run

The run $x_{2} - x_{1}$ is not zero, so we may divide by it:

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

The $d$ 's cancelled, and $A$ and $B$ were any two points of the line. So every pair of points gives the same rise-over-run ratio, and that ratio is $m .$

Any equation $a x + b y + c = 0$ with $b \neq 0$ can be put in this form. Move $a x$ and $c$ to the right-hand side, then divide by $b ⁣ :$

From standard form

y = - \frac{a}{b} x - \frac{c}{b}

(13.4)

where

$m$: $- \frac{a}{b},$ the slope
$d$: $- \frac{c}{b},$ the $y$ -intercept

ExampleNCERT p. 99

Reading the slope and the intercept

Write $4 x - 2 y = 6$ in slope-intercept form. What do $m$ and $d$ tell you? Draw the line from a small table of values.

Try it yourself first. Then reveal the solution one step at a time.

The sign of $m$ sorts lines into kinds, as Table 13.1 and Fig. 13.7 show. A horizontal line such as $y = - 2$ never rises, so $m = 0.$ A vertical line such as $x = - 3$ is the odd one out: between any two of its points the run is 0, and we cannot divide by 0. So a vertical line has no slope (its slope is undefined), and it is the one kind of line that $y = m x + d$ cannot describe.

Table 13.1Four kinds of line

Slope	The line	Example
$m > 0$	rises to the right	$y = \frac{1}{2} x + 1$
$m < 0$	falls to the right	$y = - 2 x + 3$
$m = 0$	horizontal	$y = - 2$
none	vertical	$x = - 3$

5/5The four kinds of line together.

Fig. 13.7Rising, falling, flat and upright lines.NCERT Figs. 13.12–13.13 · p. 100

A grid from minus 5 to 5 on both axes with a legend on the right. Step 1 shows the solid line y equals one half x plus 1, which rises: m greater than 0. Step 2 shows the dashed line y equals minus 2x plus 3, which falls: m less than 0. Step 3 shows the dotted horizontal line y equals minus 2: m equals 0. Step 4 shows the dash-dot vertical line x equals minus 3, which has no slope. Step 5 shows all four lines together with the full legend.

Quick check

1 of 4

Find the slope of the line through

(- 1, 4)

and

(3, - 2) .

Show answer

Answer -1.5

Rise

= - 2 - 4 = - 6

and run

= 3 - (- 1) = 4,

m = \frac{- 6}{4} = - 1.5.

The minus sign says the line falls.

Practice · 4 questions

Practice 13.3: questions with answers

1
Short answerEasyNCERT Exercise Set 13.3, Q1
Three ramps at a skate park are described in the table. Put them in order of steepness, from least steep to steepest.
Ramp Rise Run
P 1.2 m 2 m
Q 1.5 m 3 m
R 2 m 2.5 m
1. Ramp P
2. Ramp R
3. Ramp Q
Show answer
Answer
Ramp Q
Ramp P
Ramp R
Slopes: Q $= \frac{1.5}{3} = 0.5,$ P $= \frac{1.2}{2} = 0.6,$ R $= \frac{2}{2.5} = 0.8.$ Q rises more than P but over a longer run, so it is gentler.
Short answerEasyNCERT Exercise Set 13.3, Q2
A tin roof rises 1.8 m over a horizontal run of 6 m. What is the slope of the roof?
Show answer
Answer 0.3
$\frac{1.8}{6} = 0.3 ⁣ :$ the roof rises 0.3 m for every metre across.
3
Short answerMediumNCERT Exercise Set 13.3, Q3
Find the slope of the line through each pair of points.
1. $(- 3, 2)$ and $(4, 2)$
  Show answer
  Answer 0
  Rise $= 0,$ run $= 7,$ so $m = 0 ⁣ :$ a horizontal line.
2. $(- 2, 5)$ and $(2, - 3)$
  Show answer
  Answer -2
  Rise $= - 8,$ run $= 4,$ so $m = - 2.$
3. $(- 4, - 2)$ and $(4, 2)$
  Show answer
  Answer 0.5
  Rise $= 4,$ run $= 8,$ so $m = \frac{1}{2} .$
4. $(- 1, - 3)$ and $(- 1, 4)$
  07undefined
  Show answer
  Answer (C) undefined
  The run is $- 1 - (- 1) = 0.$ The line is vertical and its slope is undefined, not 0.
Short answerMediumNCERT Exercise Set 13.3, Q4
A school is building a wheelchair ramp beside a flight of 4 steps, each 15 cm high. The guideline is a rise of 1 cm for every 12 cm of horizontal run. How long must the run of the ramp be, in metres?
m
Show answer
Answer 7.2 m
Total rise $= 4 \times 15 = 60$ cm. The run is 12 times the rise: $60 \times 12 = 720$ cm $= 7.2$ m.

Ramp	Rise	Run
P	1.2 m	2 m
Q	1.5 m	3 m
R	2 m	2.5 m

13.4 Pair of Linear Equations in Two Variables

A single equation in two letters has endless solutions, as Ishaan's bill showed. To pin down one answer we need a second, separate piece of information. Here is a number puzzle: I am thinking of two numbers. Their sum is 31 and their difference is 7.

You could guess and check, as in Table 13.2. Every guess there already has the right sum, and we adjust it until the difference is right as well.

Table 13.2Guess and check for the number puzzle

First number	Second number	Sum	Difference
16	15	31	1
20	11	31	9
19	12	31	7

Guessing works for small whole numbers, but it soon gets slow. Algebra is quicker. Calling the numbers $x$ and $y,$ the two facts become $x + y = 31$ and $x - y = 7.$ Two facts written about the same two unknowns give a pair of linear equations in two variables (रैखिक समीकरण युग्म). An ordered pair solves the pair only if it makes both equations true at once; fitting just one is not enough.

3/3The lines cross at $(19, 12),$ the only pair that fits both.

Fig. 13.8Each clue alone has a whole line of answers; only the crossing point fits both.NCERT p. 104

A grid from 0 to 32 on both axes, with gridlines every 5. Step 1 draws the solid line x plus y equals 31 and marks (10, 21) and (25, 6), two pairs with sum 31. Step 2 adds the dashed line x minus y equals 7 with (10, 3) and (31, 24), two pairs with difference 7. Step 3 rings the crossing point (19, 12), the only pair that fits both clues.

Everyday problems lead to pairs just as naturally. At a planetarium, a family of 2 adults and 3 children pays ₹410 for tickets, and another family of 1 adult and 4 children pays ₹380. If an adult ticket costs ₹ $x$ and a child's ticket ₹ $y,$ then $2 x + 3 y = 410$ and $x + 4 y = 380.$ Solving this pair gives both prices; you will do that in the quick check of Section 13.5.

