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NCERT Solutions · Class 10 Maths · Chapter 3

Chapter 3: Pair of Linear Equations in Two Variables (Algebra)

Step-by-step NCERT solutions for Class 10 Maths Chapter 3, Pair of Linear Equations in Two Variables (2026-27 reprint): Exercise 3.1 graphical method and consistency by comparing ratios, Exercise 3.2 substitution method and word problems, and Exercise 3.3 elimination method, with neat graphs. All 13 questions are answered, with the key answer highlighted.

Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-10-maths/chapter-3-pair-of-linear-equations-in-two-variables

For $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ : if $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ the lines intersect (unique solution, consistent); if $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ they coincide (infinitely many solutions, dependent and consistent); if $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ they are parallel (no solution, inconsistent).

Exercise 3.1

1 (i)

10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz (form the equations and solve graphically).

Solution

Let the number of girls be $x$ and of boys be $y$ :

$x + y = 10, x - y = 4$

x + y = 10			x − y = 4
x	4	10	x	4	10
y	6	0	y	0	6

The lines meet at (7, 3)

The lines intersect at $(7, 3)$ .

7 girls and 3 boys.

1 (ii)

5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and one pen (graphically).

Solution

Let a pencil cost ₹ $x$ and a pen ₹ $y$ :

$5 x + 7 y = 50, 7 x + 5 y = 46$

5x + 7y = 50				7x + 5y = 46
x	3	10	−4	x	3	8	−2
y	5	0	10	y	5	−2	12

The lines meet at (3, 5)

A pencil costs ₹3 and a pen costs ₹5.

Comparing the ratios a₁/a₂, b₁/b₂ and c₁/c₂, find whether the lines intersect at a point, are parallel or coincident: (i) 5x – 4y + 8 = 0, 7x + 6y – 9 = 0 (ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0 (iii) 6x – 3y + 10 = 0, 2x – y + 9 = 0

Solution

$\frac{a _{1}}{a _{2}} = \frac{5}{7}$ , $\frac{b _{1}}{b _{2}} = \frac{- 4}{6} = - \frac{2}{3}$ . These are unequal, so the lines intersect at a point.
$\frac{9}{18} = \frac{3}{6} = \frac{12}{24} = \frac{1}{2}$ . All three ratios are equal, so the lines are coincident.
$\frac{6}{2} = \frac{- 3}{- 1} = 3$ , but $\frac{c _{1}}{c _{2}} = \frac{10}{9} \neq = 3$ , so the lines are parallel.

(i) Intersecting (ii) Coincident (iii) Parallel

Comparing the ratios, find whether the pairs are consistent or inconsistent: (i) 3x + 2y = 5, 2x – 3y = 7 (ii) 2x – 3y = 8, 4x – 6y = 9 (iii) 3x/2 + 5y/3 = 7, 9x – 10y = 14 (iv) 5x – 3y = 11, –10x + 6y = –22 (v) 4x/3 + 2y = 8, 2x + 3y = 12

Solution

$\frac{3}{2} \neq = \frac{2}{- 3}$ : unique solution, consistent.
$\frac{2}{4} = \frac{- 3}{- 6} = \frac{1}{2}$ but $\frac{8}{9} \neq = \frac{1}{2}$ : parallel, inconsistent.
$\frac{3/2}{9} = \frac{1}{6}$ , $\frac{5/3}{- 10} = - \frac{1}{6}$ : unequal, so a unique solution, consistent.
$\frac{5}{- 10} = \frac{- 3}{6} = \frac{11}{- 22} = - \frac{1}{2}$ : coincident, consistent (dependent).
$\frac{4/3}{2} = \frac{2}{3}$ , $\frac{2}{3}$ , $\frac{8}{12} = \frac{2}{3}$ : coincident, consistent (dependent).

(i) Consistent (ii) Inconsistent (iii) Consistent (iv) Consistent (v) Consistent

Which pairs are consistent or inconsistent? If consistent, solve graphically: (i) x + y = 5, 2x + 2y = 10 (ii) x – y = 8, 3x – 3y = 16 (iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0 (iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

Solution

(i) $\frac{1}{2} = \frac{1}{2} = \frac{5}{10}$ : the lines coincide, so it is consistent with infinitely many solutions: every point of $y = 5 - x$ , e.g. $(0, 5), (2, 3), (5, 0)$ .

