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Precalculus : Solve Nonlinear Systems of Equations

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Example Questions

Example Question #1 : Solve Nonlinear Systems Of Equations

Solve the following system of nonlinear equations:

-x^2+y+3x=7

5=-y-x^2+7x

Possible Answers:

(5,2),(7,3)

(-2,5),(-3,7)

(-2,-5),(-3,-7)

(2,-5),(3,-7)

(2,5),(3,7)

Correct answer:

(2,5),(3,7)

Explanation:

We can start by rearranging each equation so that y is on one side:

$-x^2+y+3x=7\rightarrow y=x^2-3x+7$

$5=-y-x^2+7x\rightarrow y=-x^2+7x-5$

Now that both equations are equal to y, we can set the right sides equal to each other and solve for x:

$x^2-3x+7=-x^2+7x-5\rightarrow 2x^2-10x+12=0$

$2(x^2-5x+6)=0\rightarrow 2(x-2)(x-3)=0$

x=2,x=3

Our last step is to plug these values of x into either equation to find the y values of our solutions:

y=x^2-3x+7=(2)^2-3(2)+7=5

y=x^2-3x+7=(3)^2-3(3)+7=7

So our solutions are the folloiwing two points:

(2,5),(3,7)

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Example Question #2 : Solve Nonlinear Systems Of Equations

Solve the following system of nonlinear equations:

y+x^2-2x=13

15+8x+y=x^2

Possible Answers:

(-5,2),(22,-7)

(-2,5),(7,-22)

(-2,-5),(-7,-22)

(5,-2),(-22,7)

(2,-5),(-7,22)

Correct answer:

(-2,5),(7,-22)

Explanation:

Our first step is to rearrange each equation so that the left side is just y:

$y+x^2-2x=13\rightarrow y=-x^2+2x+13$

$15+8x+y=x^2\rightarrow y=x^2-8x-15$

Now that both equations are equal to y, we can see that the right sides of each equation are equal to each other, so we set this up below and solve for x:

$-x^2+2x+13=x^2-8x-15\rightarrow 2x^2-10x-28=0$

$2(x^2-5x-14)=0\rightarrow 2(x+2)(x-7)=0$

x=-2,x=7

Our last step is to plug these values of x into either equation to solve for the y values of our solutions:

y=x^2-8x-15=(-2)^2-8(-2)-15=5

y=x^2-8x-15=(7)^2-8(7)-15=-22

So the solutions to the system are the following points:

(-2,5),(7,-22)

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Example Question #3 : Solve Nonlinear Systems Of Equations

Solve the following system:

x^2+y^2=117

y-x=3

Possible Answers:

(6,9), (-9,-6)

x=6,-9

(6,-9), (-9,6)

x=-6,9

Correct answer:

(6,9), (-9,-6)

Explanation:

Our first step is to solve the bottom equation for

y=x+3

We can now substitute it into the top equation:

x^2+(x+3)^2=117

and solve for our values:

x^2+x^2+6x+9=117

2x^2+6x+9-117=0

2x^2+6x-108=0

2(x^2+3x-54)=0

2(x+9)(x-6)=0

our values are then: x=6,-9

We can now plug our values into the bottom equation we had solved for and arrive at our solutions:

y=6+3=9 y=-9+3=-6

So, our solutions are (6,9), (-9,-6)

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Example Question #4 : Solve Nonlinear Systems Of Equations

Solve the following system:

x^2+y^2=5

x^2+3x^2y=x^4

Possible Answers:

(2,1), (-2,1)

$(2,1), (-2,1), (\sqrt{11},-4), (-\sqrt{11},-4)$

$(2,1), (\sqrt{11},-4)$

$(\sqrt{11},-4), (-\sqrt{11},-4)$

Correct answer:

(2,1), (-2,1)

Explanation:

Our first step is to solve the bottom equation for x^2 :

x^2(1+3y)=x^4

$(1+3y)=\frac{x^4}{x^2}$

(1+3y)=x^2

so we can substitute it into the top equation:

(1+3y)+y^2=5

1+3y+y^2-5=0

y^2+3y-4=0

(y-1)(y+4)=0

y=1,-4

Now we can plug in our y-values into the bottom equation to find our x-values:

(1+3(1))=x^2

4=x^2

$x=\pm 2$

(1+3(-4))=x^2

1-12=x^2

-11=x^2

Remember we cannot take a square root of a negative number without getting an imaginary number. As such, we'll just focus on the $x=\pm 2$ values.

