NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 3: Pair of Linear Equations in Two Variables - Exercise 3.3
Question 11

Question. 11

By the graphical method, decide consistency and solve:
(i). \(3x + y + 4 = 0\) and \(6x - 2y + 4 = 0\) (ii). \(x - 2y = 6\) and \(3x - 6y = 0\) (iii). \(x + y = 3\) and \(3x + 3y = 9\)

Answer:

(i). Consistent with a unique solution: \(x = -1\), \(y = -1\).

(ii). Inconsistent (parallel). No solution.

(iii). Consistent and dependent: infinitely many solutions (the line \(x + y = 3\)).

Detailed Answer with Explanation:

(i).

First equation: \(3x + y + 4 = 0\).

Move terms: \(y = -3x - 4\).

This is a straight line with slope \(-3\).

Second equation: \(6x - 2y + 4 = 0\).

Rearrange: \(-2y = -6x - 4\).

Divide by \(-2\): \(y = 3x + 2\).

This line has slope \(3\).

Since slopes are different, the two lines will intersect at one point.

To find the point, set the two expressions for \(y\) equal:

\(-3x - 4 = 3x + 2\).

Bring terms together: \(-3x - 3x = 2 + 4\).

\(-6x = 6\).

So, \(x = -1\).

Put \(x = -1\) in \(y = -3x - 4\):

\(y = -3(-1) - 4 = 3 - 4 = -1\).

Solution is \((x, y) = (-1, -1)\).

Since they intersect, the system is consistent and has a unique solution.

(ii).

First equation: \(x - 2y = 6\).

Second equation: \(3x - 6y = 0\).

Notice: if we multiply the first equation by 3, we get:

\(3x - 6y = 18\).

But the second equation is \(3x - 6y = 0\).

Left-hand sides are the same, but right-hand sides are different (18 vs 0).

This means the two lines are parallel and never meet.

So, the system is inconsistent (no solution).

(iii).

First equation: \(x + y = 3\).

Second equation: \(3x + 3y = 9\).

Divide the second equation by 3:

\(x + y = 3\).

This is exactly the same as the first equation.

So both equations represent the same line.

That means there are infinitely many solutions (all points on the line \(x + y = 3\)).

The system is consistent and dependent.

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 3: Pair of Linear Equations in Two Variables – Exercise 3.3 | Detailed Answers