Nonlinear Systems of Equations

13.4 Nonlinear Systems
In this final section, we want to learn how to solve systems of equations that are not necessarily
all linear. We call these non-linear systems of equations.
Definition: Non-linear system of equations
A system of equations where one or more equations involved is not a line.
We primarily use the substitution method to solve a non-linear system. However, sometimes
elimination will work as well.
Also, just as before, the solution to a non-linear system is all the points of intersection of the
graphs of the equations. Therefore, since we now have more that just lines, we can have a
variety of numbers of solutions. The graph can intersect once, twice, several times or not at all.
Therefore, we should always verify the solution(s) to a system by looking at the graph.
Example 1:
Solve the system.
2 2
x + y = 100
a.
y − x = 2
2 2
x + y = 25
b.
2
c.
y
= x + 5
x
2
+
2 2
y
xy = 4
= 12
Solution:
a. We will solve this system by substitution. So we start by solving the bottom equation for
y and then substitute it into the top equation. We get
y − x = 2 ⇒ y = x + 2
x
2
2
2
+ y = 100 ⇒ x + x
x
2
2
+ x
2x
2 ( + 2)
= 100
2 ( x + 2x
− 48)
2
+ 4x
+ 4 = 100
+ 4x
− 96 = 0
( x + 8)(
x − 6)
= 0
= 0
x = −8,
6
So we have two different x values. This means we should have two points of
intersection. Lets now find the y values and verify with a graph.
y = x + 2
y = x + 2
y = −8
+ 2
2
2
y = 6 + 2
y = −6
y = 8
So our solutions are ( − 8, − 6)
and ( 6 , 8)
. We clearly have a circle and a line, thus we
can easily graph them together and get

. Again we will use substitution to solve. This time notice that the bottom equation is
already solved for y 2 and we have a y 2 in the top equation. Thus, that is the substitution
we will make. We get
x
y
2
2
+ y
2
= 25
= x + 5
Now we solve for x and then solve for y.
x
x
2
2
+
2
⇒ x + ( x + 5)
= 25
( x + 5)
+ x − 20 = 0
( x + 5)(
x − 4)
= 25
= 0
x = −5,
4
We substitute these back in to get y. We get
y
y
2
2
= −5
+ 5
= 0
y = 0
y
y
2
2
= 4 + 5
= 9
y = ± 3
5,
0 , 3
So we have three different solutions ( − ) , ( 4 ) and ( , 3)
graph. We have here a parabola and a circle. We get
4 − . Lets verify with a
c. Again, we will use substitution to solve. We need to decide which variable to solve for
first. It seems that x or y on the bottom equation would be easiest. So we will solve for x.
We get
4
xy = 4 ⇒ x =
y
Now we substitute that into the first equation and solve. We have
x
2
2 ⎛ 4 ⎞
+ 2y
= 12 ⇒ ⎜ ⎟ + 2y
⎝ y ⎠
2
2
16 2
= 12
+ 2y
= 12
2
y
We will have to clear the fractions and solve as we did in chapter 10, that is, using a
substitution.

Example 2:
2 ⎛ 16 2 ⎞
y ⎜ + 2y
=
2 ⎟ y
⎝ y ⎠
2y
2u
2
2
4
16 + 2y
−12y
2
2
+ 16 = 0
−12u
+ 16 = 0
4
2 ( u − 6u
+ 8)
( u − 4)(
u − 2)
y
2
= 4
y = ± 2
= 12y
= 0
= 0
u =
4,
( 12)
2
2
2
y
2
Substitute u = y 2
= 2
y = ±
2
Re-substitute u = y 2
So since we have four different y values we will have to find x for each one. We
substitute these in to
4
x =
2
= 2
x
4
x =
− 2
= −2
4
y
= to get
x =
=
= 2
4
⋅
2
4 2
2
2
2
2
4
x = ⋅
− 2
= −
= −2
4 2
2
So we have four solutions, ( 2, 2)
, ( − 2,
− 2)
, ( 2 2,
2)
and ( 2 2,
− 2)
− .
4
Lets verify with a graph. We have an ellipse and a basic function ( xy = 4 ⇒ y = ).
x
So we have
Solve the system.
x
y = 3
y
= log ( x + 1)
a. b.
y = 3
2x −
2
2
y = 5 − log
2
( x − 3)
2
2
2