With letters standing for all six numbers, every such pair has the shape below. In each equation, at least one of the two coefficients must be non-zero.

A pair of linear equations

\begin{aligned} a_{1} x + b_{1} y + c_{1} & = 0 \\ a_{2} x + b_{2} y + c_{2} & = 0 \end{aligned}

(13.5)

Quick check

1 of 3

A pen costs ₹

p

and a pencil ₹

q .

Three pens and two pencils cost ₹47, and a pen costs ₹4 more than a pencil. Which pair describes this?

3 p + 2 q = 47

and

p = q + 4

3 p + 2 q = 47

and

q = p + 4

2 p + 3 q = 47

and

p = q + 4

3 p + 2 q = 47

and

p + q = 4

Show answer

Answer (A) $3 p + 2 q = 47$ and $p = q + 4$

The pen is the dearer item, so

p = q + 4,

and the first fact gives

3 p + 2 q = 47.

Writing

q = p + 4

would make the pencil dearer. (Solving gives

q = 7,

p = 11

Practice · 3 questions

Practice 13.4: questions with answers

For each situation, use the letters given and turn the two facts into two equations. Solving them comes later.

Short answerEasyNCERT Exercise Set 13.4, Q1
A bakery sells cakes at ₹ $x$ each and boxes of cookies at ₹ $y$ each. An order of 3 cakes and 5 boxes costs ₹1,400, and an order of 2 cakes and 1 box costs ₹700.
Show solution
Answer $3 x + 5 y = 1400$ and $2 x + y = 700.$
The first order gives $3 x + 5 y = 1400$ and the second gives $2 x + y = 700.$ (Solving later gives $x = 300$ and $y = 100$ .)
Short answerEasyNCERT Exercise Set 13.4, Q2
An auto-rickshaw charges a fixed amount of ₹ $x$ plus ₹ $y$ for every kilometre. Neha's 4 km ride cost ₹86 and Arjun's 9 km ride cost ₹161.
Show solution
Answer $x + 4 y = 86$ and $x + 9 y = 161.$
Each fare is the fixed part plus the distance times the rate: $x + 4 y = 86$ and $x + 9 y = 161.$ (They give $y = 15$ and $x = 26$ .)
Short answerMediumNCERT Exercise Set 13.4, Q3
Tickets for a school play cost ₹80 for adults and ₹50 for students. In all, 320 tickets were sold and ₹19,600 was collected. Let $x$ adult tickets and $y$ student tickets be sold.
Show solution
Answer $x + y = 320$ and $80 x + 50 y = 19600.$
Counting tickets: $x + y = 320.$ Counting money: $80 x + 50 y = 19600.$ (Solving gives 120 adult and 200 student tickets.)

13.5 Finding Solutions to a Pair of Linear Equations

Every method for a pair rests on one trick: turn two equations in two letters into a single equation in one letter, which you already know how to solve. Once one letter is known, putting it back gives the other. The two methods in this section are two ways of making a letter disappear.

13.5.1 Algebraic Methods for Solving a Pair of Linear Equations

Did you spot the shortcut? Halving Board 1 gives 'kite + top = ₹130'. Board 2 contains exactly 'kite + top', so we can replace it by 130: then 2 drums cost $210 - 130 = 80$ rupees, and a drum costs ₹40. Board 3 now gives 2 kites = ₹140, so a kite costs ₹70 and a top ₹60. Replacing something by an amount known to be equal is the heart of both methods below: the substitution method and the elimination method.

The substitution method

In the substitution method (प्रतिस्थापन विधि) we use one equation to write one letter in terms of the other, and put that expression into the other equation. The steps are:

Choose the equation in which a letter is easiest to isolate, and make that letter the subject.
Substitute this expression into the other equation and solve the one-letter equation.
Put the value back to find the second letter.
Check the pair in both original equations.

Example 13.7NCERT Example 7 · p. 106

Substitution with a fraction answer

Solve this pair by substitution: equation (1) is $4 x + 9 y = 5$ and equation (2) is $x - 2 y = 3.$

Try it yourself first. Then reveal the solution one step at a time.

Example 13.8NCERT Example 8 · p. 107

Angles on a straight line

Two angles form a linear pair, so together they make $180^{\circ} .$ The larger angle is $30^{\circ}$ less than twice the smaller one. Find both angles.

Try it yourself first. Then reveal the solution one step at a time.

The elimination method

In the elimination method (विलोपन विधि) we make the coefficients of one letter equal in size, and then add or subtract the equations so that this letter cancels. It often saves work when no letter has a coefficient of 1.

Multiply one or both equations by suitable numbers so that one letter has coefficients of the same size in both.
Subtract the equations if those coefficients have the same sign; add them if the signs are opposite. One letter is eliminated.
Solve for the remaining letter, then substitute back to find the other.
Check in both original equations.

4/4(1) gives $15 + 2 y = 19,$ so $y = 2;$ check in (2): $25 - 8 = 17.$

Fig. 13.9Eliminating

y

from

3 x + 2 y = 19

and

5 x - 4 y = 17.

NCERT p. 108

Equations written one under another with tags on the left. Step 1 shows (1) 3x plus 2y equals 19 and (2) 5x minus 4y equals 17. Step 2 adds (3), equation (1) times 2, 6x plus 4y equals 38, with the terms minus 4y and plus 4y highlighted and a note to make the y-parts match. Step 3 draws a rule and adds (2) and (3) to get 11x equals 55, so x equals 5. Step 4 puts x equals 5 into (1), giving 15 plus 2y equals 19, so y equals 2, and checks it in (2): 25 minus 8 equals 17.

Example 13.9NCERT Example 9 · p. 108

Incomes and expenses

The monthly incomes of two friends are in the ratio $5 : 3,$ and their monthly expenses are in the ratio $2 : 1.$ Each of them saves ₹2,500 a month. Find their monthly incomes.

Try it yourself first. Then reveal the solution one step at a time.

Example 13.10NCERT Example 10 · p. 109

A number and its reverse

A two-digit number is added to the number formed by reversing its digits, and the total is 121. The two digits differ by 3. What could the number be? Find every possibility.

Try it yourself first. Then reveal the solution one step at a time.

When elimination removes both letters

So far every pair had exactly one solution. Now look at $2 x + 3 y = 7$ (1) and $6 x + 9 y = 21$ (2). To eliminate $x,$ multiply (1) by 3. It becomes $6 x + 9 y = 21,$ which is equation (2) exactly, so subtracting leaves $0 = 0.$ Both letters have vanished, and what remains is always true.

