## 17Calculus Precalculus - Solving Nonlinear Systems of Equations

Types of Systems of Equations

 Basics of linear systems with the same number of equations as unknowns - covered on the main linear systems page Linear with fewer equations than unknowns - covered on the dependent systems page Non-Linear - covered on this page

Getting Started

Solving systems of nonlinear equations can be done using techniques you have already learned with linear equations, substitution and elimination. Knowing which one to use is based on the form of the equations and, if you carefully look at the systems, usually one of the techniques will seem to work best. Here are some ideas to get you started.

1. If you have terms that with powers, try elimination first. It is best not to solve for a variable under a power. For example, if one of your equations is $$y=x^2$$, do not try to solve for x in order to use substitution. This will introduce a complexity that can lead to incomplete and incorrect answers.
2. If there is an obvious substitution, try to substitute cautiously. Sometimes substitution can get messy. Other times it can simplify the equations significantly.
3. You may end up with a quadratic. In this case, completing the square will help.

Difference From Linear Systems

Although we use the same techniques (substitution and elimination), we may end up with more than one solution. If you think about what is going on, this makes sense. For example, if we have a parabola and a line, intersection of the two curves may occur at two points. It is difficult to know just from the equations, how many points solve the system.

If you are allowed to, it helps to plot a graph on your calculator or computer and get a feel for what the graphs look like. This will help you know what to do when solving the equations.

Okay, now try your hand at some practice problems.

Practice

Unless otherwise instructed, find all the solutions to these nonlinear systems of equations, giving your answers in exact form.

Basic

$$y=x^2-9 \text{ and } y=9-x^2$$

Problem Statement

$$y=x^2-9 \text{ and } y=9-x^2$$

Solution

### 1777 video

video by MIP4U

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$$x^2-5y=6 \text{ and } x^2+y=18$$

Problem Statement

$$x^2-5y=6 \text{ and } x^2+y=18$$

Solution

### 1778 video

video by MIP4U

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$$2x^2-y^2=23 \text{ and } x^2+2y^2=34$$

Problem Statement

$$2x^2-y^2=23 \text{ and } x^2+2y^2=34$$

Solution

### 1779 video

video by MIP4U

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$$2x^2-10y^2=8 \text{ and } x^2-3y^2=6$$

Problem Statement

$$2x^2-10y^2=8 \text{ and } x^2-3y^2=6$$

Solution

### 1780 video

video by Brightstorm

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$$x^2+4x-y=7 \text{ and } 2x-y=-1$$

Problem Statement

$$x^2+4x-y=7 \text{ and } 2x-y=-1$$

Solution

### 1781 video

video by mattemath

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$$-x+y=4 \text{ and } x^2+y=3$$

Problem Statement

$$-x+y=4 \text{ and } x^2+y=3$$

Solution

### 1782 video

video by mattemath

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$$-2x+y=5 \text{ and } x^2+3x-y=1$$

Problem Statement

$$-2x+y=5 \text{ and } x^2+3x-y=1$$

Solution

### 1783 video

video by mattemath

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$$y=x^2 \text{ and } 3x+y=10$$

Problem Statement

$$y=x^2 \text{ and } 3x+y=10$$

Solution

### 1789 video

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$$x^2+y^2=25 \text{ and } 3x+4y=0$$

Problem Statement

$$x^2+y^2=25 \text{ and } 3x+4y=0$$

Solution

### 1790 video

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$$y-x=4 \text{ and } x^2+8x=y-16$$

Problem Statement

$$y-x=4 \text{ and } x^2+8x=y-16$$

Solution

### 1791 video

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$$6x-y=5 \text{ and } xy=1$$

Problem Statement

$$6x-y=5 \text{ and } xy=1$$

Solution

### 1792 video

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$$y=x/2 \text{ and } 2x^2-y^2=7$$

Problem Statement

$$y=x/2 \text{ and } 2x^2-y^2=7$$

Solution

### 1784 video

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$$y=x+1 \text{ and } x^2+y^2=25$$

Problem Statement

$$y=x+1 \text{ and } x^2+y^2=25$$

Solution

### 1785 video

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$$y=-x^2+6 \text{ and }y=-2x-2$$

Problem Statement

$$y=-x^2+6 \text{ and }y=-2x-2$$

Solution

### 1786 video

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$$y=2x^2+3x-6 \text{ and } y=-x^2$$

Problem Statement

$$y=2x^2+3x-6 \text{ and } y=-x^2$$

Solution

### 1787 video

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$$y=2(x-4)^2+3 \text{ and } y=-x^2+2x-2$$

Problem Statement

$$y=2(x-4)^2+3 \text{ and } y=-x^2+2x-2$$

Solution

### 1788 video

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Intermediate

$$x^3 + 9x^2y = 10$$ and $$y^3 + xy^2 = 2$$

Problem Statement

$$x^3 + 9x^2y = 10$$ and $$y^3 + xy^2 = 2$$

Hint

Expand out $$(x + 3y)^3$$.

Problem Statement

$$x^3 + 9x^2y = 10$$ and $$y^3 + xy^2 = 2$$

Hint

Expand out $$(x + 3y)^3$$.

Solution

### 3784 video

video by blackpenredpen

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Really UNDERSTAND Precalculus

 systems of linear equations solving by substitution solving by elimination

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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 Getting Started Difference From Linear Systems Practice

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