Question

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

Answer 1

Checking if $x=\sqrt{2}andy=4\sqrt{2}$ is the solution of $x-2y=4$or not:

We know that if the given value of the variable $x=\sqrt{2}andy=4\sqrt{2}$ satisfies the given equation $x-2y=4$then that value will be the solution of the equation.

Step1:Putting the value of $x=\sqrt{2}andy=4\sqrt{2}$in the given equation $x-2y=4$:

$$\begin{gathered} L.H.S \\ \Rightarrow \sqrt{2}-2\times 4\sqrt{2}=1\sqrt{2}-8\sqrt{2} \\ \Rightarrow \sqrt{2}-2\times 4\sqrt{2}=-7\sqrt{2} \\ \Rightarrow L.H.S=-7\sqrt{2} \end{gathered}$$

Since R.H.S $4$

Step2: Comparing Both sides:

We can see that $L.H.S\ne R.H.S$

Hence, $x=\sqrt{2}andy=4\sqrt{2}$ is not a solution of $x-2y=4$.