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.

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

(a)$\left(\frac{8}{7},\frac{6}{7}\right)$

(b)$\left(1,\frac{3}{4}\right)$

Short answer

Part (a) An ordered pair $\left(\frac{8}{7},\frac{6}{7}\right)$is a solution .

Part (b) An ordered pair $\left(1,\frac{3}{4}\right)$is not a solution.

Step-by-step solution

Part (a) Step 1. Given information

Consider the system of equations.

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

Part (a) Step 2. Determine whether an ordered pair 87,67 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 $\frac{8}{7}$for $x$and $\frac{6}{7}$for $y$into $x+y=2$.

$\begin{array}{rcl}\frac{8}{7}+\frac{6}{7}& =& 2\\ \frac{8+6}{7}& =& 2\\ \frac{14}{7}& =& 2\\ 2& =& 2\left(\text{True}\right)\end{array}$

Substitute $\begin{array}{rcl}& & \frac{8}{7}\end{array}$for $x$and role="math" localid="1647513827253" $\begin{array}{rcl}& & \frac{6}{7}\end{array}$for $y$into $y=\frac{3}{4}x$.

$\begin{array}{rcl}\frac{6}{7}& =& \frac{3}{4}·\frac{8}{7}\\ \frac{6}{7}& =& 3·\frac{2}{7}\\ \frac{6}{7}& =& \frac{6}{7}\left(\text{True}\right)\end{array}$

Conclude that the ordered pair $\left(\frac{8}{7},\frac{6}{7}\right)$made both equations true.

Thus, $\left(\frac{8}{7},\frac{6}{7}\right)$is a solution to the given system.

Part (b) Step 1. Determine whether an ordered pair 1,34 is a solution to the given system of two equations.

Substitute $1$for $x$and localid="1647514284269" $\frac{3}{4}$for localid="1647514277458" $y$into $x+y=2$.

$\begin{array}{rcl}1+\frac{3}{4}& =& 2\\ \frac{4+3}{4}& =& 2\\ \frac{7}{4}& =& 2\left(\text{False}\right)\end{array}$

Substitute $1$for $x$and $\frac{3}{4}$for localid="1647514271923" $y$into localid="1647514115243" $y=\frac{3}{4}x$.

localid="1647514827572" $$\begin{gathered} \frac{3}{4}=\frac{3}{4}·1 \\ \frac{3}{4}=\frac{3}{4}\left(\begin{array}{rcl}& & \text{True}\end{array}\right) \end{gathered}$$

Conclude that the ordered pair $\left(1,\frac{3}{4}\right)$made one equation true and other equation false.

Thus, $\left(1,\frac{3}{4}\right)$is not a solution to the given system.

Hence, $\left(\frac{8}{7},\frac{6}{7}\right)$is a solution and $\left(1,\frac{3}{4}\right)$is not a solution to the given system of equations.