Chapter 1.1, Problem 47E

Here is another method to solve a system of linear equations: Solve one of the equations for one of the variables, and substitute the result into the other equations.Repeat this process until you run out of variables orequations. Consider the example discussed on page 2:

$\left|\begin{array}{c}x+2y+3z=39\\ x+3y+2z=34\\ 3x+2y+z=26\end{array}\right|$ .

We can solve the first equation for x: $x=39-2y-3z$ .

Then we substitute this equation into the other equations:

$\left|\begin{array}{c}\left(39-2y-3z\right)+3y+2z=34\\ 3\left(39-2y-3z\right)+2y+z=26\end{array}\right|$ .

We can simplify:

$\left|\begin{array}{c}y-z=-5\\ -4y-8z=-91\end{array}\right|$ .

Now, $y=z-5$ , so that $-4\left(z-5\right)-8z=-91$ , or

$-12z=-111$ .

We find that $z=\frac{111}{12}=9.25$ . Then

$y=z-5=4.25$

and

$x=39-2y-3z=2.75$ .

Explain why this method is essentially the same as the method discussed in this section; only the bookkeeping is different.