Q. 283 - Intermediate Algebra

Question

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

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

Step-by-Step Solution

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.

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

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

$$\begin{gathered} y<\frac{3}{2}x+3 \\ -1<\frac{3}{2}\times (-4)+3 \\ -1<-6+3 \\ -1<-3 \end{gathered}$$

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

$y<\frac{3}{2}x+3$:

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

This is also true.

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