Week 12 – Solving Systems of Equations Algebraically

This is a linear-quadratic system:

y = x + 4

y = x² + x

To solve it algebraically, make sure a variable is isolated, and then substitute the y value of one equation into the other. The variable y is already isolated.

Let’s substitute y = x + 4 in the second equation.

x + 4 = x² + x

Then, move the x + 4 to the right side and factor.

0 = x² – 4

(x – 2)(x + 2)

So, x = 2 or x = -2

Substitue each value of x in the first equation.

y = 2 + 4

y = 6

One solution is (2, 6).

y = -2 + 4

y = 2

The other solution is (-2, 2)

You can also find the solutions of a quadratic-quadratic system.

For example:

y = x² + 4

y = -x² + 12

First substitute.

x² + 4 = -x² + 12

Move to one side and simplify.

x² + 4 + x² – 12 = 0

2x² – 8 = 0

Factor.

2(x – 2)(x + 2)

So, x = 2 or x = -2

Plug in the x values into the first equation.

y = (2)² + 4

y = 4 + 4

y = 8

y = (-2)² + 4

y = 4 + 4

y = 8

The solutions are (2, 8) and (-2, 8).