NCERT Solutions for Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables — Complete Guide

A pair of linear equations in two variables $x$ and $y$ has the general form:

where $a_1, b_1, c_1, a_2, b_2, c_2$ are real numbers and $a_1^2 + b_1^2 \neq 0$, $a_2^2 + b_2^2 \neq 0$ (i.e., both equations are genuinely linear, not degenerate).

Each equation represents a straight line in the $xy$-plane. Solving the system means finding the point(s) where these two lines meet.

The relationship between the two lines determines the nature of the solution:

Case 1: Intersecting Lines ($\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$)
- The system has exactly one (unique) solution — the coordinates of the intersection point.
- The system is called consistent.

Case 2: Coincident Lines ($\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$)
- The two equations represent the same line.
- The system has infinitely many solutions — every point on the line is a solution.
- The system is called consistent (and dependent).

Case 3: Parallel Lines ($\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$)
- The lines never meet.
- The system has no solution.
- The system is called inconsistent.

This ratio test is the fastest way to determine the nature of a system without actually solving it.

1. Substitution Method: Express one variable in terms of the other from one equation, then substitute into the second equation. Best when one variable has coefficient $1$ or $-1$.

2. Elimination Method: Multiply the equations by suitable constants to make the coefficients of one variable equal (or negatives), then add or subtract to eliminate that variable. Best for equations with integer coefficients.

3. Cross-Multiplication Method: For $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

4. Graphical Method: Plot both lines on a graph and read off the intersection point. Required for Exercise 3.2 but rarely used in board exams for computation (too slow and imprecise).

While any method works for any system (as long as a unique solution exists), choosing wisely saves time:

• If one equation already has $x$ or $y$ isolated (like $y = 3x + 2$), use substitution.
  - If coefficients are small integers, use elimination — it's the fastest.
  - If the problem says "use cross-multiplication," use the formula.
  - For MCQs, the ratio test (consistency check) is often enough — you don't need to solve!

In board exams, elimination is the most popular method among toppers because it's fast and less error-prone than substitution.

Problem: Aftab tells his daughter: "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be." Represent this algebraically and graphically.

Solution:
Let Aftab's present age $= x$ years, daughter's present age $= y$ years.

Seven years ago: Aftab was $(x - 7)$, daughter was $(y - 7)$.

Three years hence: Aftab will be $(x + 3)$, daughter will be $(y + 3)$.

From the ratio test: $\frac{1}{1} \neq \frac{-7}{-3}$, i.e., $1 \neq \frac{7}{3}$.

Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the system has a unique solution (intersecting lines).

Solving (1) - (2): $-4y + 48 = 0 \implies y = 12$, then $x = 7(12) - 42 = 42$.

Answer: Aftab is $42$ years old, daughter is $12$ years old.

Problem: The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, he buys another bat and 3 more balls for Rs 1300. Represent algebraically and check consistency.

Solution:
Let cost of a bat $= x$, cost of a ball $= y$.

Wait — let's recheck. From (1): $x + 2y = 1300$. From (2): $x + 3y = 1300$.

Subtracting: $-y = 0 \implies y = 0$? That gives a ball costing Rs 0, which doesn't make sense.

Actually, let me re-read: the second purchase is 1 bat and 3 balls for Rs 1300. And equation (1) after dividing by 3 gives $x + 2y = 1300$.

Subtracting (1) from (2): $y = 0$. Then $x = 1300$.

So each bat costs Rs 1300 and each ball costs Rs 0. The system is consistent with a unique solution, but the physical answer suggests the problem data may be contrived.

Ratio test: $\frac{1}{1} \neq \frac{2}{3}$. Unique solution — lines are intersecting.

Solution:
Let cost of pencil $= x$, cost of eraser $= y$.

Ratio test: $\frac{2}{4} = \frac{3}{6} = \frac{9}{18} = \frac{1}{2}$.

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident — infinitely many solutions.

Equation (2) is just $2 \times$ equation (1). Any $(x, y)$ satisfying $2x + 3y = 9$ is a solution.

Solution:
Find points for each line:

Line 1: $x + 3y = 6$
- $x = 0 \Rightarrow y = 2$: Point $(0, 2)$
- $y = 0 \Rightarrow x = 6$: Point $(6, 0)$
- $x = 3 \Rightarrow y = 1$: Point $(3, 1)$

Line 2: $2x - 3y = 12$
- $x = 0 \Rightarrow y = -4$: Point $(0, -4)$
- $y = 0 \Rightarrow x = 6$: Point $(6, 0)$
- $x = 3 \Rightarrow y = -2$: Point $(3, -2)$

Both lines pass through $(6, 0)$. The intersection point is $(6, 0)$.

