# Thread: Linear Algebra System Basic Proof

1. ## Linear Algebra System Basic Proof

Given a system of the form

mx + y = c
nx + y = d

where c, d, m and n are constants:

A. Show that the system will have a unique solution if m does not equal n.
B. If m = n, show the system will only be consistent if c = d.
C. Give a geometric interpretation of parts A and B

I have no idea where to begin with that. I tried using elementary row operations on the augmented matrix, but I couldn't gather anything from it and I fear I may have done it incorrectly.

Any help would be greatly appreciated.

2. Originally Posted by WTFsandwich
Given a system of the form

mx + y = c
nx + y = d

where c, d, m and n are constants:

A. Show that the system will have a unique solution if m does not equal n.
B. If m = n, show the system will only be consistent if c = d.
C. Give a geometric interpretation of parts A and B

I have no idea where to begin with that. I tried using elementary row operations on the augmented matrix, but I couldn't gather anything from it and I fear I may have done it incorrectly.

Any help would be greatly appreciated.
A. These are linear functions.

You should know that you can express a linear function in the form $\displaystyle y = ax + b$, with $\displaystyle a$ as the gradient and $\displaystyle b$ as the $\displaystyle y$ intercept.

Rearranging the equations gives:

$\displaystyle y = -mx + c$
$\displaystyle y = - nx + d$.

If $\displaystyle m = n$ then the gradients will be the same and the lines will be parallel. Therefore there will not be a solution.

For $\displaystyle m \neq n$ we have

$\displaystyle -mx + c = -nx + d$

$\displaystyle c - d = (m - n)x$

$\displaystyle x = \frac{c - d}{m - n}$.

Substituting back into one or both of the equations, you will see there is only one solution.

B. If $\displaystyle m = n$, like I said, the lines are parallel.

However if $\displaystyle c = d$ as well, the equations are in fact, identically equal. In other words, they are the same line.

C. Done in A and B - not sure if it's what you wanted...