The two equations are really one equation written twice, so every solution of $2 x + 3 y = 7$ also solves $6 x + 9 y = 21.$ The pair has infinitely many solutions. The test is whether one equation is a scaled copy of the other: is there a number $k \neq 0$ with $a_{1} = k a_{2},$ $b_{1} = k b_{2}$ and $c_{1} = k c_{2}$ ? If $a_{2},$ $b_{2}$ and $c_{2}$ are all non-zero, that is the same as saying that three ratios are equal:

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}

Now change just one number: $2 x + 3 y = 7$ (1) and $6 x + 9 y = 20$ (2). Multiplying (1) by 3 gives $6 x + 9 y = 21.$ Now the two equations claim that the very same expression, $6 x + 9 y,$ equals 21 and also equals 20. That is impossible (subtracting gives the false statement $0 = 1$ ), so no $(x, y)$ satisfies both, and the pair has no solution. Here $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{1}{3},$ but $\frac{c_{1}}{c_{2}} = \frac{7}{20} .$

Putting the three outcomes together gives a quick test for any pair (with $a_{2},$ $b_{2},$ $c_{2}$ non-zero), even before you start solving:

Table 13.3The three possible outcomes for a pair of linear equations

Solutions	Ratios	Elimination ends with
exactly one	$a_{1} / a_{2} \neq b_{1} / b_{2}$	one letter left
infinitely many	$a_{1} / a_{2} = b_{1} / b_{2} = c_{1} / c_{2}$	$0 = 0$
none	$a_{1} / a_{2} = b_{1} / b_{2} \neq c_{1} / c_{2}$	a false statement, such as $0 = 1$

From historyNCERT A Pinch of History · p. 112

A very old method

Ishaan's bill asks which whole numbers fit one equation like $a x + b y = c .$ Āryabhaṭa (499 CE) gave a step-by-step method for exactly this, the kuṭṭaka or 'pulveriser'. About 130 years later, Brahmagupta (628 CE) handled several unknowns at once by giving each its own colour name, much as we now use $x$ and $y .$

People were solving equations in pairs long before that. Babylonian scribes, nearly 4,000 years ago, set problems with two unknowns on clay tablets. The Chinese classic Nine Chapters on the Mathematical Art, about 2,000 years old, devotes a chapter called fangcheng to them. Its procedure, carried out with bamboo counting rods on a board, is the elimination method of this section: take multiples of one equation away from another until an unknown drops out.

Quick check

1 of 5

To eliminate

y

from

2 x + 3 y = 11

and

5 x - 2 y = 18,

which is the best move?

Multiply the first by 2 and the second by 3, then addMultiply the first by 2 and the second by 3, then subtractMultiply the first by 5 and the second by 2, then addAdd the two equations as they are

Show answer

Answer (A) Multiply the first by 2 and the second by 3, then add

This makes the

y

-terms

+ 6 y

and

- 6 y,

and adding removes them:

19 x = 76,

x = 4

and

y = 1.

Subtracting would leave

12 y .

Multiplying by 5 and 2 matches the

x

-terms, not the

y

-terms.

13.6 Graphical Method for Solving a Pair of Linear Equations in Two Variables

Each equation of a pair has a straight line as its graph, and the points of that line are exactly its solutions. A solution of the pair must therefore lie on both lines. So if we draw the two lines on the same axes, the solutions are simply the points the lines share.

Your four drawings show the only three ways two lines can sit in a plane, pictured in Fig. 13.10:

Intersecting lines (प्रतिच्छेदी रेखाएँ) meet at exactly one point, so the pair has exactly one solution: that point.
Parallel lines (समांतर रेखाएँ) never meet, so the pair has no solution.
Coincident lines (संपाती रेखाएँ) lie on top of each other, so every point is shared and the pair has infinitely many solutions.

1/3 $x + y = 5$ and $2 x - y = 4$ meet at $(3, 2) ⁣ :$ one solution.

Fig. 13.10Intersecting, parallel and coincident lines.NCERT p. 113

A grid from minus 1 to 6 across and minus 2 to 5 up, with the two equations and a verdict on the right. Step 1 shows x plus y equals 5 and the dashed line 2x minus y equals 4 crossing at the ringed point (3, 2): they meet once, so there is one solution. Step 2 shows 2x plus y equals 4 and the dashed 4x plus 2y equals 2, two parallel lines that never meet: no solution. Step 3 shows x minus 2y equals 2 as a wide pale band with the dashed 3x minus 6y equals 6 running along it: the same line, so infinitely many solutions.

Example 13.11NCERT Example 11 · p. 113

Solving by drawing

Solve the pair $x + 2 y = 8$ and $3 x - 2 y = 0$ graphically, and say what kind of solution it has.

Try it yourself first. Then reveal the solution one step at a time.

3/3They cross at $(2, 3) .$

Fig. 13.11The two lines of Example 13.11 meet at

(2, 3) .

NCERT Fig. 13.14 · p. 114

A grid from minus 1 to 9 across and minus 1 to 7 up. Step 1 draws the solid line x plus 2y equals 8 through (0, 4) and (8, 0). Step 2 adds the dashed line 3x minus 2y equals 0 through the origin and (4, 6). Step 3 rings the crossing point (2, 3) and notes that the lines meet there.

Example 13.12NCERT Example 12 · p. 114

Predict, then draw

Decide whether the lines $x - 2 y + 2 = 0$ and $2 x - 4 y - 8 = 0$ intersect, are parallel or coincide. Then check by drawing them.

Try it yourself first. Then reveal the solution one step at a time.

3/3Both go 2 across, 1 up, and stay 3 units apart vertically.

Fig. 13.12Equal slopes and different intercepts: the lines of Example 13.12 never meet.NCERT Fig. 13.15 · p. 115

A grid from minus 3 to 7 across and minus 3 to 5 up. Step 1 draws the solid line x minus 2y plus 2 equals 0 through (minus 2, 0), (0, 1) and (4, 3). Step 2 adds the dashed line 2x minus 4y minus 8 equals 0 through (0, minus 2), (4, 0) and (6, 1). Step 3 draws a step of 2 across and 1 up on each line and a double arrow showing the constant vertical gap of 3 units between them, with the note that they never meet.

PropertyEqual slopes, parallel lines

Two different lines with the same slope never meet: they are parallel.

Example 13.13NCERT Example 13 · p. 115

A purchase that adds nothing new

Meera buys 3 pencils and 2 erasers for ₹24. Her brother buys 6 pencils and 4 erasers of the same kind for ₹48. Write the pair of equations and describe its graph. Can you find the prices?

Try it yourself first. Then reveal the solution one step at a time.

Try any pair you like in the explorer below. Change the numbers in each equation and watch the two lines, their ratios, their slopes and the number of solutions. Then switch to Make it and build intersecting, parallel and coincident pairs yourself.