Answer: $x = 6, y = 0$.

Verification: $6 + 3(0) = 6$ $\checkmark$ and $2(6) - 3(0) = 12$ $\checkmark$.

Solution:
Let cost of pencil $= x$, cost of pen $= y$.

Ratio test: $\frac{5}{7} \neq \frac{7}{5}$. So a unique solution exists.

Solving by elimination:
Multiply (1) by 5: $25x + 35y = 250$
Multiply (2) by 7: $49x + 35y = 322$
Subtract: $24x = 72 \implies x = 3$
From (1): $15 + 7y = 50 \implies y = 5$

Answer: Pencil costs Rs $3$, pen costs Rs $5$.

For the graph, plot both lines using the points above and confirm they intersect at $(3, 5)$.

Solution:
From equation (1): $x = 14 - y \quad \cdots (*)$

Substitute in equation (2):

From (*): $x = 14 - 5 = 9$

Answer: $x = 9, y = 5$

Verification: $9 + 5 = 14$ $\checkmark$, $9 - 5 = 4$ $\checkmark$

Solution:
From equation (1): $s = t + 3 \quad \cdots (*)$

Substitute in equation (2):

Multiply through by 6:

From (*): $s = 6 + 3 = 9$

Answer: $s = 9, t = 6$

Verification: $9 - 6 = 3$ $\checkmark$, $\frac{9}{3} + \frac{6}{2} = 3 + 3 = 6$ $\checkmark$

Solution:
From equation (1): $y = 3x - 3 \quad \cdots (*)$

Substitute in equation (2):

This is always true! The two equations are dependent (equation (2) is $3 \times$ equation (1)).

Answer: Infinitely many solutions. Every point on the line $y = 3x - 3$ is a solution.

Examples: $(1, 0)$, $(2, 3)$, $(0, -3)$, etc.

Solution:
Multiply both equations by 10 to clear decimals:

From (1): $x = \frac{13 - 3y}{2} \quad \cdots (*)$

Substitute in (2):

From (*): $x = \frac{13 - 9}{2} = 2$

Answer: $x = 2, y = 3$

Solution:
From equation (1): $x = -\frac{\sqrt{3}}{\sqrt{2}} y \quad \cdots (*)$

Substitute in equation (2):

From (*): $x = 0$

Answer: $x = 0, y = 0$ (the trivial solution).

Note: When both constant terms are $0$ (homogeneous system), the system always has the trivial solution $(0, 0)$. It has non-trivial solutions only if the lines are coincident.

Problem: The taxi charges in a city consist of a fixed charge together with a charge per km. For 10 km, the charge is Rs 105. For 15 km, the charge is Rs 155. Find the fixed charge and per-km rate.

Solution:
Let fixed charge $= x$, per-km charge $= y$.

From (1): $x = 105 - 10y$. Substitute in (2):

Answer: Fixed charge $=$ Rs $5$, charge per km $=$ Rs $10$.

For 25 km: charge $= 5 + 10 \times 25 =$ Rs $255$.

Solution:
Multiply equation (1) by 3:

Add (1') and (2):

From (1): $\frac{19}{5} + y = 5 \implies y = 5 - \frac{19}{5} = \frac{6}{5}$

Answer: $x = \frac{19}{5}, y = \frac{6}{5}$

Solution:
Multiply equation (2) by 2:

Add equation (1) and (2'):

From (2): $2(2) - 2y = 2 \implies 4 - 2y = 2 \implies y = 1$

Answer: $x = 2, y = 1$

Verification: $3(2) + 4(1) = 10$ $\checkmark$, $2(2) - 2(1) = 2$ $\checkmark$

Solution:
Rewrite in standard form:

Multiply (1) by 3: $9x - 15y = 12$
Subtract (2): $(9x - 15y) - (9x - 2y) = 12 - 7$

From (1): $3x = 4 + 5 \times (-\frac{5}{13}) = 4 - \frac{25}{13} = \frac{52 - 25}{13} = \frac{27}{13}$

Answer: $x = \frac{9}{13}, y = -\frac{5}{13}$

Solution:
Multiply equation (1) by 6: $3x + 4y = -6 \quad \cdots (1')$
Multiply equation (2) by 3: $3x - y = 9 \quad \cdots (2')$

Subtract (2') from (1'): $(3x + 4y) - (3x - y) = -6 - 9$

From (2'): $3x - (-3) = 9 \implies 3x = 6 \implies x = 2$

Answer: $x = 2, y = -3$

Problem (a): The sum of a two-digit number and the number formed by reversing its digits is 66. If the digits differ by 2, find the number.

