This week we started our final unit in math, linear and non-linear systems. It took me a few practice equations to finally get the hang of how to answer some of the equations. However, now I feel more confident solving linear systems by substitution.

We must first look at our equations to find out the solution. Our equations are 4x – 5y = 1, y = x + 2.Our life is easier because there is already an equation written in slope intercept form. We can replace any y’s in the other equation with the answer from the slope intercept form. Our new equation will look like 4x – 5 (x + 2) = 1. Now by simply solving this equation we will get x = -11. We can then put that x in the place of x in our slope intercept form. Our next equation will look like y = -11 + 2. If we add -11 and 2 it will give us y = -9. So, our coordinates for our solution is (-11, -9).