Question

Determine whether an ordered pair is a solution of a system of equations. In the following exercises, determine if the following points are solutions to the given system of equations.

$\left\{\begin{array}{l}x+3y=9\\ y=\frac{2}{3}x-2\end{array}\right.$

(a)$(-6,5)$

(b)$\left(5,\frac{4}{3}\right)$

Short answer

Part (a) An ordered pair $(-6,5)$is not a solution.

Part (b) An ordered pair $\left(5,\frac{4}{3}\right)$is a solution.

Step-by-step solution

Part (a) Step 1. Given information

Consider the system of equations.

$\left\{\begin{array}{l}x+3y=9\\ y=\frac{2}{3}x-2\end{array}\right.$

Part (a) Step 2. Determine whether an ordered pair (-6,5) is a solution to the given system of two equations.

Substitute the values of the variables into each equation to determine whether an ordered pair is a solution to a given system of two equations. It is a solution to the system if the ordered pair makes both equations true.

Substitute $-6$for $x$and $5$for $y$into $x+3y=9$.

$\begin{array}{rcl}-6+3\left(5\right)& =& 9\\ -6+15& =& 9\\ 9& =& 9\left(\text{True}\right)\end{array}$

Substitute $-6$for $x$and $5$for $y$into $y=\frac{2}{3}x-2$.

$\begin{array}{rcl}5& =& \frac{2}{3}(-6)-2\\ 5& =& 2(-3)-2\\ 5& =& -6-2\\ 5& =& -8\text{(False})\end{array}$

Conclude that the ordered pair $(-6,5)$made one equation true and the other false.

Thus, $(-6,5)$is not a solution to the given system.

Part (b) Step 1. Determine whether an ordered pair 5,43 is a solution to the given system of two equations.

Substitute $5$for $x$and $\frac{4}{3}$for $y$into $x+3y=9$.

$\begin{array}{rcl}5+3\left(\frac{4}{3}\right)& =& 9\\ 5+4& =& 9\\ 9& =& 9\left(\text{True}\right)\end{array}$

Substitute $5$for $x$and $\frac{4}{3}$for $y$into $y=\frac{2}{3}x-2$.

$\begin{array}{rcl}\frac{4}{3}& =& \frac{2}{3}\left(5\right)-2\\ \frac{4}{3}& =& \frac{10}{3}-2\\ \frac{4}{3}& =& \frac{10-6}{3}\\ \frac{4}{3}& =& \frac{4}{3}\left(\text{True}\right)\end{array}$

Conclude that the ordered pair $\left(5,\frac{4}{3}\right)$made both equations true.

Thus, $\left(5,\frac{4}{3}\right)$is a solution to the given system.

Hence, $\left(5,\frac{4}{3}\right)$is a solution and $(-6,5)$is not a solution to the given system of equations.