Chapter 6.2, Problem 16PPE

To determine

To solve: The system of linear equations using substitution.

Answer to Problem 16PPE

The solution is x = $\frac{31}{4}$ and y = − $\frac{1}{8}$ .

Explanation of Solution

Given information:

The equations $3x+2y=23$ and $\frac{1}{2}x-4=y$

Calculation:

Substitute $\frac{1}{2}x-4$ for y in equation $3x+2y=23$

  $3x+2\left(\frac{1}{2}x-4\right)=23$

Now simplify above equation.

  $\begin{array}{l}3x+2\left(\frac{1}{2}x-4\right)=23\\ \Rightarrow 3x+x-8=23\\ \Rightarrow 4x-8=23\\ \Rightarrow 4x=31\\ \Rightarrow x=\frac{31}{4}\end{array}$

Use above value ofx in equation $\frac{1}{2}x-4=y$

  $\begin{array}{c}y=\frac{1}{2}\left(\frac{31}{4}\right)-4\\ =\frac{31}{8}-4\\ =\frac{31-32}{8}\\ =-\frac{1}{8}\end{array}$

Therefore the solution is x = $\frac{31}{4}$ and y = − $\frac{1}{8}$ .

Check

  $\begin{array}{l}y=\frac{1}{2}x-43x+2y=23\\ -\frac{1}{8}=\frac{1}{2}\left(\frac{31}{4}\right)-43\left(\frac{31}{4}\right)+2\left(-\frac{1}{8}\right)=23\\ -\frac{1}{8}=\frac{31}{8}-4\frac{93}{4}-\frac{1}{4}=23\\ -\frac{1}{8}=\frac{31-32}{8}\frac{92}{4}=23\\ -\frac{1}{8}=\frac{-1}{8}23=23\end{array}$