InteractiveTwo-line explorer

Two linear equations in the form a x plus b y plus c equals 0, each with plus and minus buttons for a, b and c from minus 9 to 9, drawn as a solid and a dashed line on a grid from minus 8 to 8. A ring marks the crossing point with its coordinates. A status box says whether the lines intersect (with the exact solution), are parallel (no solution) or coincide (infinitely many), and tables show the ratios a1 over a2, b1 over b2, c1 over c2 and the slope of each line. Preset buttons load an intersecting, a parallel and a coincident pair. A Make it mode fixes line 1 and asks you to change line 2 to make the lines parallel, the same line, or meet at a given point, with hints and a score.

Fig. 13.13Change

a,

b

and

c

for each line and see whether the lines intersect, are parallel or coincide.

Quick check

1 of 4

The graphs of a pair of linear equations are parallel lines. How many solutions does the pair have?

exactly onenoneinfinitely manytwo

Show answer

Answer (B) none

Solutions are the common points, and parallel lines have none. Infinitely many solutions come from coincident lines, not parallel ones.

Practice · 11 questions

Practice 13.5: questions with answers

1
Short answerMediumNCERT Exercise Set 13.5, Q1 (i)–(iv)
Write two equations for each problem, then solve them.
1. The sum of two integers is $- 4$ and their difference is 18. Find them.
  Show solution
  Answer7 and $- 11.$
  $x + y = - 4$ and $x - y = 18.$ Adding: $2 x = 14,$ so $x = 7$ and $y = - 11.$ Check: $7 - (- 11) = 18.$
2. Two numbers differ by 45, and one of them is four times the other. Find them.
  Show solution
  Answer60 and 15.
  $x - y = 45$ and $x = 4 y .$ Substituting: $3 y = 45,$ so $y = 15$ and $x = 60.$
3. A hockey club buys 4 sticks and 10 balls for ₹5,100, and later 3 sticks and 4 balls for ₹3,300. Find the price of a stick and of a ball.
  Show solution
  AnswerStick ₹900, ball ₹150.
  $4 s + 10 b = 5100$ and $3 s + 4 b = 3300.$ Multiply by 3 and 4: $12 s + 30 b = 15300$ and $12 s + 16 b = 13200.$ Subtracting, $14 b = 2100,$ so $b = 150,$ and then $3 s = 3300 - 600,$ so $s = 900.$
4. A courier charges a fixed fee plus a rate per kilogram. A 3 kg parcel costs ₹110 and an 8 kg parcel costs ₹210. Find the fee and the rate. What does a 12 kg parcel cost?
  Show solution
  AnswerFee ₹50, ₹20 per kg; a 12 kg parcel costs ₹290.
  $f + 3 r = 110$ and $f + 8 r = 210.$ Subtracting, $5 r = 100,$ so $r = 20$ and $f = 50.$ For 12 kg: $50 + 12 \times 20 = 290.$
2
Short answerHardNCERT Exercise Set 13.5, Q1 (v)–(vii)
Write two equations for each problem, then solve them.
1. A fraction becomes $\frac{1}{2}$ when 1 is added to both its numerator and denominator, and $\frac{3}{5}$ when 4 is added to both. Find the fraction.
  Show solution
  Answer $\frac{5}{11}$
  $2 (x + 1) = y + 1$ gives $2 x - y = - 1;$ $5 (x + 4) = 3 (y + 4)$ gives $5 x - 3 y = - 8.$ From the first, $y = 2 x + 1;$ then $5 x - 6 x - 3 = - 8,$ so $x = 5,$ $y = 11.$ Check: $\frac{6}{12} = \frac{1}{2}$ and $\frac{9}{15} = \frac{3}{5} .$
2. If 3 is subtracted from the numerator of a fraction, it becomes $\frac{1}{3} .$ If 4 is added to the denominator instead, it becomes $\frac{1}{2} .$ Find the fraction.
  Show solution
  Answer $\frac{5}{6}$
  $3 (x - 3) = y$ and $2 x = y + 4.$ Substituting $y = 3 x - 9 ⁣ :$ $2 x = 3 x - 5,$ so $x = 5$ and $y = 6.$ Check: $\frac{2}{6} = \frac{1}{3}$ and $\frac{5}{10} = \frac{1}{2} .$
3. Six years ago Kavya was four times as old as her cousin. In four years she will be twice as old as him. Find their present ages.
  Show solution
  AnswerKavya 26, cousin 11.
  $x - 6 = 4 (y - 6)$ gives $x - 4 y = - 18;$ $x + 4 = 2 (y + 4)$ gives $x - 2 y = 4.$ Subtracting, $2 y = 22,$ so $y = 11$ and $x = 26.$ Check: $20 = 4 \times 5$ and $30 = 2 \times 15.$
3
Short answerHardNCERT Exercise Set 13.5, Q1 (viii)–(x)
Write two equations for each problem, then solve them.
1. The digits of a two-digit number add up to 12, and 7 times the number equals 4 times the number with its digits reversed. Find the number.
  Show solution
  Answer48
  $x + y = 12$ and $7 (10 x + y) = 4 (10 y + x),$ which simplifies to $66 x = 33 y,$ so $y = 2 x .$ Then $3 x = 12,$ $x = 4,$ $y = 8.$ Check: $7 \times 48 = 336 = 4 \times 84.$
2. Rohit withdraws ₹4,100 from a bank in ₹200 and ₹500 notes only, and gets 13 notes in all. How many of each does he get?
  Show solution
  Answer8 notes of ₹200 and 5 notes of ₹500.
  $x + y = 13$ and $200 x + 500 y = 4100.$ Substituting $x = 13 - y ⁣ :$ $2600 + 300 y = 4100,$ so $y = 5$ and $x = 8.$
3. A cycle shop charges a fixed amount for the first 2 hours and a set amount for each extra hour. A 6-hour hire costs ₹140 and a 9-hour hire costs ₹200. Find the fixed amount and the hourly charge.
  Show solution
  Answer₹60 fixed, ₹20 per extra hour.
  A 6-hour hire has 4 extra hours and a 9-hour hire has 7: $x + 4 y = 140$ and $x + 7 y = 200.$ Subtracting, $3 y = 60,$ so $y = 20$ and $x = 60.$
DiagramMediumNCERT Exercise Set 13.5, Q2
