17Calculus Precalculus - Solving Nonlinear Systems of Equations
17Calculus
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 |
Topics You Need To Understand For This Page |
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systems of linear equations solving by substitution solving by elimination |
Recommended Books on Amazon (affiliate links) | ||
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What Does It Mean to 'Solve'?
Like you did with linear systems, solving non-linear systems means to find the points that solve the both equations. In terms of graphs, these points are where the graphs intersect. We use the term non-linear when at least one of the equations is not a straight line.
Here is a great video where the instructor explains this in more detail.
Michel vanBiezen - What Does It Mean to "Solve..."?
video by Michel vanBiezen |
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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.
These practice problems are divided into these sections.
1. Basic
2. Intermediate
3. Word Problems
4. Inequalities
Basic
\( x^2 + y^2 = 13 \) and \( 2x + y = -1 \)
Problem Statement |
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\( x^2 + y^2 = 13 \) and \( 2x + y = -1 \)
Final Answer |
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\( ( 1.2, -3.4 ) \) and \( ( -2, 3 ) \)
Problem Statement
\( x^2 + y^2 = 13 \) and \( 2x + y = -1 \)
Solution
Michel vanBiezen - 4116 video solution
video by Michel vanBiezen |
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Final Answer
\( ( 1.2, -3.4 ) \) and \( ( -2, 3 ) \)
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\( x^2+4x-y=7\) and \( 2x-y=-1 \)
Problem Statement |
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\( x^2+4x-y=7\) and \( 2x-y=-1 \)
Final Answer |
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\( (-4, -7) \) and \( (2, 5) \)
Problem Statement
\( x^2+4x-y=7\) and \( 2x-y=-1 \)
Solution
mattemath - 1781 video solution
video by mattemath |
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Final Answer
\( (-4, -7) \) and \( (2, 5) \)
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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
mattemath - 1782 video solution
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
mattemath - 1783 video solution
video by mattemath |
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\( x^2 - y^2 = 1 \) and \( x^2 + y^2 = 49 \)
Problem Statement |
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\( x^2 - y^2 = 1 \) and \( x^2 + y^2 = 49 \)
Final Answer |
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\( ( 5, 2\sqrt{6} ) \), \( ( 5, -2\sqrt{6} ) \), \( ( -5, 2\sqrt{6} ) \) and \( ( -5, -2\sqrt{6} ) \)
Problem Statement
\( x^2 - y^2 = 1 \) and \( x^2 + y^2 = 49 \)
Solution
Michel vanBiezen - 4117 video solution
video by Michel vanBiezen |
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Final Answer
\( ( 5, 2\sqrt{6} ) \), \( ( 5, -2\sqrt{6} ) \), \( ( -5, 2\sqrt{6} ) \) and \( ( -5, -2\sqrt{6} ) \)
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\(\displaystyle{ \frac{x^2}{4} + \frac{y^2}{9} = 1 }\) and \( x^2 + y^2 = 9 \)
Problem Statement |
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\(\displaystyle{ \frac{x^2}{4} + \frac{y^2}{9} = 1 }\) and \( x^2 + y^2 = 9 \)
Final Answer |
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\( ( 0, 3 ) \) and \( ( 0, -3 ) \)
Problem Statement
\(\displaystyle{ \frac{x^2}{4} + \frac{y^2}{9} = 1 }\) and \( x^2 + y^2 = 9 \)
Solution
Michel vanBiezen - 4118 video solution
video by Michel vanBiezen |
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Final Answer
\( ( 0, 3 ) \) and \( ( 0, -3 ) \)
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\( 2x^2 - y^2 = 1 \) and \( xy = -1 \)
Problem Statement |
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\( 2x^2 - y^2 = 1 \) and \( xy = -1 \)
Final Answer |
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\( (1,-1) \) and \( ( -1,1 ) \)
Problem Statement
\( 2x^2 - y^2 = 1 \) and \( xy = -1 \)
Solution
Michel vanBiezen - 4119 video solution
video by Michel vanBiezen |
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Final Answer
\( (1,-1) \) and \( ( -1,1 ) \)
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\( x^2 + 2xy - 2y^2 = 6 \) and \( -x^2 + 3xy + 2y^2 = -6 \)
Problem Statement |
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\( x^2 + 2xy - 2y^2 = 6 \) and \( -x^2 + 3xy + 2y^2 = -6 \)
Final Answer |
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\( ( \sqrt{6}, 0 ) \) and \( ( -\sqrt{6}, 0 ) \)
Problem Statement
\( x^2 + 2xy - 2y^2 = 6 \) and \( -x^2 + 3xy + 2y^2 = -6 \)
Solution
Michel vanBiezen - 4120 video solution
video by Michel vanBiezen |
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Final Answer
\( ( \sqrt{6}, 0 ) \) and \( ( -\sqrt{6}, 0 ) \)
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\(y=x^2-9 \) and \( y=9-x^2\)
Problem Statement
\(y=x^2-9 \) and \( y=9-x^2\)
Solution
MIP4U - 1777 video solution
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
MIP4U - 1778 video solution
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
MIP4U - 1779 video solution
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
Brightstorm - 1780 video solution
video by Brightstorm |
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\( y=x^2 \text{ and } 3x+y=10 \)
Problem Statement
\( y=x^2 \text{ and } 3x+y=10 \)
Solution
Your Math Gal - 1789 video solution
video by Your Math Gal |
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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
Your Math Gal - 1790 video solution
video by Your Math Gal |
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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
Your Math Gal - 1791 video solution
