Not all systems are strictly linear. When one of the equations is not linear, you need to be prepared to solve quadratic equations. You can use factoring or the quadratic formula to solve quadratic equations algebraically.

Example:  Solve using substitution:  

Solution:

The first equation is already solved for y. That fact will make it easy to substitute.

Set each factor equal to 0 because If ab = 0, then a = 0 or b = 0

x – 2 = 0          or         x + 1 = 0

Solve each equation for x

x = 2            or               x = -1 

Substitute each x-value into the first equation  to find the corresponding y-value

y = 7 – (2) = 5
y = 7 – (-1) = 7 + 1 = 8     

The solutions are (2, 5) and (-1, 8).

If we look at the graphs on our calculator, we will see that the line intersects the parabola in two places. And if we look at the table, we can verify our solutions.

The area of a rectangular garden is 240 square feet. The length of the garden is 4 feet less than twice the width. What are the dimensions of the garden?

Solution:

Draw a figure to represent the rectangular garden and label it.

Area is length times width so lw = 240
                                                           
The length is 4 less than twice the width so  l = 2w – 4

Since the second equation is solved for l, substitution is pretty easy.

lw = ( 2w – 4 ) w= 2w2 – 4w =	240 240 240	The first equation Substitute ( 2w – 4 ) into the equation for l Distribute w
2w2 – 4w – 240 =	240-240	Subtract 240 from both sides so the quadratic equation equals 0
2w2 – 4w – 240 = 2(w2 – 2w – 120) =	0 0	Simplify Factor the GCF of 2
2(w – 12)(w + 10) =	0	Factor
w – 12 = or w + 10 =	0   0	Set each factor equal to 0 and solve for w.
w = or w =	12   -10	(a negative width makes no sense)

Substitute 12 into the second equation for w and simplify.

l = 2w – 4 = 2 (12) – 4 = 20                

The length of the garden is 20 feet and its width is 12 feet.