(ii) $\frac{1}{3} = \frac{- 1}{- 3}$ but $\frac{8}{16} = \frac{1}{2}$ : parallel lines, inconsistent.

(iii) $\frac{2}{4} \neq = \frac{1}{- 2}$ : consistent with a unique solution.

2x + y = 6				4x − 2y = 4
x	0	3	2	x	0	1	2
y	6	0	2	y	−2	0	2

The lines meet at (2, 2)

The solution is $x = 2$ , $y = 2$ .

(iv) $\frac{2}{4} = \frac{- 2}{- 4}$ but $\frac{- 2}{- 5} = \frac{2}{5} \neq = \frac{1}{2}$ : parallel lines, inconsistent.

(i) Consistent, infinitely many solutions (y = 5 – x) (ii) Inconsistent (iii) Consistent, x = 2, y = 2 (iv) Inconsistent

Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Solution

Let the length be $x$ m and the width $y$ m:

$x = y + 4, x + y = 36$

Substituting: $2 y + 4 = 36$ , so $y = 16$ and $x = 20$ . (Graphically, the lines $x - y = 4$ and $x + y = 36$ meet at $(20, 16)$ .)

Length 20 m, width 16 m.

Given the equation 2x + 3y – 8 = 0, write another linear equation so that the pair represents (i) intersecting lines (ii) parallel lines (iii) coincident lines.

Solution

Choose ratios with $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , e.g. $3x + 2y - 7 = 0$ .
Keep $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}}$ but change the constant, e.g. $2x + 3y - 12 = 0$ (or $4 x + 6 y - 5 = 0$ ).
Multiply the whole equation by a constant, e.g. $4x + 6y - 16 = 0$ .

(i) 3x + 2y – 7 = 0 (ii) 2x + 3y – 12 = 0 (iii) 4x + 6y – 16 = 0 (other answers are possible)

Draw the graphs of x – y + 1 = 0 and 3x + 2y – 12 = 0. Find the vertices of the triangle formed by these lines and the x-axis, and shade the triangle.

Solution

x − y + 1 = 0				3x + 2y − 12 = 0
x	−1	0	2	x	4	0	2
y	0	1	3	y	0	6	3

The triangle ABC formed by the two lines and the x-axis (shaded)

$x - y + 1 = 0$ meets the $x$ -axis ( $y = 0$ ) at $A (- 1, 0)$ .
$3 x + 2 y - 12 = 0$ meets the $x$ -axis at $B (4, 0)$ .
The lines meet each other at $C (2, 3)$ .

The vertices are (–1, 0), (4, 0) and (2, 3).

Exercise 3.2

Solve by substitution: (i) x + y = 14, x – y = 4 (ii) s – t = 3, s/3 + t/2 = 6 (iii) 3x – y = 3, 9x – 3y = 9 (iv) 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3 (v) √2x + √3y = 0, √3x – √8y = 0 (vi) 3x/2 – 5y/3 = –2, x/3 + y/2 = 13/6

Solution

(i) From $x - y = 4$ : $x = y + 4$ . Then $2 y + 4 = 14$ , so $y = 5$ , $x = 9$ .

(ii) $s = t + 3$ . Then $\frac{t + 3}{3} + \frac{t}{2} = 6 \Rightarrow 2 t + 6 + 3 t = 36 \Rightarrow t = 6$ , so $s = 9$ .

(iii) $y = 3 x - 3$ . Substituting: $9 x - 3 (3 x - 3) = 9 \Rightarrow 9 = 9$ , which is always true. The equations are the same line, so there are infinitely many solutions: $y = 3 x - 3$ for any real $x$ .

