Question

Check which of the following are solutions of the equation $x-2y=4$

• A. $(0,2)$
• B. $(2,0)$
• C. $(4,0)$
• D. $(\sqrt{2},4\sqrt{2})$

Answer 1

The correct option is C

$(4,0)$

The explanation for the correct option.

Option(C) $\left(4,0\right)$

Check whether $\left(4,0\right)$ is a solution of the equation $x-2y=4$ or not.

Substitute $x=4$ and $y=0$ in the equation $x-2y=4$ and check whether LHS is equal to RHS or not:

$\begin{array}{rcl}\mathrm{LHS}& =& x-2y\\ & =& 4-2\left(0\right)\\ & =& 4\\ & =& \mathrm{RHS}\end{array}$

Thus, $\left(4,0\right)$ is a solution of the equation $x-2y=4$.

Hence, the option $C$ is correct.

The explanation for the incorrect options.

Option(A) $\left(0,2\right)$

Check whether $\left(0,2\right)$ is a solution of the equation $x-2y=4$ or not.

Substitute $x=0$ and $y=2$ in the equation $x-2y=4$ and check whether LHS is equal to RHS or not:

$\begin{array}{rcl}\mathrm{LHS}& =& x-2y\\ & =& 0-2\left(2\right)\\ & =& -4\\ & \ne & \mathrm{RHS}\end{array}$

Thus, $\left(0,2\right)$ is not a solution of the equation $x-2y=4$.

Option(B) $\left(2,0\right)$

Check whether $\left(2,0\right)$ is a solution of the equation $x-2y=4$ or not.

Substitute $x=2$ and $y=0$ in the equation $x-2y=4$ and check whether LHS is equal to RHS or not:

$\begin{array}{rcl}\mathrm{LHS}& =& x-2y\\ & =& 2-2\left(0\right)\\ & =& 2\\ & \ne & \mathrm{RHS}\end{array}$

Thus, $\left(2,0\right)$ is not a solution of the equation $x-2y=4$.

Option(D) $\left(\sqrt{2},4\sqrt{2}\right)$

Check whether $\left(\sqrt{2},4\sqrt{2}\right)$ is a solution of the equation $x-2y=4$ or not.

Substitute $x=\sqrt{2}$ and $y=4\sqrt{2}$ in the equation $x-2y=4$ and check whether LHS is equal to RHS or not:

$\begin{array}{rcl}\mathrm{LHS}& =& x-2y\\ & =& \sqrt{2}-2\left(4\sqrt{2}\right)\\ & =& \sqrt{2}-8\sqrt{2}\\ & =& \sqrt{2}\left(1-8\right)\\ & =& -7\sqrt{2}\\ & \ne & \mathrm{RHS}\end{array}$

Thus, $\left(\sqrt{2},4\sqrt{2}\right)$ is not a solution of the equation $x-2y=4$.

Hence, the options $A,B,\mathrm{and}D$ are incorrect.

Hence, Option (C) is correct option.