Question

In the following exercises, determine if the following points are solutions to the given system of equations.

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

localid="1644921239282" $$\begin{gathered} \left(\mathrm{a}\right)\left(\frac{8}{7},\frac{6}{7}\right) \\ \left(\mathrm{b}\right)\left(1,\frac{3}{4}\right) \end{gathered}$$

Short answer

For (a) The ordered pair, $\left(\frac{8}{7},\frac{6}{7}\right)$is a solution of the system of linear equations.

For (b) The ordered pair$\left(1,\frac{3}{4}\right)$is not a solution of the system of linear equations.

Step-by-step solution

Part (a) Step 1. Given information.

The given the system of equations is,

$\left\{\begin{array}{l}x+y=2\\ y=\frac{3}{4}x\end{array}\right.$and the ordered pair is$(\frac{8}{7},\frac{6}{7})$

Part (a) Step 2. Checking the values.

Substituting the values, we get,

$\begin{array}{rcl}x+y& =& 2\\ \frac{8}{7}+\frac{6}{7}& =& 2\\ \frac{14}{7}& =& 2\\ 2& =& 2\end{array}$

Part (a) Step 3. Checking the values.

Substituting for another equation, we get,

$\begin{array}{rcl}y& =& \frac{3}{4}\\ \frac{6}{7}& =& \frac{3}{4}\left(\frac{8}{7}\right)\\ 6·4& =& 3·8\\ 24& =& 24\end{array}$

Part (b) Step 1: Given information

Given the system of equation$\left\{\begin{array}{l}x+y=2\\ y=\frac{3}{4}x\end{array}\right.$and ordered pair$(1,\frac{3}{4})$

Part (b) Step 2: Checking the value

Substituting the value, we get

$$\begin{gathered} 1+\frac{3}{4}=2 \\ \frac{7}{4}=2 \end{gathered}$$

It is not true

Part (b) Step 3: Checking the value

Substituting, we get

$$\begin{gathered} \frac{3}{4}=\frac{3}{4}\left(1\right) \\ \frac{3}{4}=\frac{3}{4} \end{gathered}$$

It is true