Q. 337 - Intermediate Algebra

Question

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

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

Step-by-Step Solution

Answer

The solution for the system of linear equations is $(4,5)$. 

Step 1. Given information.

The system of linear equations,

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

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

$y=\frac{1}{2}x+3$

Now substituting this in another equation we get

$$\begin{gathered} 3x-2\left(\frac{1}{2}x+3\right)=2 \\ 3x-x-6=2 \\ 2x=8 \\ x=4 \end{gathered}$$

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

$$\begin{gathered} 3\left(4\right)-2y=2 \\ 12-2y=2 \\ 2y=10 \\ y=5 \end{gathered}$$

Step 4. Checking the result.

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

$$\begin{gathered} 3x-2y=2 \\ 3\left(4\right)-2\left(5\right)=2 \\ 12-10=2 \\ 2=2 \end{gathered}$$

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

$$\begin{gathered} y=\frac{1}{2}x+3 \\ 5=\frac{1}{2}\left(4\right)+3 \\ 5=2+3 \\ 5=5 \end{gathered}$$

Hence, this is true.