Lesson 7.4, page 746
Nonlinear Systems of Equations
Objective: To solve a nonlinear
system of equations.
Review – What?



1.
2.
3.
4.
System – 2 or more equations
together
Solution of system – any ordered
pair that makes all equations true
Possible solutions:
One point
More than one point
No solution
Infinite solutions
Review - How?

What methods have we used to
solve linear systems of equations?
1. Graphing
2. Substitution
3. Elimination
Review
Steps for using SUBSTITUTION
1.
Solve one equation for one variable. (Hint:
2.
Substitute into the other equation.
Solve this equation to find a value for the
variable.
Substitute again to find the value of the
other variable.
Check.
3.
4.
5.
Look for an equation already solved for a variable or
for a variable with a coefficient of 1 or -1.)
Review
Solve using Substitution.
2x – y = 6
y = 5x
Review
STEPS for ELIMINATION
Review
Solve using elimination.
3x + 5y = 11
 2x + 3y = 7

What’s New?




A non-linear system is one in which
one or more of the equations has a
graph that is not a line.
With non-linear systems, the solution could
be one or more points of intersection or
no point of intersection.
We’ll solve non-linear systems using
substitution or elimination.
A graph of the system will show the points
of intersection.
An Example…
 Solve
the following system of
equations:
x  y 9
2x  y  3
2
2
(1) The graph is a circle.
(2) The graph is a line.
An Example…

We use the substitution method.
First, we solve equation (2) for y.
2x  y  3
 y  2 x  3
y  2x  3
An Example…

Next, we substitute
y = 2x  3 in equation (1) and solve
for x: x 2  (2 x  3) 2  9
x  4 x  12 x  9  9
2
2
5 x  12 x  0
x (5 x  12)  0
2
12
x  0 or x 
5
An Example… …


Now, we substitute these numbers for x in equation
(2) and solve for y.
x
y
y
y
=
=
=
=
0
2x  3
2(0)  3
3
x = 12 / 5
y  2x  3
y  2( 125 )  3
9
y
5
SOLUTIONS
(0, 3) and  12 , 9 


5
5


An Example…
Check: (0, 3)

x2  y 2  9
2x  y  3
0 3 9
99
2(0)  (3)  3
33
2
3
 12 9 
 , 
5 5

x2  y 2  9
2x  y  3
Check:
 
12 2
5


9 2
5
9
99
2( 125 )  ( 95 )  3
33
Visualizing the
Solution
See Example 1, page 747
Check Point 1: Solve by substitution.
 x2 = y – 1
 4x – y = -1
See Example 2, page 748
Check Point 2: Solve by substitution.
x + 2y = 0
(x – 1)2 + (y – 1)2 = 5

Another example to watch…

Solve the following system of
equations:
xy = 4
3x + 2y = 10
xy  4
4
y
x
Solve xy = 4 for y.
3 x  2 y  10
Substitute into
3x + 2y = 10.
3 x  2( 4x )  10
3 x  8x  10
x  3x 
8
x
  10( x)
3 x  8  10 x
2
3 x  10 x  8  0
2

Use the quadratic
formula (or factor)
to solve:
b  b 2  4ac
x
2a
10  102  4(3  8)
x
2(3)
x
3 x  10 x  8  0
2
x
x
x
10  100  96
6
10  4 10  2

6
6
10  2
10  2
and
6
6
4
and  2
3
Substitute values of x to find y.
3x + 2y = 10
x = 4/3
x = 2
3  34   2 y  10
3  2   2 y  10
4  2 y  10
6  2 y  10
2 y  6
2 y  4
y  3
y  2
The solutions are
(4/3, 3) and (2, 2).

Visualizing the
Solution
Need to watch another one?

Solve the system of equations:
5 x  2 y  13
2
2
3 x  4 y  39
2
2
5 x 2  2 y 2  13
3 x 2  4 y 2  39

Solve by elimination.
Multiply equation (1) by 2 and
add to eliminate the y2 term.
10 x  4 y  26
2
2
3x  4 y  39
2
13x
2
 13
2
x 1
2
x  1
Substituting x = 1 in equation (2) gives us:
x=1
x = -1
3 x 2  4 y 2  39
3 x 2  4 y 2  39
3(1) 2  4 y 2  39
3( 1) 2  4 y 2  39
4 y 2  36
4 y 2  36
y2  9
y  3
y2  9
y  3

The possible solutions are
(1, 3), (1, 3), (1, 3) and (1, 3).

All four pairs check,
so they are the
solutions.
Visualizing the Solution
See Example 3, page 749
Check Point 3: Solve by elimination.
 3x2 + 2y2 = 35
 4x2 + 3y2 = 48
See Example 4, page 750
Check Point 4: Solve by elimination.
 y = x2 + 5
 x2 + y2 = 25