A relay squad of 12 runners has 2 more boys than girls. Form a pair of equations and find the numbers of girls and boys by drawing graphs.
Show solution
Answer5 girls and 7 boys.
$x + y = 12$ passes through $(0, 12),$ $(6, 6)$ and $(12, 0) .$ $y = x + 2$ passes through $(0, 2),$ $(4, 6)$ and $(5, 7) .$ The lines cross at $(5, 7),$ so there are 5 girls and 7 boys. Check: $5 + 7 = 12$ and $7 - 5 = 2.$
5
Short answerMediumNCERT Exercise Set 13.5, Q3
Without drawing, compare $\frac{a_{1}}{a_{2}},$ $\frac{b_{1}}{b_{2}}$ and $\frac{c_{1}}{c_{2}}$ to classify each pair of lines as intersecting, parallel or coincident.
1. $3 x + 5 y - 7 = 0$ and $4 x - 2 y + 1 = 0$
  intersectparallelcoincide
  Show answer
  Answer (A) intersect
  $\frac{3}{4} \neq \frac{5}{- 2},$ so the lines intersect.
2. $4 x - 6 y + 10 = 0$ and $6 x - 9 y + 15 = 0$
  intersectparallelcoincide
  Show answer
  Answer (C) coincide
  $\frac{4}{6} = \frac{- 6}{- 9} = \frac{10}{15} = \frac{2}{3},$ so the lines coincide.
3. $5 x + 2 y - 3 = 0$ and $10 x + 4 y + 7 = 0$
  intersectparallelcoincide
  Show answer
  Answer (B) parallel
  $\frac{5}{10} = \frac{2}{4} = \frac{1}{2},$ but $\frac{- 3}{7} \neq \frac{1}{2},$ so the lines are parallel.
6
DiagramMediumNCERT Exercise Set 13.5, Q4
Decide first whether each pair can be solved at all. Then draw the graphs and read off any solutions.
1. $2 x - y = 3$ and $4 x - 2 y = 6$
  Show solution
  AnswerInfinitely many: every point of $y = 2 x - 3.$
  All three ratios equal $\frac{1}{2},$ so the lines coincide. Every point on $y = 2 x - 3,$ such as $(0, - 3),$ $(1, - 1)$ or $(3, 3),$ is a solution.
2. $x + 2 y = 6$ and $3 x + 6 y = 10$
  Show solution
  AnswerNo solution.
  $\frac{1}{3} = \frac{2}{6}$ but $\frac{6}{10} \neq \frac{1}{3},$ so the lines are parallel and never meet.
3. $x - y = 1$ and $2 x + y = 8$
  Show solution
  Answer $(3, 2)$
  $\frac{1}{2} \neq \frac{- 1}{1},$ so the lines intersect. $x - y = 1$ passes through $(1, 0)$ and $(4, 3);$ $2 x + y = 8$ passes through $(4, 0)$ and $(2, 4) .$ They cross at $(3, 2) ⁣ :$ $3 - 2 = 1$ and $6 + 2 = 8.$
4. $3 x + y - 2 = 0$ and $6 x + 2 y - 9 = 0$
  Show solution
  AnswerNo solution.
  $\frac{3}{6} = \frac{1}{2}$ but $\frac{- 2}{- 9} = \frac{2}{9} \neq \frac{1}{2},$ so the lines are parallel.
Short answerMediumNCERT Exercise Set 13.5, Q5
The perimeter of a rectangular plot is 58 m, and its length is 1 m less than twice its width. Find its length and width.
Show solution
AnswerLength 19 m, width 10 m.
$l + w = 29$ and $l = 2 w - 1.$ Substituting: $3 w - 1 = 29,$ so $w = 10$ and $l = 19.$ Check: $2 (19 + 10) = 58.$
8
Short answerMediumNCERT Exercise Set 13.5, Q6
Start from the line $3 x - 4 y + 5 = 0.$ Make up a partner equation whose graph, together with this line, gives
1. intersecting lines
  Show solution
  AnswerFor example $x + y - 2 = 0.$
  Any equation with $\frac{3}{a_{2}} \neq \frac{- 4}{b_{2}}$ works, such as $x + y - 2 = 0,$ since $\frac{3}{1} \neq \frac{- 4}{1} .$
2. parallel lines
  Show solution
  AnswerFor example $6 x - 8 y + 1 = 0.$
  Double the $x$ - and $y$ -coefficients but not the constant: $\frac{3}{6} = \frac{- 4}{- 8} = \frac{1}{2},$ while $\frac{5}{1} \neq \frac{1}{2} .$
3. coincident lines
  Show solution
  AnswerFor example $6 x - 8 y + 10 = 0.$
  Multiply the whole equation by any non-zero number, here 2, so all three ratios equal $\frac{1}{2} .$
Short answerHardNCERT Exercise Set 13.5, Q7
Mahāvīrācārya's Gaṇitasārasaṅgraha (9th century CE) has fruit-price puzzles in which the two purchases swap their numbers. Here is a new one in that style: 8 guavas and 5 pomegranates cost 131 coins, while 5 guavas and 8 pomegranates cost 116 coins. Find the price of each fruit without using substitution or the usual elimination steps.
Show solution
AnswerA guava costs 12 coins and a pomegranate 7 coins.
Let a guava cost $x$ coins and a pomegranate $y ⁣ :$ $8 x + 5 y = 131$ and $5 x + 8 y = 116.$ Adding gives $13 x + 13 y = 247,$ so $x + y = 19.$ Subtracting gives $3 x - 3 y = 15,$ so $x - y = 5.$ Then $2 x = 24,$ so $x = 12$ and $y = 7.$ The swap makes the sum and the difference of the equations come out as tidy multiples of $x + y$ and $x - y .$
Short answerMediumNCERT Exercise Set 13.5, Q8
6 rulers and 4 glue sticks cost ₹168, while 3 rulers and 5 glue sticks cost ₹138. Find the cost of a ruler and of a glue stick.
Show solution
AnswerRuler ₹16, glue stick ₹18.
$6 r + 4 g = 168$ and $3 r + 5 g = 138.$ Doubling the second: $6 r + 10 g = 276.$ Subtracting the first: $6 g = 108,$ so $g = 18,$ and $6 r = 168 - 72 = 96,$ so $r = 16.$
Short answerMediumNCERT Exercise Set 13.5, Q9
A flower pot stands on top of a shelf, and the top of the pot is 130 cm above the floor. When the pot is moved to the floor beside the shelf, the top of the shelf is 50 cm higher than the top of the pot. How tall are the shelf and the pot?
Show solution
AnswerShelf 90 cm, pot 40 cm.
$s + p = 130$ and $s - p = 50.$ Adding: $2 s = 180,$ so $s = 90$ and $p = 40.$

Practice · 16 questions

Practice questions with answers (NCERT “End-of-chapter exercises”)

