Question

In the following exercise, determine whether each ordered pair is a solution to the system.

$\left\{\begin{array}{l}y<\frac{3}{2}x+3\\ \frac{3}{4}x-2y<5\end{array}\right.$

Short answer

Part (a). The ordered pair $\left(-4,-1\right)$is not the solution of the given system of inequalities.

Part (b). The ordered pair $\left(8,3\right)$is the solution to the given system of inequalities.

Step-by-step solution

Part (a) Step 1 . Substitute -4,-1 in the given system of inequalities.

Substituting $(-4,-1)$in the inequality $y<\frac{3}{2}x+3$:

role="math" localid="1644514660111" $$\begin{gathered} y<\frac{3}{2}x+3 \\ -1<\frac{3}{2}\times (-4)+3 \\ -1<-6+3 \\ -1<-3 \end{gathered}$$

This is not true.

There is no need to check further as $\left(-4,-1\right)$is not a solution.

Part (b) Step 1. Substitute 8,3 in the given system of inequalities.

Substituing $\left(8,3\right)$in the given inequality

$\frac{3}{4}x-2y<5$:

$$\begin{gathered} \frac{3}{4}x-2y<5 \\ \frac{3}{4}\times 8-2\times 3<5 \\ 6-6<5 \\ 0<5 \end{gathered}$$

This is true.

Substituting $\left(8,3\right)$in the inequality

role="math" localid="1644514974268" $y<\frac{3}{2}x+3$:

$$\begin{gathered} y<\frac{3}{2}x+3 \\ 3<\frac{3}{2}\times 8+3 \\ 3<15 \end{gathered}$$

This is also true.

Thus, the ordered pair$\left(8,3\right)$is the solution.