236
Section 11.6 – Solving System of Nonlinear Equations
When solving systems of nonlinear equations, we will generally employ one
of three different techniques: 1) Substitution, 2) Elimination, & 3) Graphing.
Which technique works best will depend on the system of equations itself. It
will be important to always check all solutions with the original equations to
ensure that we have not picked up any extraneous solutions.
Objective #1 Solving a System of Nonlinear Equations Using Substitution.
Solve the following using substitution:
Ex. 1
1)
y = 2x + 1
2)
2x2 + y2 = 1
Solution:
Replace y in equation #2 by 2x +1 and solve:
2x2 + y2 = 1
(substitute y = (2x + 1))
2
2
2x + (2x + 1) = 1
(FOIL)
2
2
2x + 4x + 4x + 1 = 1 (combine like terms)
6x2 + 4x + 1 = 1
(subtract 1 from both sides)
2
6x + 4x = 0
(factor)
2x(3x + 2) = 0
(solve)
x = 0 or x = –
2
3
Plug the x-values into y = 2x + 1:
x = 0:
y = 2(0) + 1 = 1
⇒
x=–
2
3€
y = 2(–
Check:
(0, 1):
1)
2)
€
€
2
1
(– , – ):
3
3
)+1=–
1
3
3
(–
⇒
2
3
,–
2(–
3
2 2
)
3
+ (–
So, the solutions are (0, 1) and (–
€
€
€
Ex. 2
(0, 1)
1 = 2(0) + 1
⇒
0=0√
2
2
2(0) + (1) = 1 ⇒
1=1√
€ 1
€
€
2
1)
– = 2(– ) + 1 ⇒
–
2)
€
2
3
2
3
1 2
)
3
,–
=1
1
).
€3
⇒
€
€
1)
y2 – x =€45
2)
x =3–y
€
€
Solution:
Square each side of equation #2 and simplify:
€
1
)
3
1
3
=–
1
3
√
1=1√
237
€
( x )2 = (3 – y)2
(FOIL)
2
x = 9 – 6y + y
In equation #1, replace x with 9 – 6y + y2:
y2 – x = 45
y2 – (9 – 6y + y2) = 45
(simplify)
2
2
y – 9 + 6y – y = 45
6y – 9 = 45
(solve)
6y = 54
y=9
Since x = 9 – 6y + y2 = 9 – 6(9) + (9)2 = 9 – 54 + 81 = 36
Check:
(36, 9)
1)
(9)2 – (36) = 45 ⇒
45 = 45 √
2)
⇒
6=–6 X
36 = 3 – (9)
Since the second equation does not check, then (36, 9) is not a
solution. Thus, our system does not have a solution.
Objective #2: € Solving a system of nonlinear equations by elimination.
Solve the following using elimination:
Ex. 3
1)
x2 + y2 = 25
2)
x2 – y = 13
Solution:
Let's eliminate the x2 by multiplying equation #2 by – 1 and adding
the result to equation #1:
1)
x2 + y2 = 25
– 1•2)
– x2 + y = – 13
y2 + y = 12
(subtract 12 from both sides)
2
y + y – 12 = 0
(factor)
(y – 3)(y + 4) = 0
(solve)
y = 3 or y = – 4
Plug the y-values into equation #2:
y = 3:
x2 – (3) = 13
⇒
x2 = 16
⇒
x=±4
y=–4
Check:
(4, 3):
(– 4, 3):
x2 – (– 4) = 13
1)
2)
1)
2)
⇒
(4)2 + (3)2 = 25
(4)2 – (3) = 13
(– 4)2 + (3)2 = 25
(– 4)2 – (3) = 13
x2 = 9
⇒
⇒
⇒
⇒
⇒
25 = 25 √
13 = 13 √
25 = 25 √
13 = 13 √
x=±3
238
1)
(3)2 + (– 4)2 = 25
⇒
25 = 25 √
2
2)
(3) – (– 4) = 13
⇒
13 = 13 √
2
2
(– 3, – 4): 1)
(– 3) + (– 4) = 25
⇒
25 = 25 √
2
2)
(– 3) – (– 4) = 13
⇒
13 = 13 √
All four answers check so our solution is (4, 3), (– 4, 3), (3, – 4), and
(– 3, – 4).
The graph of the first equation
is a circle of radius 5 centered
on the origin. If we solve the
second equation for y, we get
y = x2 – 13 so its graph is a
parabola shifted down by 13 units.
