What are the coordinates $\left(x,y\right)$ at which the graphs of $2x+3y=12$ and $2x-y=4$ intersect?

$\begin{array}{l}\text{A.}\left(3,2\right)\\ \text{B.}\left(2,3\right)\\ \text{C.}\left(1,-2\right)\\ \text{D.}\left(-3,6\right)\end{array}$

Equation of line in slope intercept form is expressed below.

$y=mx+c$

Where m is the slope and c is the intercept of y-axis.

Consider the first equation $2x+3y=12$.

Rewrite the equation in form of slope-intercept form.

Subtract $2x$from both the sides.

$\begin{array}{c}2x+3y-2x=12-2x\\ 3y=12-2x\end{array}$

Divide both sides by 3.

$\begin{array}{c}y=4-\frac{2}{3}x\\ y=-\frac{2}{3}x+4\end{array}$

Now, the equation is in the form $y=mx+c$.

Now, consider the second equation $2x-y=4$.

Add the term y on both the sides.

$\begin{array}{c}2x-y+y=4+y\\ 2x=4+y\\ y=2x-4\end{array}$

Now, the equation is in the form $y=mx+c$.

Plot the equations on the same plane and the point where both the equations intersect is the solution of the system of the equations.

The red line denotes the equation $2x+3y=12$ and blue line denotes the equation $2x-y=4$.

Therefore, the point of intersection is $\left(3,2\right)$.

The point of intersection is solution of system of equations if the point satisfies both the equation.

Substitute the point $\left(3,2\right)$in the equation $2x+3y=12$.

Substitute x as 3 and y as 2 and check whether right hand side is equal to left hand side of the equation.

$\begin{array}{c}2\left(3\right)+3\left(2\right)=12\\ 6+6=12\\ 12=12\end{array}$

Since, this is true so the point $\left(3,2\right)$satisfy the equation $2x+3y=12$.

Substitute the point $\left(3,2\right)$in the equation $2x-y=4$.

Substitute x as 3 and y as 2 and check whether right hand side is equal to left hand side of the equation.

$\begin{array}{c}2\left(3\right)-2=4\\ 6-2=4\\ 4=4\end{array}$

Since, this is true so the point $\left(3,2\right)$satisfy the equation $2x-y=4$.

Hence, the solution of the system of equations is $\left(3,2\right)$.