Solution:
Let tens digit $= x$, units digit $= y$.
Number $= 10x + y$, reversed $= 10y + x$.

$(10x + y) + (10y + x) = 66 \implies 11(x + y) = 66 \implies x + y = 6 \quad \cdots (1)$

$|x - y| = 2$, so $x - y = 2$ or $x - y = -2 \quad \cdots (2)$

Case 1: $x + y = 6$ and $x - y = 2$. Adding: $2x = 8 \implies x = 4, y = 2$. Number $= 42$.

Case 2: $x + y = 6$ and $x - y = -2$. Adding: $2x = 4 \implies x = 2, y = 4$. Number $= 24$.

Answer: The number is 42 or 24.

Problem (b): 5 pencils and 7 pens cost Rs 50; 7 pencils and 5 pens cost Rs 46. Find costs.

Solution:

Multiply (1) by 5 and (2) by 7:
$25x + 35y = 250$
$49x + 35y = 322$

Subtract: $24x = 72 \implies x = 3$. From (1): $15 + 7y = 50 \implies y = 5$.

Answer: Pencil $=$ Rs $3$, pen $=$ Rs $5$.

Solution:
Here $a_1 = 2, b_1 = 3, c_1 = -46$ and $a_2 = 3, b_2 = 5, c_2 = -74$.

Verification: $2(8) + 3(10) = 16 + 30 = 46$ $\checkmark$, $3(8) + 5(10) = 24 + 50 = 74$ $\checkmark$

Solution:
$a_1 = 3, b_1 = 1, c_1 = -5$; $a_2 = 1, b_2 = -1, c_2 = -5$.

Answer: $x = \frac{5}{2}, y = -\frac{5}{2}$

Solution:
$a_1 = 1, b_1 = 2, c_1 = 1$; $a_2 = 2, b_2 = -3, c_2 = -12$.

Answer: $x = 3, y = -2$

Problem: For what value of $k$ will the system $kx + 3y - (k-3) = 0$ and $12x + ky - k = 0$ have no solution?

Solution:
For no solution (parallel lines): $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

Check $\frac{c_1}{c_2}$:
For $k = 6$: $\frac{c_1}{c_2} = \frac{-(6-3)}{-6} = \frac{-3}{-6} = \frac{1}{2}$ and $\frac{a_1}{a_2} = \frac{6}{12} = \frac{1}{2}$.
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2}$, the lines are coincident (infinitely many solutions), not parallel. So $k = 6$ gives dependent, not inconsistent.

For $k = -6$: $\frac{a_1}{a_2} = \frac{-6}{12} = -\frac{1}{2}$, $\frac{b_1}{b_2} = \frac{3}{-6} = -\frac{1}{2}$, $\frac{c_1}{c_2} = \frac{-(-6-3)}{-(-6)} = \frac{9}{6} = \frac{3}{2}$.
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel. No solution.

Answer: $k = -6$

Solution:
Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$.

Multiply (1) by 3 and (2) by 2:
$9u + 6v = 36$
$4u + 6v = 26$

Subtract: $5u = 10 \implies u = 2$.
From (1): $6 + 2v = 12 \implies v = 3$.

$x = \frac{1}{u} = \frac{1}{2}$, $y = \frac{1}{v} = \frac{1}{3}$.

Answer: $x = \frac{1}{2}, y = \frac{1}{3}$

Solution:
Let $u = \frac{1}{\sqrt{x}}$ and $v = \frac{1}{\sqrt{y}}$.

Multiply (1) by 3: $6u + 9v = 6$.
Add (2): $6u + 9v + 4u - 9v = 6 + (-1)$
$10u = 5 \implies u = \frac{1}{2}$.

From (1): $1 + 3v = 2 \implies v = \frac{1}{3}$.

$\frac{1}{\sqrt{x}} = \frac{1}{2} \implies \sqrt{x} = 2 \implies x = 4$.
$\frac{1}{\sqrt{y}} = \frac{1}{3} \implies \sqrt{y} = 3 \implies y = 9$.

Answer: $x = 4, y = 9$

Solution:
Let $u = \frac{1}{x}$. The system becomes:

Multiply (1) by 4 and (2) by 3:
$16u + 12y = 56$
$9u - 12y = 69$

Add: $25u = 125 \implies u = 5$.
So $x = \frac{1}{5}$.

From (1): $20 + 3y = 14 \implies y = -2$.

Answer: $x = \frac{1}{5}, y = -2$

Solution:
Let $u = \frac{1}{x-1}$ and $v = \frac{1}{y-2}$.

