NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 3: Pair of Linear Equations in Two Variables - Exercise 3.1
Question 10

Question.  10

10. A pair of linear equations which has a unique solution \(x = 2\), \(y = -3\) is

(A)

\(x + y = -1\)
\(2x - 3y = -5\)

(B)

\(2x + 5y = -11\)
\(4x + 10y = -22\)

(C)

\(2x - y = 1\)
\(3x + 2y = 0\)

(D)

\(x - 4y - 14 = 0\)
\(5x - y - 13 = 0\)

Detailed Answer with Explanation:

Step 1: Check option (A).

Equation 1: \(x + y = -1\). Substituting \(x = 2, y = -3\):

\(2 + (-3) = -1\) ✓ satisfied.

Equation 2: \(2x - 3y = -5\). Substituting:

\(2(2) - 3(-3) = 4 + 9 = 13 \ne -5\).

So, option (A) is not correct.

Step 2: Check option (B).

Equation 1: \(2x + 5y = -11\). Substituting:

\(2(2) + 5(-3) = 4 - 15 = -11\) ✓ satisfied.

Equation 2: \(4x + 10y = -22\). Substituting:

\(4(2) + 10(-3) = 8 - 30 = -22\) ✓ satisfied.

But notice: the second equation is exactly 2 × (first equation). So both equations represent the same line ⇒ infinitely many solutions, not a unique one.

Step 3: Check option (C).

Equation 1: \(2x - y = 1\). Substituting:

\(2(2) - (-3) = 4 + 3 = 7 \ne 1\).

So, option (C) is not correct.

Step 4: Check option (D).

Equation 1: \(x - 4y - 14 = 0\). Substituting:

\(2 - 4(-3) - 14 = 2 + 12 - 14 = 0\) ✓ satisfied.

Equation 2: \(5x - y - 13 = 0\). Substituting:

\(5(2) - (-3) - 13 = 10 + 3 - 13 = 0\) ✓ satisfied.

Both equations are satisfied and they are not multiples of each other ⇒ two distinct lines intersecting at exactly one point.

Conclusion. Option (D) gives a pair of equations with a unique solution at \((2, -3)\).

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 3: Pair of Linear Equations in Two Variables – Exercise 3.1 | Detailed Answers