NCERT Solutions
Class 10 - Mathematics - Chapter 3: PAIR OF LINEAR EQUATIONS IN TWO VARIABLES - Exercise 3.1
Question 4

Question. 4

Which of the following pairs of linear equations are consistent or inconsistent? If consistent, obtain the solution 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) \(2x - 2y - 2 = 0\); \(4x - 4y - 5 = 0\)

Answer:

Use the criteria based on \( \dfrac{a_1}{a_2}, \dfrac{b_1}{b_2}, \dfrac{c_1}{c_2} \).

(i) \(x + y = 5\) and \(2x + 2y = 10\)

Second equation simplifies by dividing through by 2:

\[ 2x + 2y = 10 \Rightarrow x + y = 5. \]

Thus both equations represent the same line. Here

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

All ratios are equal, so the pair is consistent with infinitely many solutions.

Graphical solution: Both equations give the same line \(x + y = 5\). Every point on this line, such as \((0,5), (5,0), (2,3)\), is a solution.

(ii) \(x - y = 8\) and \(3x - 3y = 16\)

Write in standard form: \(x - y - 8 = 0\); \(3x - 3y - 16 = 0\).

Ratios:

\[ \dfrac{a_1}{a_2} = \dfrac{1}{3}, \quad \dfrac{b_1}{b_2} = \dfrac{-1}{-3} = \dfrac{1}{3}, \quad \dfrac{c_1}{c_2} = \dfrac{-8}{-16} = \dfrac{1}{2}. \]

We have \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}\); therefore the lines are parallel and distinct.

The pair is inconsistent (no solution).

(iii) \(2x + y - 6 = 0\) and \(4x - 2y - 4 = 0\)

Rewrite the second equation:

\[ 4x - 2y - 4 = 0. \]

Here \(a_1 = 2, b_1 = 1, c_1 = -6;\) \(a_2 = 4, b_2 = -2, c_2 = -4.\)

Ratios:

\[ \dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}, \quad \dfrac{b_1}{b_2} = \dfrac{1}{-2} = -\dfrac{1}{2}. \]

Since \(\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}\), the lines intersect at a unique point, so the pair is consistent.

Solving algebraically for the point of intersection (which will match the graphical solution):

From \(2x + y - 6 = 0\), \(y = 6 - 2x\).

Substitute into \(4x - 2y - 4 = 0\):

\[ 4x - 2(6 - 2x) - 4 = 0 \Rightarrow 4x - 12 + 4x - 4 = 0 \Rightarrow 8x - 16 = 0 \Rightarrow x = 2. \]

Then \(y = 6 - 2x = 6 - 4 = 2\).

Graphically, the lines intersect at \((2,2)\).

(iv) \(2x - 2y - 2 = 0\) and \(4x - 4y - 5 = 0\)

Here \(a_1 = 2, b_1 = -2, c_1 = -2;\) \(a_2 = 4, b_2 = -4, c_2 = -5.\)

Ratios:

\[ \dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}, \quad \dfrac{b_1}{b_2} = \dfrac{-2}{-4} = \dfrac{1}{2}, \quad \dfrac{c_1}{c_2} = \dfrac{-2}{-5} = \dfrac{2}{5}. \]

Since \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}\), the lines are parallel and distinct.

The pair is inconsistent (no solution).

Detailed Answer with Explanation:

Use the ratio test to label each pair as intersecting, parallel, or coincident. For intersecting pairs, either plot the lines or solve one equation for a variable and substitute to find the intersection point.

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