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# 8.5: Solving Nonlinear Systems

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LEARNING OBJECTIVES

Identify nonlinear systems.

Solve nonlinear systems using the substitution method.

Nonlinear Systems

A system of equations where at least one equation is not linear is called a nonlinear system. In this section we will use the substitution method to solve nonlinear systems. Recall that solutions to a system with two variables are ordered pairs (x,y)(x,y) that satisfy both equations.

Example 1

Solve: {x+2y=0x2+y2=5{x+2y=0x2+y2=5.

Solution:

In this case we begin by solving for x in the first equation.

{x+2y=0x2+y2=5  =⇒=  x=−2y{x+2y=0  ⇒  x=−2yx2+y2=5

Substitute x=−2yx=−2y into the second equation and then solve for y.

(−2y)2+y24y2+y25y2y2y=====5551±1(−2y)2+y2=54y2+y2=55y2=5y2=1y=±1

Here there are two answers for y; use x=−2yx=−2y to find the corresponding x-values.

Using y=−1y=−1

Using y=1y=1

x===−2y−2(−1)2x=−2y=−2(−1)=2

x===−2y−2(1)−2x=−2y=−2(1)=−2

This gives us two ordered pair solutions, (2,−1)(2,−1) and (−2,1).(−2,1).

In the previous example, the given system consisted of a line and a circle. Graphing these equations on the same set of axes, we can see that the two ordered pair solutions correspond to the two points of intersection.

If we are given a system consisting of a circle and a line, then there are 3 possibilities for real solutions—two solutions as pictured above, one solution, or no solution.

Example 2

Solve: {x+y=3x2+y2=2{x+y=3x2+y2=2.

Solution:

Solve for y in the first equation.

{x+y=3x2+y2=2  =⇒=  y=3−x{x+y=3  ⇒  y=3−xx2+y2=2

Next, substitute y=3−xy=3−x into the second equation and then solve for x.

x2+(3−x)2x2+9−6x+x22x2−6x+92x2−6x+7====2220x2+(3−x)2=2x2+9−6x+x2=22x2−6x+9=22x2−6x+7=0

The resulting equation does not factor. Furthermore, using a=2a=2, b=−6b=−6, and c=7c=7 we can see that the discriminant is negative:

b2−4ac===(−6)2−4(2)(7)36−56−20b2−4ac=(−6)2−4(2)(7)=36−56=−20

We conclude that there are no real solutions to this equation and thus no solution to the system.

Try this! Solve: {x−y=5x2+(y+1)2=8{x−y=5x2+(y+1)2=8.

(click to see video)

If given a circle and a parabola, then there are 5 possibilities for solutions.

When using the substitution method, we can perform the substitution step using entire algebraic expressions. The goal is to produce a single equation in one variable that can be solved using the techniques learned up to this point in our study of algebra.

Example 3

Solve: {x2+y2y−x2==2−2{x2+y2=2y−x2=−2.

Solution:

We can solve for x2x2 in the second equation.

{x2+y2=2y−x2=−2    ⇒  y+2=x2{x2+y2=2y−x2=−2    ⇒  y+2=x2

Substitute x2=y+2x2=y+2 into the first equation and then solve for y.

y+2+y2y2+yy(y+1)y====2000       or      y=−1y+2+y2=2y2+y=0y(y+1)=0y=0       or      y=−1

Back substitute into x2=y+2x2=y+2 to find the corresponding x-values.

Using y=−1y=−1

Using y=0y=0

x2x2x2x====y+2−1+21±1x2=y+2x2=−1+2x2=1x=±1

x2x2x2x====y+20+22±2–√x2=y+2x2=0+2x2=2x=±2

This leads us to four solutions, (±1,−1)(±1,−1) and (±2–√,0).(±2,0).

Example 4

Solve: {(x−1)2−2y2=4x2+y2=9{(x−1)2−2y2=4x2+y2=9.

Solution:

We can solve for y2y2 in the second equation,

{(x−1)2−2y2=4x2+y2=9  =⇒=  y2=9−x2{(x−1)2−2y2=4x2+y2=9  ⇒  y2=9−x2

Substitute y2=9−x2y2=9−x2 into the first equation and then solve for x.

(x−1)2−2(9−x2)x2−2x+1−18+2x23x2−2x−21(3x+7)(x−3)3x+7x======40000−73  or  x−3x==03(x−1)2−2(9−x2)=4x2−2x+1−18+2x2=03x2−2x−21=0(3x+7)(x−3)=03x+7=0  or  x−3=0x=−73x=3

Back substitute into y2=9−x2y2=9−x2 to find the corresponding y-values.

Using x=−73x=−73

Using x=3x=3

y2y2y2y====9−(−73)291−499329±32√3=±42√3y2=9−(−73)2y2=91−499y2=329y=±323=±423

y2y2y===9−(3)200y2=9−(3)2y2=0y=0

This leads to three solutions, (−73,±42√3)(−73,±423) and (3,0).(3,0).

Example 5

Solve: {x2+y2=2xy=1{x2+y2=2xy=1.

Solution:

Solve for yy in the second equation.

{x2+y2=2xy=1  =⇒=  y=1x{x2+y2=2xy=1  ⇒  y=1x

Substitute y=1xy=1x into the first equation and then solve for x.

x2+(1x)2x2+1x2==22x2+(1x)2=2x2+1x2=2

This leaves us with a rational equation. Make a note that x≠0x≠0 and multiply both sides by x2.x2.

x2(x2+1x2)x4+1x4−2x2+1(x2−1)(x2−1)====2⋅x22x200x2(x2+1x2)=2⋅x2x4+1=2x2x4−2x2+1=0(x2−1)(x2−1)=0

At this point we can see that both factors are the same. Apply the zero product property.

x2−1x2x===01±1x2−1=0x2=1x=±1

Back substitute into y=1xy=1x to find the corresponding y-values.

Using x=−1x=−1

Using x=1x=1

y===1x1−1−1y=1x=1−1=−1

y===1x111y=1x=11=1

Try this! Solve: ⎧⎩⎨1x+1y=41x2+1y2=40{1x+1y=41x2+1y2=40.