Simultaneous Linear Equations — Assertion-Reason Type Questions

Assertion (A): A solution of x - y = 1, 2x + y = $\dfrac{7}{2}$ is x = $\dfrac{3}{2}$, y = $\dfrac{1}{2}$.

Reason (R): One of the methods of solving a pair of linear equations is elimination method.

1. Assertion (A) is true, Reason (R) is false.

2. Assertion (A) is false, Reason (R) is true.

3. Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Answer

One of the methods of solving a pair of linear equations is elimination method.

∴ Reason (R) is true.

Given, x - y = 1 .................(1)

2x + y = $\dfrac{7}{2}$ ..................(2)

Adding equation (1) and (2) we get

$\Rightarrow (x - y) + (2x + y) = 1 + \dfrac{7}{2}\\[1em] \Rightarrow x - y + 2x + y = \dfrac{2}{2} + \dfrac{7}{2}\\[1em] \Rightarrow 3x = \dfrac{9}{2}\\[1em] \Rightarrow x = \dfrac{9}{2 \times 3}\\[1em] \Rightarrow x = \dfrac{3}{2}$

Substituting the value of x in equation (1), we get :

⇒ x - y = 1

$\Rightarrow \dfrac{3}{2} - y = 1\\[1em] \Rightarrow \dfrac{3}{2} - 1 = y\\[1em] \Rightarrow \dfrac{3 - 2}{2} = y\\[1em] \Rightarrow y = \dfrac{1}{2}.$

So, x = $\dfrac{3}{2}$, y = $\dfrac{1}{2}$

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason (or explanation) for Assertion (A).

Hence, option 3 is the correct option.

Assertion (A): Solving $\sqrt{2}x - \sqrt{3}y$ = 0, $\sqrt{3}x + \sqrt{2}y$ = 5 yields x = $\sqrt{3}$, y = $\sqrt{2}$.

Reason (R): We can use cross-multiplication method to solve a pair of linear equations.

1. Assertion (A) is true, Reason (R) is false.

2. Assertion (A) is false, Reason (R) is true.

3. Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Answer

We can use cross-multiplication method to solve a pair of linear equations.

∴ Reason (R) is true.

Given, equations can be written as :

$\sqrt{2}x - \sqrt{3}y - 0$ = 0 .................(1)

$\sqrt{3}x + \sqrt{2}y$ - 5 = 0................(2)

By cross multiplication method,

$\therefore \dfrac{x}{-\sqrt{3} \times (-5) - \sqrt{2} \times 0} = \dfrac{y}{0 \times \sqrt{3} - (-5) \times \sqrt{2}} = \dfrac{1}{\sqrt{2} \times \sqrt{2} - \sqrt{3} \times (-\sqrt{3})}\\[1em] \Rightarrow \dfrac{x}{5\sqrt{3} - 0} = \dfrac{y}{0 + 5 \sqrt{2}} = \dfrac{1}{2 + 3}\\[1em] \Rightarrow \dfrac{x}{5\sqrt{3}} = \dfrac{y}{5 \sqrt{2}} = \dfrac{1}{5}\\[1em] \Rightarrow \dfrac{x}{5\sqrt{3}} = \dfrac{1}{5} \text{ and } \dfrac{y}{5 \sqrt{2}} = \dfrac{1}{5} \\[1em] \Rightarrow x = \dfrac{5\sqrt{3}}{5} \text{ and } y = \dfrac{5\sqrt{2}}{5}\\[1em] \Rightarrow x = \sqrt{3} \text{ and } y = \sqrt{2}.$

So, the solution are x = $\sqrt{3}$, y = $\sqrt{2}$

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.