Chapter 7.3, Problem 38E

To determine

To solve: The system of linear equations.

Answer to Problem 38E

Every ordered triple of the form $(\frac{1}{2}a-\frac{3}{2},a,\frac{10}{3}-\frac{4}{3}a)$ is a solution of the system.

Explanation of Solution

Given information: The system of linear equations,

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

Calculation:

Rewriting the system,

  $\begin{array}{l}2x+3y+3z=7\to (1)\\ 4x+18y+15z=44\to (2)\end{array}$

There is more number of unknowns than the number of equations. So, the system has infinitely many solutions.

A new second equation can be written by,

  $eqn(2)\to 2eqn(1)-eqn(2)$

  $\begin{array}{l}2x+3y+3z=7\to (1)\\ 12y+9z=30\to (2)\end{array}$

Another new second equation can be written by,

  $eqn(2)\to \frac{eqn(2)}{3}$

  $\begin{array}{l}2x+3y+3z=7\to (1)\\ 4y+3z=10\to (2)\end{array}$

From (2) solve for $z$ ,

  $\begin{array}{l}4y+3z=10\\ 3z=10-4y\\ z=\frac{10-4y}{3}\\ z=\frac{10}{3}-\frac{4y}{3}\end{array}$

Now substitute $z$ in (1)

  $\begin{array}{l}2x+3y+3(\frac{10-4y}{3})=7\\ 2x+3y+10-4y=7\\ 2x-y=7-10\\ 2x=y-3\\ x=\frac{y-3}{2}\\ x=\frac{y}{2}-\frac{3}{2}\end{array}$

Let $y=a$ , where $\text{'}a\text{'}$ is a real number.

Then, the solution is,

  $\begin{array}{l}x=\frac{a}{2}-\frac{3}{2}\\ y=a\\ z=\frac{10}{3}-\frac{4a}{3}\end{array}$

So, every ordered triple of the form

  $(\frac{a}{2}-\frac{3}{2},a,\frac{10}{3}-\frac{4a}{3})$ is a solution of the system.