video by Your Math Gal |
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\( 6x-y=5 \text{ and } xy=1 \)
Problem Statement
\( 6x-y=5 \text{ and } xy=1 \)
Solution
Your Math Gal - 1792 video solution
video by Your Math Gal |
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Find the solution(s) in the first quadrant to \( y = 2x^2 + 3 \) and \( y = -11x + 9 \)
Problem Statement |
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Find the solution(s) in the first quadrant to \( y = 2x^2 + 3 \) and \( y = -11x + 9 \)
Final Answer |
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\( ( 0.5, 3.5 ) \)
Problem Statement
Find the solution(s) in the first quadrant to \( y = 2x^2 + 3 \) and \( y = -11x + 9 \)
Solution
Michel vanBiezen - 4123 video solution
video by Michel vanBiezen |
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Final Answer
\( ( 0.5, 3.5 ) \)
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\( x^2 + y^2 = 25 \) and \( y = 4x/3 \)
Problem Statement |
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\( x^2 + y^2 = 25 \) and \( y = 4x/3 \)
Final Answer |
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\( ( 3,4 ) \) and \( ( -3,-4 ) \)
Problem Statement
\( x^2 + y^2 = 25 \) and \( y = 4x/3 \)
Solution
yaymath - 4124 video solution
video by yaymath |
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Final Answer
\( ( 3,4 ) \) and \( ( -3,-4 ) \)
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\( y = -x^2 + 4 \) and \( y = 2x + 1 \)
Problem Statement |
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\( y = -x^2 + 4 \) and \( y = 2x + 1 \)
Final Answer |
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\( ( -3,-5 ) \) and \( ( 1,3 ) \)
Problem Statement
\( y = -x^2 + 4 \) and \( y = 2x + 1 \)
Solution
yaymath - 4125 video solution
video by yaymath |
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Final Answer
\( ( -3,-5 ) \) and \( ( 1,3 ) \)
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\( x^2 + y^2 = 45 \) and \( y^2 - x^2 = 27 \)
Problem Statement |
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\( x^2 + y^2 = 45 \) and \( y^2 - x^2 = 27 \)
Final Answer |
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4 solutions, all combinations of \( ( \pm 3, \pm 6 ) \)
Problem Statement
\( x^2 + y^2 = 45 \) and \( y^2 - x^2 = 27 \)
Solution
yaymath - 4126 video solution
video by yaymath |
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Final Answer
4 solutions, all combinations of \( ( \pm 3, \pm 6 ) \)
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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
Khan Academy - 1784 video solution
video by Khan Academy |
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\( y=x+1 \) and \( x^2+y^2=25 \)
Problem Statement
\( y=x+1 \) and \( x^2+y^2=25 \)
Solution
Khan Academy - 1785 video solution
video by Khan Academy |
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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
Khan Academy - 1786 video solution
video by Khan Academy |
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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
Khan Academy - 1787 video solution
video by Khan Academy |
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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
Khan Academy - 1788 video solution
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 \( y^3 + xy^2 = 2 \)
Hint |
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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
blackpenredpen - 3784 video solution
video by blackpenredpen |
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Word Problems
The area of a rectangle is \( 8 \) and the perimeter is 12. What are the dimensions?
Problem Statement |
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The area of a rectangle is \( 8 \) and the perimeter is 12. What are the dimensions?
Final Answer |
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\( ( 2,4 ) \) and \( ( 4,2 ) \)
Problem Statement
The area of a rectangle is \( 8 \) and the perimeter is 12. What are the dimensions?
Solution
Michel vanBiezen - 4121 video solution
video by Michel vanBiezen |
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Final Answer
\( ( 2,4 ) \) and \( ( 4,2 ) \)
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The sum of the first number squared and the second number squared equals 73. The difference between second number squared and three times the first number squared equals 37. Find all possible pairs of numbers.
Problem Statement |
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The sum of the first number squared and the second number squared equals 73. The difference between second number squared and three times the first number squared equals 37. Find all possible pairs of numbers.
Final Answer |
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\( ( \pm 3, \pm 8 ) \)
Problem Statement
The sum of the first number squared and the second number squared equals 73. The difference between second number squared and three times the first number squared equals 37. Find all possible pairs of numbers.
Solution
Michel vanBiezen - 4122 video solution
video by Michel vanBiezen |
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Final Answer
\( ( \pm 3, \pm 8 ) \)
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Inequalities
\( x^2 + y^2 \leq 16 \) and \( x^2 - y^2 \gt 9 \)
For this problem, do not evaluate points. Just graph and shade the area(s) that represent your answer.
Problem Statement
\( x^2 + y^2 \leq 16 \) and \( x^2 - y^2 \gt 9 \)
For this problem, do not evaluate points. Just graph and shade the area(s) that represent your answer.
Solution
yaymath - 4127 video solution
video by yaymath |
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