Solution:
a. This time solving is a little more complicated. There are a variety of direction we could
go, however, we are going to start by noticing ( ) 2
2 x
y = 3
x
x 2
( 3 ) − 2
3
3
x = . So our system really is
y =
We can see clearly we will substitute the top equation into the bottom. That is, put y in
for 3 x . We get
2
y = y − 2
y
2
( y − 2)(
y + 1)
( )
y = y
− y − 2 = 0
= 0
y = 2,
−1
Now we substitute these values back in to get
x
2 = 3
x
−1
= 3
x = log3
2
However, the second equation is impossible (recall, exponential functions are always
positive). Thus, that solution must be omitted. So we have a solution of ( 3,
2).
b. Lastly, since these equations are both already solved for y, we can simply set them equal
to one another. Then we are left with a logarithmic equation to solve. We get
log x + 1 = 5 − log x − 3
log
2
2
− 2
2 ( )
( x + 1)
+ log 2 ( x − 3)
= 5
log 2 ( x + 1)(
x − 3)
= 5
5
( x + 1)(
x − 3)
= 2
2 ( )
x
x
2
2
− 2x
− 3 = 32
− 2x
− 35 = 0
( x − 7)(
x + 5)
= 0
log 2
x = 7,
− 5
However, -5 cannot be a solution since is doesn’t even check in the equation. Thus we
only have x = 7. Now we substitute this back into either original equation to get the y
value. We choose the first equation.
y = log 7 + 1
So the solution is .
13.4 Exercises
Solve the systems.
( 7,
3)
= log
= 3
2
2
( )
8

1. 2. 3.
2
2
2
2
=
+
=
+
y
x
y
x
1
25
2
2
=
−
=
+
x
y
y
x
15
3
5
225
9
25
2
2
=
+
=
+
y
x
y
x
4. 5. 6.
6
2
3
36
4
9
2
2
=
+
=
+
y
x
y
x
4
2
3
2
+
=
+
=
x
y
x
y
2
3
2
+
=
=
y
x
x
y
7. 8. 9.
1
2
16
2
2
=
−
=
−
y
x
y
x
7
2
25
4 2
2
=
+
=
+
y
x
y
x
3
2
18
2
2
=
+
=
+
y
x
y
x
10. 11.
2
12.
3
2
3
2
=
−
=
−
y
x
y
x
2
2
20
x
y
y
x
=
=
+
3
3
2
2
2
−
=
=
−
x
y
y
x
13. 14. 15.
0
3
4
2
2
=
−
=
−
−
y
x
y
x
x
6
2
8
2
2
2
2
=
+
=
+
−
y
x
y
x
x
1
13
2
2
2
−
=
=
+
x
y
y
x
16. 17. 18.
25
5
2
2
2
=
+
=
−
y
x
y
x
5
3
2
25
2
2
2
2
=
−
=
+
y
x
y
x
144
16
9
4
2
2
2
2
=
+
=
+
y
x
y
x
19. 20. 21.
16
13
2
2
2
2
−
=
−
=
+
y
x
y
x
10
2
16
2
2
2
2
=
−
=
+
y
y
y
x
12
20
2
2
2
2
−
=
−
=
+
y
x
y
x
22. 23.
4
14
2
2
2
2
=
−
=
+
y
x
y
x
6
2
3
5
9
=
+
−
=
y
x
xy
24.
7
6
−
=
−
=
+
xy
y
x
25. 26. 27.
16
4
2
2
2
−
=
−
−
=
y
x
x
y
5
25
2
2
2
+
=
=
+
x
y
y
x
xy
x
y
xy
5
6
1
=
+
=
28.
xy
y
x
xy
7
12
1
=
+
=
29. 30.
5
2
7
2 2
2
=
−
=
+
+
y
x
y
xy
x
2
27
3 2
2
=
−
=
+
−
y
x
y
xy
x
31. 32. 33.
2
5
2
2
=
=
+
xy
y
x
8
20
2
2
=
=
+
xy
y
x
8
3
2
34
3
2
2
=
−
=
+
x
xy
x
xy
34. 35.
4
2
3
7
3
2
2
2
=
−
=
+
y
xy
y
xy
3
1
1
5
1
1
−
=
−
=
+
y
x
y
x
36.
2
1
1
4
1
1
−
=
+
=
−
y
x
y
x

37.
1
2
3
3
5
2
2
2
2
2
=
−
=
+
y
x
y
x
38.
35
1
2
14
3
1
2
2
2
2
=
+
=
−
y
x
y
x
39.
0
1
3
1
2
4
=
+
−
−
=
x
y
y
x
40. 41.
0
16
0
3
=
−
=
−
xy
y
x
( ) 4
3
2
2
=
+
−
−
=
y
x
x
y
42.
( ) 1
2
2
2
=
+
+
=
y
x
x
y
43. 44. 45.
12
2
2
2 −
=
=
x
x
y
y
1
5
5
2 −
=
=
x
x
y
y
6
2
4
+
=
=
x
x
e
y
e
y
46. 47.
1
2 2
−
=
=
x
x
e
y
e
y ( )
( 1
log
2
4
log
2
2
+
−
= )
+
=
x
y
x
y
48.
( )
1
log
log
6
6
+
−
=
=
x
y
x
y
49.
( )
2
1
9
9
log
1
log
+
=
+
=
x
y
x
y
50.
( )
( ) 2
1
16
16
1
log
3
log
+
−
=
+
=
x
y
x
y