1
Short answerMediumNCERT End-of-chapter Exercises, Q1
A phone charges steadily from empty. Its battery level is $y$ per cent after $x$ minutes, and the graph of $y = 1.5 x$ passes through $(0, 0)$ and $(4, 6) .$
1. What is the slope of the line, and what does it mean here? What is the $y$ -intercept, and what does it mean?
  Show solution
  AnswerSlope 1.5; intercept 0.
  Slope $= \frac{6 - 0}{4 - 0} = 1.5 ⁣ :$ the battery gains 1.5 per cent each minute. The $y$ -intercept is 0: at $x = 0$ the battery is empty.
2. How long does the battery take to reach 90 per cent?
  min
  Show answer
  Answer 60 min
  $1.5 x = 90$ gives $x = 60$ minutes.
3. Does the point $(10, 16)$ lie on this line?
  yesno
  Show answer
  Answer (B) no
  At $x = 10$ the line gives $y = 1.5 \times 10 = 15,$ not 16.
4. Where does this line meet the line $x = 6$ ?
  Show solution
  Answer $(6, 9)$
  Every point of $x = 6$ has $x = 6;$ on $y = 1.5 x$ that gives $y = 9.$ So they meet at $(6, 9) .$
2
Short answerMediumNCERT End-of-chapter Exercises, Q2
Temperatures in Fahrenheit and Celsius are linked by the linear equation $F = \frac{9}{5} C + 32.$
1. Find two points to draw the graph with $C$ across and $F$ up. What are its slope and intercept?
  Show solution
  AnswerFor example $(0, 32)$ and $(100, 212);$ slope $\frac{9}{5},$ intercept 32.
  $C = 0$ gives $F = 32$ and $C = 100$ gives $F = 212.$ The slope $\frac{9}{5}$ says $F$ rises 9 degrees for every 5 degrees of $C;$ the intercept says water freezes at 32 °F.
2. A pleasant day is 25 °C. What is that in Fahrenheit?
  °F
  Show answer
  Answer 77 °F
  $\frac{9}{5} \times 25 + 32 = 45 + 32 = 77.$
3. A child's temperature is 104 °F. What is that in Celsius?
  °C
  Show answer
  Answer 40 °C
  $104 - 32 = 72$ and $72 \times \frac{5}{9} = 40.$
4. At what temperature do both scales show the same number?
  Show answer
  Answer -40
  Put $F = C ⁣ :$ $C = \frac{9}{5} C + 32$ gives $- \frac{4}{5} C = 32,$ so $C = - 40.$
DiagramEasyNCERT End-of-chapter Exercises, Q3
Solve the pair $x + y = 7$ and $2 x - y = 2$ by drawing graphs.
Show solution
Answer $(3, 4)$
$x + y = 7$ passes through $(0, 7)$ and $(7, 0) .$ $2 x - y = 2$ passes through $(0, - 2)$ and $(1, 0) .$ The lines cross at $(3, 4) .$ Check: $3 + 4 = 7$ and $6 - 4 = 2.$
Short answerMediumNCERT End-of-chapter Exercises, Q4
Line $l$ passes through $(0, 6)$ and $(6, 0) .$ Line $n$ passes through $(0, - 2)$ and $(4, 6) .$ Find the point where they meet.
Show solution
Answer $(\frac{8}{3}, \frac{10}{3})$
Line $l$ has slope $- 1$ and intercept 6: $x + y = 6.$ Line $n$ has slope $\frac{8}{4} = 2$ and intercept $- 2 ⁣ :$ $y = 2 x - 2.$ Substituting: $x + 2 x - 2 = 6,$ so $x = \frac{8}{3}$ and $y = \frac{10}{3} .$ A graph shows the crossing between grid lines, which is why the algebra is needed.
Short answerMediumNCERT End-of-chapter Exercises, Q5
Where does the line $y = m x + d$ cross the $x$ -axis? When is there no such point?
Show solution
AnswerAt $(- \frac{d}{m}, 0)$ when $m \neq 0.$
Put $y = 0 ⁣ :$ $m x + d = 0,$ so $x = - \frac{d}{m},$ provided $m \neq 0.$ If $m = 0$ the line $y = d$ is flat: it never meets the $x$ -axis when $d \neq 0,$ and it is the $x$ -axis itself when $d = 0.$
Short answerMediumNCERT End-of-chapter Exercises, Q6
Cab app X charges ₹40 plus ₹12 per km. Cab app Y charges ₹70 plus ₹9 per km. For what distance do they cost the same? When is each one cheaper?
Show solution
AnswerEqual at 10 km; X is cheaper below 10 km, Y above.
$40 + 12 d = 70 + 9 d$ gives $3 d = 30,$ so $d = 10$ km, where both cost ₹160. X has the smaller fixed charge, so it is cheaper for shorter rides ( $d < 10$ ). Y has the smaller rate per km, so it wins for longer rides ( $d > 10$ ).
7
Short answerEasyNCERT End-of-chapter Exercises, Q7
How many lines are there
1. with slope $- 2$ ?
  Show solution
  AnswerInfinitely many.
  $y = - 2 x + d$ is a line of slope $- 2$ for every value of $d ⁣ :$ a whole family of parallel lines.
2. with slope $- 2$ that pass through $(1, 3)$ ?
  Show solution
  AnswerExactly one: $y = - 2 x + 5.$
  Put $(1, 3)$ into $y = - 2 x + d ⁣ :$ $3 = - 2 + d,$ so $d = 5.$ Only one value of $d$ works.
Short answerMediumNCERT End-of-chapter Exercises, Q8
The pair $6 x + k y - 3 = 0$ and $3 x - 2 y + 5 = 0$ has exactly one solution for most values of $k .$ Which value of $k$ must be avoided?
Show solution
Answer $k = - 4$ (every other value gives exactly one solution).
$\frac{6}{3} = 2,$ so we need $\frac{k}{- 2} \neq 2,$ that is $k \neq - 4.$ (When $k = - 4$ the lines are parallel, since $\frac{- 3}{5} \neq 2$ .)
Short answerHardNCERT End-of-chapter Exercises, Q9
Choose $a$ and $b$ so that $(2 a - b) x + (a + b) y = 4 a + 1$ and $2 x + 3 y = 7$ describe the same line.
Show solution
Answer $a = 5,$ $b = 4.$
We need $\frac{2 a - b}{2} = \frac{a + b}{3} = \frac{4 a + 1}{7} .$ The first and third give $6 a - 7 b = 2;$ the second and third give $- 5 a + 7 b = 3.$ Adding, $a = 5,$ and then $b = 4.$ Check: the first equation becomes $6 x + 9 y = 21,$ which is 3 times $2 x + 3 y = 7.$
Short answerHardNCERT End-of-chapter Exercises, Q10
For one value of $k,$ the lines $2 x + 3 y = 4$ and $(k - 1) x + (k + 2) y = 2 k - 2$ lie exactly on top of each other. Find it.
Show answer
Answer 7
$3 (k - 1) = 2 (k + 2)$ gives $k = 7.$ The second equation is then $6 x + 9 y = 12,$ exactly 3 times the first, so the lines coincide.
11
Short answerHardNCERT End-of-chapter Exercises, Q11
Two delivery drones fly in straight lines over a field marked with a grid.
1. Drone A starts at $(0, 0)$ and moves 3 units up for every 2 units right. Drone B starts at $(8, 0)$ and moves 3 units up for every 1 unit right. Do their paths cross? Where?
  Show solution
  AnswerYes, at $(16, 24) .$
  A: $y = \frac{3}{2} x$ for $x \geq 0.$ B: $y = 3 (x - 8)$ for $x \geq 8.$ Setting them equal: $1.5 x = 3 x - 24,$ so $x = 16$ and $y = 24.$ Both drones pass $x = 16,$ so the paths really cross there.
2. Drone C starts at $(2, 0)$ and moves 1 unit up for every 1 unit right. Drone D starts at $(6, 0)$ and moves 1 unit down for every 2 units right. Do their paths cross?
  Show solution
  AnswerNo.
  The full lines $y = x - 2$ and $y = - \frac{1}{2} (x - 6)$ meet at $(\frac{10}{3}, \frac{4}{3}) .$ But D only moves rightwards from $x = 6,$ and $\frac{10}{3} < 6,$ so that point is behind D's start. The paths never cross.
Short answerHardNCERT End-of-chapter Exercises, Q12
At one moment a clock shows $p$ minutes past 4. Twenty minutes later it shows $q$ minutes past 5. If $p$ is five times $q,$ what time was the second reading?
Show solution
Answer10 minutes past 5 (5:10).
$p + 20 = 60 + q$ and $p = 5 q .$ Substituting: $4 q = 40,$ so $q = 10$ and $p = 50.$ The readings were 4:50 and 5:10.
Short answerMediumNCERT End-of-chapter Exercises, Q13
The digits of a two-digit number add up to 13. Swapping the digits gives a number 27 less than the original. Find the number.
Show answer
Answer 85
$x + y = 13$ and $9 (x - y) = 27,$ so $x - y = 3.$ Then $x = 8,$ $y = 5 ⁣ :$ the number is 85, and $85 - 58 = 27.$
Short answerHardNCERT End-of-chapter Exercises, Q14
In a cyclic quadrilateral $P Q R S,$ $∠ P = (2 x + y)^{\circ},$ $∠ Q = (3 y + 25)^{\circ},$ $∠ R = (5 x + y + 10)^{\circ}$ and $∠ S = (4 x + y + 15)^{\circ} .$ Find all four angles.
Show solution
Answer $∠ P = 55^{\circ},$ $∠ Q = 70^{\circ},$ $∠ R = 125^{\circ},$ $∠ S = 110^{\circ} .$
$∠ P + ∠ R = 180$ gives $7 x + 2 y = 170.$ $∠ Q + ∠ S = 180$ gives $4 x + 4 y = 140,$ so $y = 35 - x .$ Then $7 x + 70 - 2 x = 170,$ so $x = 20$ and $y = 15.$ The angles are 55°, 70°, 125° and 110°, which add up to 360°.
Short answerHardNCERT End-of-chapter Exercises, Q15
A bus covers a fixed distance at a steady speed. If it went 10 km/h faster, the trip would take 2 hours less; if it went 10 km/h slower, it would take 3 hours more. Find the speed, the time and the distance.
Show solution
Answer50 km/h for 12 hours: 600 km.
$(v + 10) (t - 2) = v t$ simplifies to $- v + 5 t = 10.$ $(v - 10) (t + 3) = v t$ simplifies to $3 v - 10 t = 30.$ From the first, $v = 5 t - 10;$ then $15 t - 30 - 10 t = 30,$ so $t = 12$ and $v = 50.$ Distance $= 600$ km. Check: 60 km/h takes 10 h and 40 km/h takes 15 h.
Short answerHardNCERT End-of-chapter Exercises, Q16
A father is three times as old as the sum of his two children's ages. In 5 years he will be twice as old as the sum of their ages. How old is he now?
years
Show answer
Answer 45 years
With father $F$ and children's sum $S ⁣ :$ $F = 3 S$ and $F + 5 = 2 (S + 10) .$ Then $3 S + 5 = 2 S + 20,$ so $S = 15$ and $F = 45.$

