IS-Richmond Lesson 13: Solving Systems by Substitution

1. x+2y=8

2. x=−5

We can substitute x=−5 into the first equation:

−5+2y=8

-5 + 2y = 8

+5 +5

0 + 2y = 8 + 5

2y = 8 + 5

Now, solve for y

2y = 8 + 5

2y=13 (We have to eliminate the 2 in front of the y, we do that by dividing both sides by 2)

2y = 13

2 2

y = 13/2

So, the solution to the system of equations is:

x=−5

y=13/ 2

This means the point of intersection for the two lines is ( -5, 13/2 )