Multiply (1) by 3: $15u + 3v = 6$.
Add (2): $21u = 7 \implies u = \frac{1}{3}$.

From (1): $\frac{5}{3} + v = 2 \implies v = \frac{1}{3}$.

$x - 1 = 3 \implies x = 4$.
$y - 2 = 3 \implies y = 5$.

Answer: $x = 4, y = 5$

Problem: Ten years hence, the age of Nuri will be three times the age of Sonu. Five years ago, Nuri's age was five times Sonu's. Find their present ages.

Solution:
Let Nuri's present age $= x$, Sonu's present age $= y$.

Ten years hence: $x + 10 = 3(y + 10) \implies x - 3y = 20 \quad \cdots (1)$

Five years ago: $x - 5 = 5(y - 5) \implies x - 5y = -20 \quad \cdots (2)$

Subtract (2) from (1): $(-3y) - (-5y) = 20 - (-20)$
$2y = 40 \implies y = 20$

From (1): $x = 3(20) + 20 = 80$.

Wait — $x = 80$ means Nuri is 80? Let me recheck.
From (1): $x - 3(20) = 20 \implies x = 80$.

Check: Ten years hence: Nuri $= 90$, Sonu $= 30$. $90 = 3 \times 30$ $\checkmark$.
Five years ago: Nuri $= 75$, Sonu $= 15$. $75 = 5 \times 15$ $\checkmark$.

Answer: Nuri is $50$ years old, Sonu is $10$ years old.

Actually let me redo: $x - 3y = 20$ and $x - 5y = -20$.
Subtract: $2y = 40 \implies y = 20$, $x = 80$.

Hmm, the check works with $x = 80, y = 20$, so that must be correct despite seeming old.

Answer: Nuri is $80$ years, Sonu is $20$ years.

Problem: A train covered a certain distance at a uniform speed. If the train had been 10 km/h faster, it would have taken 2 hours less. If the speed were 10 km/h slower, it would have taken 3 more hours. Find the distance.

Solution:
Let the speed $= s$ km/h and time $= t$ hours. Distance $= st$.

Condition 1: $(s + 10)(t - 2) = st$
$st - 2s + 10t - 20 = st$
$-2s + 10t = 20 \implies -s + 5t = 10 \quad \cdots (1)$

Condition 2: $(s - 10)(t + 3) = st$
$st + 3s - 10t - 30 = st$
$3s - 10t = 30 \quad \cdots (2)$

Multiply (1) by 2: $-2s + 10t = 20 \quad \cdots (1')$
Add (2) and (1'): $s = 50$.

From (1): $-50 + 5t = 10 \implies t = 12$.

Distance $= 50 \times 12 = 600$ km.

Answer: Distance $= 600$ km, speed $= 50$ km/h.

Problem: A fraction becomes $\frac{1}{3}$ when 1 is subtracted from the numerator and becomes $\frac{1}{4}$ when 8 is added to the denominator. Find the fraction.

Solution:
Let the fraction be $\frac{x}{y}$.

$\frac{x - 1}{y} = \frac{1}{3} \implies 3x - 3 = y \implies 3x - y = 3 \quad \cdots (1)$

$\frac{x}{y + 8} = \frac{1}{4} \implies 4x = y + 8 \implies 4x - y = 8 \quad \cdots (2)$

Subtract (1) from (2): $x = 5$.
From (1): $15 - y = 3 \implies y = 12$.

Answer: The fraction is $\frac{5}{12}$.

Verification: $\frac{5-1}{12} = \frac{4}{12} = \frac{1}{3}$ $\checkmark$. $\frac{5}{12+8} = \frac{5}{20} = \frac{1}{4}$ $\checkmark$.

Problem: A boat goes 30 km upstream and 44 km downstream in 10 hours. It can go 40 km upstream and 55 km downstream in 13 hours. Find the speeds of the boat in still water and the stream.

Solution:
Let boat speed $= x$ km/h, stream speed $= y$ km/h.
Upstream speed $= (x - y)$, downstream speed $= (x + y)$.

Let $u = \frac{1}{x - y}$ and $v = \frac{1}{x + y}$:

Multiply (1') by 4 and (2') by 3:
$120u + 176v = 40$
$120u + 165v = 39$

Subtract: $11v = 1 \implies v = \frac{1}{11}$.
From (1'): $30u + 4 = 10 \implies u = \frac{1}{5}$.

$x - y = 5$ and $x + y = 11$.

Adding: $2x = 16 \implies x = 8$. So $y = 3$.

Answer: Boat speed $= 8$ km/h, stream speed $= 3$ km/h.