91Ó°ÊÓ

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 $8x-3y=-3$.

Rewrite the equation in form of slope-intercept form.

Subtract both sides by $8x$.

$\begin{array}{c}8x−3y−8x=−3−8x\\ −3y=−3−8x\end{array}$

Next, divide both sides by $-3$and rearrange the terms.

Now, the equation is in the form $y=mx+c$. Here slope m of the line is $\frac{8}{3}$ and intercept of y-axis c is 1.

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

Rewrite the equation in form of slope-intercept form.

Subtract both sides by $4x$.

Next, divide both sides by $-2$and rearrange the terms.

Now, the equation is in the form $y=mx+c$. Here slope m of the line is 2 and intercept of y-axis c is 2.

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 $8x-3y=-3$ and blue line denotes the equation $4x-2y=-4$.

Therefore, the point of intersection is $\left(\frac{3}{2},5\right)$.

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

Substitute the point $\left(\frac{3}{2},5\right)$in the equation $8x-3y=-3$.

Substitute x as $\frac{3}{2}$ and y as 5 and check whether right hand side is equal to left hand side of the equation.

Since, this is true so the point $\left(\frac{3}{2},5\right)$satisfy the equation $8x-3y=-3$.

Substitute the point $\left(\frac{3}{2},5\right)$ in the equation $4x-2y=-4$.

Substitute x as $\frac{3}{2}$ and y as $5$ and check whether right hand side is equal to left hand side of the equation.

Since, this is true so the point $\left(\frac{3}{2},5\right)$satisfy the equation $4x-2y=-4$.

Hence, the solution of the system of equations is $\left(\frac{3}{2},5\right)$.