Solving a system of equations with fewer equations than variables

Question

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?

Answer 1

Are you familiar with Gaussian elimination for non-square linear systems? You want to construct the augmented matrix

$$ \begin{bmatrix} 3 & 4 & 7 & -3 & | & 3 \\ 0 & 5 & 3 & 0 & | & 40 \end{bmatrix}$$

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.

Answer 2

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.

Your solution won't be a finite set of points; to have that, you need an independent equation for each variable you have (i.e., no equation is a linear combination of other equations in that matrix). You'll get to that soon enough in your course. Rather, your equation will be an infinite set of points. We describe them by the final reduced row-echelon form matrix that Christopher Wong alluded to.

Answer 3

• Let $z=r.$ Then,$$5y + 3r = 40\iff 5y = 40 - 3r\iff y =\frac{40 - 3r}5.$$

• Substituting $y$ and $z$ in the first equation, we get: $$3x + 4·\frac{40 - 3r}5 + 7r - 3t = 3.$$

• Let $t=p.$ The previous equation becomes: $$3x = 3 + 3p - 32 -\frac{12r}5 - 7r\iff x =\frac{3p - \frac{47r}5 - 29}3.$$

So, we have the solution as a function of the parameters $r$ and $p$ ($r,p ∈ ℝ$).