(iv) Multiply by 10: $2 x + 3 y = 13$ and $4 x + 5 y = 23$ . From the first, $x = \frac{13 - 3 y}{2}$ . Then $2 (13 - 3 y) + 5 y = 23 \Rightarrow 26 - y = 23 \Rightarrow y = 3$ , so $x = 2$ .

(v) From the first, $x = - \frac{3}{2} y$ . Then $3 (- \frac{3}{2} y) - 8 y = 0 \Rightarrow - (\frac{3}{2} + 22) y = 0 \Rightarrow y = 0$ , so $x = 0$ .

(vi) Clear fractions: $9 x - 10 y = - 12$ and $2 x + 3 y = 13$ . From the second, $x = \frac{13 - 3 y}{2}$ . Then $\frac{9 ( 13 - 3 y )}{2} - 10 y = - 12 \Rightarrow 117 - 27 y - 20 y = - 24 \Rightarrow 47 y = 141 \Rightarrow y = 3$ , so $x = 2$ .

(i) x = 9, y = 5 (ii) s = 9, t = 6 (iii) infinitely many solutions, y = 3x – 3 (iv) x = 2, y = 3 (v) x = 0, y = 0 (vi) x = 2, y = 3

Solve 2x + 3y = 11 and 2x – 4y = –24, and hence find m for which y = mx + 3.

Solution

Subtracting the equations: $7 y = 35$ , so $y = 5$ . Then $2 x = 11 - 15 = - 4$ , so $x = - 2$ .

Since $(- 2, 5)$ lies on $y = m x + 3$ : $5 = - 2 m + 3 \Rightarrow m = - 1$ .

x = –2, y = 5; m = –1

Form the equations and solve by substitution: (i) The difference of two numbers is 26 and one is three times the other. (ii) The larger of two supplementary angles exceeds the smaller by 18°. (iii) 7 bats and 6 balls cost ₹3800; 3 bats and 5 balls cost ₹1750. (iv) Taxi fares: 10 km cost ₹105 and 15 km cost ₹155; find the fixed charge, the charge per km and the fare for 25 km. (v) A fraction becomes 9/11 if 2 is added to both terms, and 5/6 if 3 is added to both. (vi) In five years Jacob will be three times as old as his son; five years ago he was seven times as old.

Solution

(i) Let the numbers be $x > y$ : $x - y = 26$ , $x = 3 y$ . Then $2 y = 26$ , so $y = 13$ , $x = 39$ .

(ii) Let the angles be $x > y$ : $x + y = 180$ , $x - y = 18$ . Then $x = y + 18$ , $2 y = 162$ , so $y = 81°$ , $x = 99°$ .

(iii) Let a bat cost ₹ $x$ , a ball ₹ $y$ : $7 x + 6 y = 3800$ , $3 x + 5 y = 1750$ . From the second, $x = \frac{1750 - 5 y}{3}$ . Then $\frac{7 ( 1750 - 5 y )}{3} + 6 y = 3800 \Rightarrow 12250 - 35 y + 18 y = 11400 \Rightarrow y = 50$ , $x = 500$ .

(iv) Let the fixed charge be ₹ $x$ and the rate ₹ $y$ per km: $x + 10 y = 105$ , $x + 15 y = 155$ . Substituting $x = 105 - 10 y$ : $5 y = 50$ , so $y = 10$ , $x = 5$ . For 25 km: $5 + 25 \times 10 =$ ₹255.

(v) Let the fraction be $\frac{x}{y}$ : $\frac{x + 2}{y + 2} = \frac{9}{11}$ and $\frac{x + 3}{y + 3} = \frac{5}{6}$ , i.e. $11 x - 9 y = - 4$ and $6 x - 5 y = - 3$ . From the second, $x = \frac{5 y - 3}{6}$ ; then $\frac{11 ( 5 y - 3 )}{6} - 9 y = - 4 \Rightarrow 55 y - 33 - 54 y = - 24 \Rightarrow y = 9$ , $x = 7$ .

(vi) Let Jacob be $x$ and his son $y$ years: $x + 5 = 3 (y + 5)$ and $x - 5 = 7 (y - 5)$ , i.e. $x - 3 y = 10$ and $x - 7 y = - 30$ . Substituting $x = 3 y + 10$ : $- 4 y = - 40$ , so $y = 10$ , $x = 40$ .