Recap

Two Variables, One Line: key points and summary

A linear equation in two variables can be written as $a x + b y + c = 0,$ with $a$ and $b$ not both zero (Section 13.1.1).
A solution is an ordered pair $(x, y)$ that makes the equation true; the order matters (Section 13.2.1).
A single equation like $3 x + 4 y = 24$ is satisfied by infinitely many ordered pairs, since each value of $x$ gives one $y$ (Section 13.2.1).
Its graph is a straight line made up of exactly its solutions, no more and no fewer (Section 13.2.2).
The axis crossings, found by putting $x = 0$ and then $y = 0,$ are an easy way to draw the line (Example 13.2).
Slope $= \frac{rise}{run} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}},$ the same for any two points of the line (Section 13.3.1).
In $y = m x + d,$ $m$ is the slope and $d$ the $y$ -intercept, so the graph meets the $y$ -axis at $(0, d);$ from standard form, $m = - \frac{a}{b}$ (Section 13.3.2).
Positive slope rises, negative slope falls, $y = k$ has slope 0, and a vertical line $x = k$ has no slope (Table 13.1).
A pair of linear equations is solved by an ordered pair that satisfies both equations (Section 13.4).
Substitution and elimination both reduce a pair to one equation in one letter (Section 13.5.1).
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} ⁣ :$ one solution; $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} ⁣ :$ infinitely many; $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} ⁣ :$ none (Table 13.3).
Graphically, intersecting lines give one solution, coincident lines infinitely many, and parallel lines none (Section 13.6).

