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 |
Recommended Books on Amazon | ||
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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
\(y=x^2-9 \text{ and } y=9-x^2\)
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\(y=x^2-9 \text{ and } y=9-x^2\)
Solution |
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1777 video
video by MIP4U |
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\(x^2-5y=6 \text{ and } x^2+y=18\)
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\(x^2-5y=6 \text{ and } x^2+y=18\)
Solution |
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1778 video
video by MIP4U |
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\(2x^2-y^2=23 \text{ and } x^2+2y^2=34\)
Problem Statement |
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\(2x^2-y^2=23 \text{ and } x^2+2y^2=34\)
Solution |
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1779 video
video by MIP4U |
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\(2x^2-10y^2=8 \text{ and } x^2-3y^2=6\)
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\(2x^2-10y^2=8 \text{ and } x^2-3y^2=6\)
Solution |
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1780 video
video by Brightstorm |
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\(x^2+4x-y=7 \text{ and } 2x-y=-1\)
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\(x^2+4x-y=7 \text{ and } 2x-y=-1\)
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1781 video
video by mattemath |
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\(-x+y=4 \text{ and } x^2+y=3\)
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\(-x+y=4 \text{ and } x^2+y=3\)
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1782 video
video by mattemath |
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\(-2x+y=5 \text{ and } x^2+3x-y=1\)
Problem Statement |
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\(-2x+y=5 \text{ and } x^2+3x-y=1\)
Solution |
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1783 video
video by mattemath |
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\( y=x^2 \text{ and } 3x+y=10 \)
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\( y=x^2 \text{ and } 3x+y=10 \)
Solution |
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1789 video
video by Your Math Gal |
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\( x^2+y^2=25 \text{ and } 3x+4y=0 \)
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\( x^2+y^2=25 \text{ and } 3x+4y=0 \)
Solution |
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1790 video
video by Your Math Gal |
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\( y-x=4 \text{ and } x^2+8x=y-16 \)
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\( y-x=4 \text{ and } x^2+8x=y-16 \)
Solution |
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1791 video
video by Your Math Gal |
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\( 6x-y=5 \text{ and } xy=1 \)
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\( 6x-y=5 \text{ and } xy=1 \)
Solution |
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1792 video
video by Your Math Gal |
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\( y=x/2 \text{ and } 2x^2-y^2=7 \)
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\( y=x/2 \text{ and } 2x^2-y^2=7 \)
Solution |
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1784 video
video by Khan Academy |
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\( y=x+1 \text{ and } x^2+y^2=25 \)
Problem Statement |
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\( y=x+1 \text{ and } x^2+y^2=25 \)
Solution |
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1785 video
video by Khan Academy |
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\( y=-x^2+6 \text{ and }y=-2x-2 \)
Problem Statement |
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\( y=-x^2+6 \text{ and }y=-2x-2 \)
Solution |
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1786 video
video by Khan Academy |
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\( y=2x^2+3x-6 \text{ and } y=-x^2 \)
Problem Statement |
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\( y=2x^2+3x-6 \text{ and } y=-x^2 \)
Solution |
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1787 video
video by Khan Academy |
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\( y=2(x-4)^2+3 \text{ and } y=-x^2+2x-2 \)
Problem Statement |
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\( y=2(x-4)^2+3 \text{ and } y=-x^2+2x-2 \)
Solution |
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1788 video
video by Khan Academy |
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Intermediate
\( x^3 + 9x^2y = 10 \) and \( y^3 + xy^2 = 2 \)
Problem Statement |
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\( x^3 + 9x^2y = 10 \) and \( y^3 + xy^2 = 2 \)
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Expand out \( (x + 3y)^3 \).
Problem Statement |
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\( x^3 + 9x^2y = 10 \) and \( y^3 + xy^2 = 2 \)
Hint |
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Expand out \( (x + 3y)^3 \).
Solution |
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3784 video
video by blackpenredpen |
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Really UNDERSTAND Precalculus
Topics You Need To Understand For This Page
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
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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Topics Listed Alphabetically
Single Variable Calculus |
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Multi-Variable Calculus |
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Differential Equations |
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Precalculus |
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Practice Instructions
Unless otherwise instructed, find all the solutions to these nonlinear systems of equations, giving your answers in exact form.
