Chapter 7.3, Problem 44E

To determine

To calculate:Solve the system of linear equations and check any solutions algebraically.

  $\begin{array}{l}2x+3y+3z=7\\ 4x+18y+15z=44\end{array}$

Answer to Problem 44E

The solutions of the equations are $x=\frac{-3z-2}{8},y=\frac{-3z+10}{4}$ in terms of z.

Explanation of Solution

Given information:

The given pair of linear equations are $\begin{array}{l}2x+3y+3z=7\\ 4x+18y+15z=44\end{array}$

Calculation:

These equations are nonsquare and are of 3 variables. Let us write x and y in terms of z with the help of two equations.

From equation 1, we get,

  $x=\frac{7-3y-3z}{2}$

Substitute x in equation 2 we get,

  $\begin{array}{l}4\left(\frac{7-3y-3z}{2}\right)+18y+15z=44\\ 14-6y-6z+18y+15z=44\\ 12y+9z+14=44\\ 12y+9z=30\\ 4y+3z=10\end{array}$

Isolate y from the above equation we get,

  $y=\frac{10-3z}{4}$

Substitute this y value in the x equation to get it in terms of z.

  $\begin{array}{l}x=\frac{7-3y-3z}{2}\\ x=\frac{7-3\left(\frac{10-3z}{4}\right)-3z}{2}\\ x=\frac{7-\left(\frac{30-9z}{4}\right)-3z}{2}\\ x=\frac{28-30+9z-12z}{8}\\ x=\frac{-3z-2}{8}\end{array}$

Now, the solutions to the system of equations are above. By substituting different values of z we get different solutions.