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

$3x-2y=12$

(a)$(4,0)$(b)$(2,-3)$(c)$(1,6)$

We are given an equation : $3x-2y=12$and three coordinates : $(4,0),(2,-3)$and $(1,6)$.

We need to find the solution of the given equation.

To check the first coordinate : $(4,0)$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 $x=4$and $y=0$.

From the equation : $3x-2y=12$the R.H.S is $12$, we need to solve L.H.S :

$$\begin{gathered} 3x-2y=3\left(4\right)-2\left(0\right) \\ =12 \end{gathered}$$

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

So, it is a solution of the given equation.

To check the second coordinate : $(2,-3)$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 $x=2$and $y=-3$.

From the equation : $3x-2y=12$the R.H.S is $12$, we need to solve L.H.S :

$$\begin{gathered} 3x-2y=3\left(2\right)-2(-3) \\ =6+6 \\ =12 \end{gathered}$$

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

So, it is the solution of the given equation.

To check the third coordinate : $(1,6)$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 $y=6$.

From the equation : $3x-2y=12$the R.H.S is $12$, We need to solve L.H.S :

$$\begin{gathered} 3x-2y=3\left(1\right)-2\left(6\right) \\ =3-12 \\ =-9 \end{gathered}$$

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

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