# Solving a system of equations with less equations than variables

For my discrete math/linear algebra class, one of our homework problems reads as follows:

Use backsubstitution to solve the following system of equations and obtain the general
solution.
3x + 4y + 7z - 3t = 3
0x + 5y + 3z = 40


Now, I've always been taught that you need as many equations as variables to solve a system of equations. Is there a way to do this where you don't need as many equations as variables? Or would I be correct in making x and t both equal 1 so that they cancel out and then solve the system that way?

• You just made my life a hell of a lot simpler. – Nick Dell'osa Oct 28 '14 at 23:18
• This system is 'underdetermined'. There is an infinite family of solutions which in this case span a plane. You want the general relationship which these solutions satisfy. – JCW Oct 28 '14 at 23:19

$$\begin{bmatrix} 3 & 4 & 7 & -3 & | & 3 \\ 0 & 5 & 3 & 0 & | & 40 \end{bmatrix}$$
and perform back-substitution to obtain a reduced row-echelon form. You do not want to arbitrarily set $x = t = 1$, because then you are not obtaining every solution.
You have $5y+3z=40.$ Express that as $3z=40-5y \implies z=\frac{40}{3}-\frac{5}{3}y$, and then plug that value of $z$ into your first equation. That's what back substitution essentially is.