Our solution is then: (2,1), (-2,1)

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Example Question #5 : Solve Nonlinear Systems Of Equations

Solve the following system for y>0 :

4y=x^2+2x

y^2=36

Possible Answers:

(-6,6),(4,6)

(-6,6),(4,-6)

(6,6),(-4,6)

(-6,-6),(4,-6)

Correct answer:

(-6,6),(4,6)

Explanation:

The first step is to solve the bottom equation for :

$y=\sqrt{36}=\pm 6$ since our question specifies for y>0 we just focus on y=6

We now substitute this equation into the top equation:

4(6)=x^2+2x

0=x^2+2x-24

0=(x+6)(x-4)

x=-6, 4

we can now plug in our x-values into the bottom equation to find our y-values:

4y=(-6)^2+2(-6) 4y=(4)^2+2(4)

4y=36-12 4y=16+8

4y=24 4y=24

y=6 y=6

The solutions are then: (-6,6), (4,6)

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Example Question #6 : Solve Nonlinear Systems Of Equations

Solve the following system:

y^2+x^2+6x=18

y^2=x^2-2

Possible Answers:

$(2,\sqrt{2}), (2,-\sqrt{2}), (-5,\sqrt{23}),(-5,-\sqrt{23})$

x=-2,5

$(2,\sqrt{6}), (-5,\sqrt{27})$

x=2,-5

Correct answer:

$(2,\sqrt{2}), (2,-\sqrt{2}), (-5,\sqrt{23}),(-5,-\sqrt{23})$

Explanation:

We can solve this equation by using substitution since the bottom equation is already solved for y^2 . Substituting the bottom equation into the top we get:

(x^2-2)+x^2+6x=18

We then solve the equation for our values:

(x^2-2)+x^2+6x-18=0

2x^2+6x-20=0

2(x^2+3x-10)=0

2(x-2)(x+5)=0

x=2,-5

Finally, we substitute our values into the bottom equation to get our values:

y^2=(2)^2-2 y^2=(-5)^2-2

y^2=2 y^2=23

$y=\pm \sqrt{2}$ $y=\pm \sqrt{23}$

Our different solutions are then: $(2,\sqrt{2}), (2,-\sqrt{2}), (-5,\sqrt{23}),(-5,-\sqrt{23})$

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Example Question #821 : Pre Calculus

Solve the following system:

y=x^2

y^2-13x^2=-36

Possible Answers:

x=2,3

(16,4), (81,9)

$(\pm2,4), (\pm3,9)$

y=4,9

Correct answer:

$(\pm2,4), (\pm3,9)$

Explanation:

We can substitute the top equation into the bottom:

y^2-13y=-36

and solve for values:

y^2-13y+36=0

(y-4)(y-9)=0

y=4,9

Now that we have our values we can plug it into the top equation and find our values

4=x^2 9=x^2

$\pm2=x$ $\pm3=x$

So, our values are $(\pm2,4), (\pm3,9)$

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Precalculus : Solve Nonlinear Systems of Equations

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Solve Nonlinear Systems Of Equations

Example Question #2 : Solve Nonlinear Systems Of Equations

Example Question #3 : Solve Nonlinear Systems Of Equations

Example Question #4 : Solve Nonlinear Systems Of Equations

Example Question #5 : Solve Nonlinear Systems Of Equations

Example Question #6 : Solve Nonlinear Systems Of Equations

Example Question #821 : Pre Calculus

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