# How do you solve the system 4x - 3y = 1 and 12x - 9y = 3 by substitution?

May 25, 2015

Dividing both sides of the second equation by $3$ we get:

$4 x - 3 y = 1$

which is the same as the first equation.

If we attempt to solve the system by substitution, then we will find that all of the terms in $x$ and $y$ cancel out, resulting in a true, but otherwise uninformative equation of rational numbers.

For example, if we take the first equation and add $3 y$ to both sides, then divide both sides by $4$ we get:

$x = \frac{3 y + 1}{4}$

Substitute this in the second equation:

$3 = 12 x - 9 y = 12 \left(\frac{3 y + 1}{4}\right) - 9 y = 3 \left(3 y + 1\right) - 9 y = 9 y + 3 - 9 y = 3$

As a result, there are not enough constraints to determine a unique solution, but there are an infinite number of solutions.

These solutions are the points on the line which in slope intercept form is described by the equation:

$y = \frac{4}{3} x + - \frac{1}{3}$