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise Set 13.1, Q1	Practice 13.1 · Q1
NCERT Exercise Set 13.1, Q2	Practice 13.1 · Q2
NCERT Exercise Set 13.1, Q3	Practice 13.1 · Q3
NCERT Exercise Set 13.2, Q1	Practice 13.2 · Q1
NCERT Exercise Set 13.2, Q2	Practice 13.2 · Q2
NCERT Exercise Set 13.2, Q3	Practice 13.2 · Q3
NCERT Exercise Set 13.2, Q4	Practice 13.2 · Q4
NCERT Exercise Set 13.2, Q5	Practice 13.2 · Q5
NCERT Exercise Set 13.2, Q6	Practice 13.2 · Q6
NCERT Exercise Set 13.2, Q7	Practice 13.2 · Q7
NCERT Exercise Set 13.3, Q1	Practice 13.3 · Q1
NCERT Exercise Set 13.3, Q2	Practice 13.3 · Q2
NCERT Exercise Set 13.3, Q3	Practice 13.3 · Q3
NCERT Exercise Set 13.3, Q4	Practice 13.3 · Q4
NCERT Exercise Set 13.4, Q1	Practice 13.4 · Q1
NCERT Exercise Set 13.4, Q2	Practice 13.4 · Q2
NCERT Exercise Set 13.4, Q3	Practice 13.4 · Q3
NCERT Exercise Set 13.5, Q1 (i)–(iv)	Practice 13.5 · Q1
NCERT Exercise Set 13.5, Q1 (v)–(vii)	Practice 13.5 · Q2
NCERT Exercise Set 13.5, Q1 (viii)–(x)	Practice 13.5 · Q3
NCERT Exercise Set 13.5, Q2	Practice 13.5 · Q4
NCERT Exercise Set 13.5, Q3	Practice 13.5 · Q5
NCERT Exercise Set 13.5, Q4	Practice 13.5 · Q6
NCERT Exercise Set 13.5, Q5	Practice 13.5 · Q7
NCERT Exercise Set 13.5, Q6	Practice 13.5 · Q8
NCERT Exercise Set 13.5, Q7	Practice 13.5 · Q9
NCERT Exercise Set 13.5, Q8	Practice 13.5 · Q10
NCERT Exercise Set 13.5, Q9	Practice 13.5 · Q11
NCERT End-of-chapter Exercises, Q1	End-of-chapter exercises · Q1
NCERT End-of-chapter Exercises, Q2	End-of-chapter exercises · Q2
NCERT End-of-chapter Exercises, Q3	End-of-chapter exercises · Q3
NCERT End-of-chapter Exercises, Q4	End-of-chapter exercises · Q4
NCERT End-of-chapter Exercises, Q5	End-of-chapter exercises · Q5
NCERT End-of-chapter Exercises, Q6	End-of-chapter exercises · Q6
NCERT End-of-chapter Exercises, Q7	End-of-chapter exercises · Q7
NCERT End-of-chapter Exercises, Q8	End-of-chapter exercises · Q8
NCERT End-of-chapter Exercises, Q9	End-of-chapter exercises · Q9
NCERT End-of-chapter Exercises, Q10	End-of-chapter exercises · Q10
NCERT End-of-chapter Exercises, Q11	End-of-chapter exercises · Q11
NCERT End-of-chapter Exercises, Q12	End-of-chapter exercises · Q12
NCERT End-of-chapter Exercises, Q13	End-of-chapter exercises · Q13
NCERT End-of-chapter Exercises, Q14	End-of-chapter exercises · Q14
NCERT End-of-chapter Exercises, Q15	End-of-chapter exercises · Q15
NCERT End-of-chapter Exercises, Q16	End-of-chapter exercises · Q16

Quick check: MCQs on Two Variables, One Line

13.1.1 Standard Form of Linear Equations in Two Variables3 questions
13.2.2 Geometric Visualisation of Solutions4 questions
13.3.2 Slope-Intercept Form4 questions
13.4 Pair of Linear Equations in Two Variables3 questions
When elimination removes both letters5 questions
13.6 Graphical Method for Solving a Pair of Linear Equations in Two Variables4 questions

Glossary · 15

Key terms (English–Hindi)

coefficientगुणांकgunaank: The number that multiplies a variable, such as 3 in $3 x$ .First seen in the chapter →
coincident linesसंपाती रेखाएँsampaati rekhaen: Two equations whose graphs are the same line, so every point is shared.First seen in the chapter →

elimination methodविलोपन विधिvilopan vidhi: Solving a pair by matching the coefficients of one variable and adding or subtracting so that it cancels.First seen in the chapter →

graph of an equationआलेखaalekh: The set of all points whose coordinates satisfy the equation; for a linear equation it is a straight line.First seen in the chapter →

intersecting linesप्रतिच्छेदी रेखाएँpratichchhedi rekhaen: Lines that meet at exactly one point.First seen in the chapter →

linear equation in two variablesदो चरों वाला रैखिक समीकरणdo charon vaala rekhik samikaran: An equation that can be written as $a x + b y + c = 0$ , with $a$ and $b$ not both zero.First seen in the chapter →

ordered pairक्रमित युग्मkramit yugm: Two numbers written as $(x, y)$ , where the order matters: $(4, 3)$ and $(3, 4)$ are different.First seen in the chapter →

pair of linear equationsरैखिक समीकरण युग्मrekhik samikaran yugm: Two linear equations in the same two variables, considered together.First seen in the chapter →
parallel linesसमांतर रेखाएँsamaantar rekhaen: Lines in a plane that never meet; they have equal slopes.First seen in the chapter →

slopeढालdhaal: The steepness of a line: rise divided by run between any two of its points.First seen in the chapter →
slope-intercept formढाल-अंतःखंड रूपdhaal-antahkhand roop: The form $y = m x + d$ , which shows the slope $m$ and the $y$ -intercept $d$ directly.First seen in the chapter →
solutionहलhal: Values of $x$ and $y$ that, used together, make both sides of the equation equal.First seen in the chapter →
standard formमानक रूपmaanak roop: The layout $a x + b y + c = 0$ , with every term on one side and 0 on the other.First seen in the chapter →
substitution methodप्रतिस्थापन विधिpratisthaapan vidhi: Solving a pair by writing one variable in terms of the other and putting it into the other equation.First seen in the chapter →

y-intercepty-अंतःखंडy-antahkhand: The value $d$ where a line meets the $y$ -axis, at the point $(0, d)$ .First seen in the chapter →

Two Variables, One Line: Class 9 Ganita Manjari · Chapter 13

What you will learn in Two Variables, One Line

13.1 Linear Equations in Two Variables

13.1.1 Standard Form of Linear Equations in Two Variables

Practice 13.1: questions with answers

13.2 Solution of Linear Equation in Two Variables

13.2.1 Algebraic Meaning of Solution

13.2.2 Geometric Visualisation of Solutions

Practice 13.2: questions with answers

13.3 Slope-Intercept Form of a Linear Equation in Two Variables

13.3.1 Understanding Slope

13.3.2 Slope-Intercept Form

Practice 13.3: questions with answers

13.4 Pair of Linear Equations in Two Variables

Practice 13.4: questions with answers

13.5 Finding Solutions to a Pair of Linear Equations

13.5.1 Algebraic Methods for Solving a Pair of Linear Equations

The substitution method

The elimination method

When elimination removes both letters

13.6 Graphical Method for Solving a Pair of Linear Equations in Two Variables

Practice 13.5: questions with answers

Practice questions with answers (NCERT “End-of-chapter exercises”)

Two Variables, One Line: key points and summary

Find your NCERT exercise question

Quick check: MCQs on Two Variables, One Line

Key terms (English–Hindi)

You reached the end