(i) 39 and 13 (ii) 99° and 81° (iii) bat ₹500, ball ₹50 (iv) fixed ₹5, ₹10 per km, ₹255 for 25 km (v) 7/9 (vi) Jacob 40 years, son 10 years

Exercise 3.3

Solve by elimination and by substitution: (i) x + y = 5, 2x – 3y = 4 (ii) 3x + 4y = 10, 2x – 2y = 2 (iii) 3x – 5y – 4 = 0, 9x = 2y + 7 (iv) x/2 + 2y/3 = –1, x – y/3 = 3

Solution

(i) Multiply the first by 3 and add: $5 x = 19$ , so $x = \frac{19}{5}$ and $y = 5 - \frac{19}{5} = \frac{6}{5}$ . (Substitution, $x = 5 - y$ , gives the same.)

(ii) Multiply the second by 2 and add to the first: $7 x = 14$ , so $x = 2$ and $y = 1$ .

(iii) Multiply the first by 3: $9 x - 15 y = 12$ ; subtract it from $9 x - 2 y = 7$ : $13 y = - 5$ , so $y = - \frac{5}{13}$ . Then $3 x = 4 + 5 y = 4 - \frac{25}{13} = \frac{27}{13}$ , so $x = \frac{9}{13}$ .

(iv) Clear fractions: $3 x + 4 y = - 6$ and $3 x - y = 9$ . Subtracting: $5 y = - 15$ , so $y = - 3$ ; then $3 x = 9 + y = 6$ , so $x = 2$ .

(i) x = 19/5, y = 6/5 (ii) x = 2, y = 1 (iii) x = 9/13, y = –5/13 (iv) x = 2, y = –3

Form the equations and solve by elimination: (i) Adding 1 to the numerator and subtracting 1 from the denominator makes a fraction 1; adding 1 only to the denominator makes it 1/2. (ii) Five years ago Nuri was thrice as old as Sonu; ten years later Nuri will be twice as old as Sonu. (iii) The digits of a two-digit number add to 9, and nine times the number is twice the number with its digits reversed. (iv) Meena withdrew ₹2000 in ₹50 and ₹100 notes, 25 notes in all. (v) A library charges a fixed amount for the first three days and extra for each day after; Saritha paid ₹27 for 7 days and Susy ₹21 for 5 days.

Solution

(i) Let the fraction be $\frac{x}{y}$ : $x + 1 = y - 1$ and $2 x = y + 1$ , i.e. $x - y = - 2$ and $2 x - y = 1$ . Subtracting: $x = 3$ , then $y = 5$ . The fraction is $\frac{3}{5}$ .

(ii) Let Nuri be $x$ and Sonu $y$ years: $x - 5 = 3 (y - 5)$ and $x + 10 = 2 (y + 10)$ , i.e. $x - 3 y = - 10$ and $x - 2 y = 10$ . Subtracting: $y = 20$ , so $x = 50$ .

(iii) Let the tens digit be $x$ and the units digit $y$ : $x + y = 9$ and $9 (10 x + y) = 2 (10 y + x)$ , i.e. $88 x = 11 y$ , so $y = 8 x$ . Then $9 x = 9$ , so $x = 1$ , $y = 8$ . The number is 18.

(iv) Let there be $x$ ₹50 notes and $y$ ₹100 notes: $x + y = 25$ and $50 x + 100 y = 2000$ , i.e. $x + 2 y = 40$ . Subtracting: $y = 15$ , so $x = 10$ .

(v) Let the fixed charge be ₹ $x$ and the extra charge ₹ $y$ per day: $x + 4 y = 27$ and $x + 2 y = 21$ . Subtracting: $2 y = 6$ , so $y = 3$ and $x = 15$ .

(i) 3/5 (ii) Nuri 50 years, Sonu 20 years (iii) 18 (iv) 10 notes of ₹50 and 15 notes of ₹100 (v) fixed charge ₹15, ₹3 per extra day

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