In the following exercises, which order pair are solution to the given equation :

$4x-2y=8$

(a) $(3,2)$(b)$(1,4)$(c)$(0,-4)$

We are given an equation : $4x-2y=8$and three coordinates : $(3,2),(1,4)$and $(0,-4)$.

We need to find the solution of the given equation .

To check the first coordinate : $(3,2)$we need to put the value of $x$and $y$in the equation and see if L.H.S=R.H.S .

Here, we know that $x=3$and $y=2$.

From the equation : $4x-2y=8$the R.H.S is $8$, we need to solve the L.H.S :

$$\begin{gathered} 4x-2y=4\left(3\right)-2\left(2\right) \\ =12-4 \\ =8 \end{gathered}$$

Therefore, L.H.S=R.H.S .

So, it is a solution of given equation.

To check the second coordinate : $(1,4)$we need to put the value of $x$and $y$in the equation and see if L.H.S=R.H.S .

Here, we know that $x=1$and $y=4$.

From the equation $4x-2y=8$, the R.H.S is $8$, we need to solve the L.H.S :

$$\begin{gathered} 4x-2y=4\left(1\right)-2\left(4\right) \\ =4-8 \\ =-4 \end{gathered}$$

Therefore, L.H.S is not equal to R.H.S .

So, it is not the solution of the given equation.

To check the third coordinate : $(0,-4)$we need to put the value of $x$and $y$in the equation and see if L.H.S=R.H.S .

Here, we know that $x=0$and $y=-4$.

From the equation, $4x-2y=8$the R.H.S is $8$, we need to solve the L.H.S :

$$\begin{gathered} 4x-2y=4\left(0\right)-2(-4) \\ =8 \end{gathered}$$

Therefore, L.H.S=R.H.S .

So, it is the solution of the given equation.