Chapter 9, Problem 40CR

To determine

To solve the system of equations algebraically.

Answer to Problem 40CR

There are infinitely many solutions.

Explanation of Solution

Given information:

The given equations are:

  $\begin{array}{l}8\text{y}=6-2\text{x}\\ \text{x}=3-4\text{y}\end{array}$

Calculation :

Solve the first equation for $\text{y}$ :

  $\begin{array}{l}8\text{y}=6-2\text{x}\\ \text{y}=\frac{3}{4}-\frac{1}{4}\text{x }\left[\text{Divide both sides by 8}\right]\end{array}$

Substitute $\text{y}$ in second equation:

  $\begin{array}{l}\text{x}=3-4\text{y}\\ \text{x}=3-4\left(\frac{3}{4}-\frac{1}{4}\text{x}\right)\\ x=3-3+x\\ x-x=3-3\\ 0=0\end{array}$

Since $0=0$ is a true statement.

There are infinitely many solutions.