Q. 339 - Intermediate Algebra

Question

In the following exercises, solve the systems of equations by substitution. 

$\left\{\begin{array}{l}y=-2x+7\\ y=\frac{2}{3}x-1\end{array}\right.$

Step-by-Step Solution

Answer

The solution for the system of linear equations is $\left(3,1\right)$.

Step 1. Given information.

The system of linear equations, 

$\left\{\begin{array}{l}y=-2x+7\\ y=\frac{2}{3}x-1\end{array}\right.$

Step 2. Substituting the value of y from one equation into another equation.

$y=-2x+7$
Now substituting this in another equation we get 

$$\begin{gathered} -2x+7=\frac{2}{3}x-1 \\ -2x+8=\frac{2}{3}x \\ -6x+24=2x \\ 8x=24 \\ x=3 \end{gathered}$$

Step 3. Substituting the value of x in any equation.

$$\begin{gathered} y=-2\left(3\right)+7 \\ y=-6+7 \\ y=1 \end{gathered}$$

Step 4. Checking the result.

Substituting the value of $x$ and $y$ in the first equation then

$$\begin{gathered} y=-2x+7 \\ 1=-2\left(3\right)+7 \\ 1=-6+7 \\ 1=1 \end{gathered}$$

Substituting the value of $x$ and $y$ in the second equation then

$$\begin{gathered} y=\frac{2}{3}x-1 \\ 1=\frac{2}{3}\left(3\right)-1 \\ 1=2-1 \\ 1=1 \end{gathered}$$

Hence, this is true.