This week in precalculus 11, we learned how to solve systems algebraically. Solving algebraically includes substitution and elimination. In this blog post, I will be going over how to solve systems algebraically by substitution.

To substitute, you have to isolate the variable to one side of the equation. However, this is only done to one of the equations. When the variable is isolated, substitute it for the same variable on the other equation.

Ex.

x – 2y = -10  –> x = 2y – 10

3x – y = 0

3(2y – 10) – y = 0

Once it’s substitued into the other expression, there should only be one variable left in the ecpression. You then solve to find what the variable left in the espression is.

3(2y – 10) -y = 0

6y – 30 – y = 0

5y – 30 = 0

5y = 30

= 

y = 6

Now that you’ve found y, you have to substitute the answer for y back into one of the equations after the variable has been isolated.

y = 6

x = 2y – 10

x = 2(6) – 10

x = 2

The solution is (2, 6).