NCERT Solutions
Class 10 - Mathematics - Chapter 3: PAIR OF LINEAR EQUATIONS IN TWO VARIABLES - Exercise 3.1

Question 3

For a pair of linear equations in two variables

\[ a_1x + b_1y + c_1 = 0, \quad a_2x + b_2y + c_2 = 0, \]

we have:

• If \( \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} \): lines intersect; the pair is consistent with a unique solution.
• If \( \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} \): lines are parallel; the pair is inconsistent (no solution).
• If \( \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} \): lines are coincident; the pair is consistent with infinitely many solutions.

(i) \(3x + 2y = 5\) and \(2x - 3y = 7\)

Write them in standard form \(a_1x + b_1y + c_1 = 0\):

\(3x + 2y - 5 = 0\) and \(2x - 3y - 7 = 0\).

Here \(a_1 = 3, b_1 = 2, c_1 = -5;\) \(a_2 = 2, b_2 = -3, c_2 = -7\).

Ratios:

\[ \dfrac{a_1}{a_2} = \dfrac{3}{2}, \quad \dfrac{b_1}{b_2} = \dfrac{2}{-3} = -\dfrac{2}{3}. \]

Since \(\dfrac{3}{2} \neq -\dfrac{2}{3}\), the lines intersect at a point.

The system is consistent (unique solution).

(ii) \(2x - 3y = 8\) and \(4x - 6y = 9\)

Standard form:

\(2x - 3y - 8 = 0\) and \(4x - 6y - 9 = 0\).

\(a_1 = 2, b_1 = -3, c_1 = -8;\) \(a_2 = 4, b_2 = -6, c_2 = -9.\)

Ratios:

\[ \dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}, \quad \dfrac{b_1}{b_2} = \dfrac{-3}{-6} = \dfrac{1}{2}, \quad \dfrac{c_1}{c_2} = \dfrac{-8}{-9} = \dfrac{8}{9}. \]

We have \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}\) but this common value is not equal to \(\dfrac{c_1}{c_2}\). Therefore the lines are parallel, so

The system is inconsistent (no solution).

(iii) \(\dfrac{3}{2}x + \dfrac{5}{3}y = 7\) and \(9x - 10y = 14\)

First equation: multiply by 6 to clear denominators:

\[ 6 \left( \dfrac{3}{2}x + \dfrac{5}{3}y \right) = 6 \cdot 7 \Rightarrow 9x + 10y = 42. \]

So the pair becomes \(9x + 10y - 42 = 0\) and \(9x - 10y - 14 = 0\).

Here \(a_1 = 9, b_1 = 10, c_1 = -42;\) \(a_2 = 9, b_2 = -10, c_2 = -14.\)

Ratios:

\[ \dfrac{a_1}{a_2} = \dfrac{9}{9} = 1, \quad \dfrac{b_1}{b_2} = \dfrac{10}{-10} = -1. \]

Since \(1 \neq -1\), the lines intersect at one point.

The system is consistent (unique solution).

(iv) \(5x - 3y = 11\) and \(-10x + 6y = -22\)

Standard form:

\(5x - 3y - 11 = 0\); \(-10x + 6y + 22 = 0\).

Here \(a_1 = 5, b_1 = -3, c_1 = -11;\) \(a_2 = -10, b_2 = 6, c_2 = 22.\)

Ratios:

\[ \dfrac{a_1}{a_2} = \dfrac{5}{-10} = -\dfrac{1}{2}, \quad \dfrac{b_1}{b_2} = \dfrac{-3}{6} = -\dfrac{1}{2}, \quad \dfrac{c_1}{c_2} = \dfrac{-11}{22} = -\dfrac{1}{2}. \]

All three ratios are equal; hence the two equations represent the same line.

The system is consistent with infinitely many solutions.

(v) \(\dfrac{4}{3}x + 2y = 8\) and \(2x + 3y = 12\)

First equation: multiply by 3:

\[ 4x + 6y = 24 \Rightarrow 4x + 6y - 24 = 0. \]

Second equation: \(2x + 3y - 12 = 0\).

Thus \(a_1 = 4, b_1 = 6, c_1 = -24;\) \(a_2 = 2, b_2 = 3, c_2 = -12.\)

Ratios:

\[ \dfrac{a_1}{a_2} = \dfrac{4}{2} = 2, \quad \dfrac{b_1}{b_2} = \dfrac{6}{3} = 2, \quad \dfrac{c_1}{c_2} = \dfrac{-24}{-12} = 2. \]

All three are equal, so the lines are coincident.

The system is consistent with infinitely many solutions.