How do I deal with infinitely many solutions on HW 02 #3

audrey

Here's the system I have in problem #3 of HW 02: Elimination

$$
\begin{align}
-6x + 8y - 30z &= -46 \\
x - 6y + 19z &= 31 \\
-x + 2y - 7z &= -11.
\end{align}
$$

I'm pretty sure there are infinitely many solutions but I'm not sure how to enter that. Does anyone have some advice?

Student16

Let's assume you got to this system:

$$
\begin{align}
-6x + 8y - 30z &= -46 \\
4y + 12z &= 20 \\
0 &= 0
\end{align}
$$

Starting from the bottom, you set $z = z$. Then you solve y in the second equation in terms of $ z $ i.e $y = 3z - 5$. Finally using the two previous solutions, you solve for x in the first equation and that's it!

Your answer should look something like this

$$
\begin{align}
x &= 1 - z \\
y &= 5 -3z \\
z &= z
\end{align}
$$

Student36

When dealing with infinitely many solutions, one of the variables will be set equal to itself. This is interpreted as allowing the variable to be equal to any number. For this problem, let $z=z$.
Then, solve the system of equations for x and y in terms of z. Doing this allows x and y to adjust to their correct solution values whenever z changes. Here is an example of what the solution set would look like, but it is not the answer to the problem above:
$$
\begin{align}
x &= 3z+2 \\
y &= 5z-4 \\
z &= z
\end{align}
$$
Another way to answer this would be in (x,y,z) format, which is depicted below:
$$(x,y,z)=(3z+2,5z-4,z)$$