If we graph both equations on
the same coordinate system, we
can see the four points of
intersection that correspond to
our solution.
(3, – 4):
Ex. 4
Ex. 5
1)
y2 – x2 = – 4
2)
2x2 + 3y2 = 6
Solution:
Let's eliminate the x2 terms by multiplying equation #1 by 2 and
adding it to equation #2:
2•1)
2y2 – 2x2 = – 8
2)
2x2 + 3y2 = 6
5y2 = – 2
(solve for y)
2
y = – 0.4
y = ± −0.4 is not a real number
Thus, this system has no solution.
1)
2)
€
x3 – 2x2 + y2 + 3y – 4 = 0
x–2+
y2 − y
x2
=0
Solution:
First, multiply both sides of equation #2 by x2. Our system becomes:
1)
x3 – 2x2 + y2 + 3y – 4 = 0
2)
x3 – €
2x2 + y2 – y = 0
x≠0
Now, multiply equation #2 by – 1 and add it to equation #1:
239
x3 – 2x2 + y2 + 3y – 4 = 0
– x3 + 2x2 – y2 + y = 0
4y – 4 = 0 ⇒
y=1
Plug in y = 1 into equation #1 and solve for x:
x3 – 2x2 + (1)2 + 3(1) – 4 = 0
(simplify)
3
2
x – 2x = 0
(factor and solve)
2
x (x – 2) = 0
x = 0 or x = 2
Check:
(0, 1):
1)
(0)3 – 2(0)2 + (1)2 + 3(1) – 4 = 0 ⇒ 0 = 0 √
1)
– 1•2)
2)
(2, 1):
1)
2)
(0) – 2 +
(0)2
=0
⇒ Not defined = 0 X
(2)3 – 2(2)2 + (1)2 + 3(1) – 4 = 0 ⇒ 0 = 0 √
(2) – 2 +
€
Thus, (2, 1) is the only solution.
Ex. 6
(1)2 −(1)
(1)2 −(1)
(2)2
=0
⇒ 0=0√
1)
y = ln(x)
€
2)
x = ln(y)
Solution:
Rewrite equation #2 as an exponential equation:
x = ln(y)
⇒
y = ex
Thus, our system is: 1)
y = ln(x)
2)
y = ex
This is not easy to solve by
substitution or by elimination,
but it is very easy to solve by
graphing. If we graph each
equation on the same coordinate
axis, we see that the graphs do
not intersect. Thus, the system
has no solution.
y = ex
y = ln(x)
Solve the following:
Ex. 7 In a 40-mile race, the winner crosses the finish line 2 miles ahead of
the second place winner and 5 miles ahead of the third place winner. If
each of the runners maintained a constant speed throughout the race, by
how many miles did the second place winner beat the third place winner?
240
Solution:
Let v1, v2, and v3 be the speeds of the first, second, and third place
winners respectively. Let t1 and t2 be the times in hours that the first
and second place winners needed to finish the race respectively.
After t1 hours,
The first place winner went 40 miles:
⇒
1) v1t1 = 40
The second place winner went 40 – 2 = 38 miles: ⇒
2) v2t1 = 38
The third place winner went 40 – 5 = 35 miles:
⇒
3) v3t1 = 35
After t2 hours,
The second place winner went 40 miles:
⇒
4) v2t2 = 40
The third place winner went d miles:
⇒
v 3t 2 = d
The distance that the third place winner still needed to run to finish
the race a time t2 is:
40 – v3t2
(multiply the second term by t1 over t1)
( tt )
t
= 40 – 35( )
t
2
= 40 – v3t1
1
2
1
(use equation #3 to replace v3t1 by 35)
(solve equations #1 & #4 for t: t1 =
40
v1
& t2 =
40
v2
= 40 – 35(t2 ÷ t1)
and plug in the results for t1 and t2)
€
40
40
= 40 – 35
÷
(simplify)
v2
v1
€
€
€
v1
40
= 40 – 35
(solve equations #1 & #2 for v: v1 =
& v2 =
(
(v )
)
t1
2
38
t1
= 40 – 35(v1 ÷ v2)
and plug in the results for v1 and v2)
€
€
40
38
= 40 – 35
÷
(simplify)
t1
t1
€
€
€
40
= 40 – 35
≈ 3.16 miles
(
( 38 )
)
The second place winner finished approximately 3.16 miles ahead of
€ third€place winner.
the
€