Systems of Non-Linear Equations

This section covers:

Systems of Non-Linear Equations

(Note that solving trig non-linear equations can be found here).

We learned how to solve linear equations here in the Systems of Linear Equations and Word Problems Section. Sometimes we need solve systems of non-linear equations, such as those we see in conics.

We can use either Substitution or Elimination, depending on what’s easier. The main difference is that we’ll usually end up getting two (or more!) answers for a variable (since we may be dealing with quadratics or higher degree polynomials), and we need to plug in answers to get the other variable. So we’ll typically have multiple sets of answers with non-linear systems.

Here are some examples. Note that we could use factoring to solve the quadratics, but sometimes we will need to use the Quadratic Formula.

Problem and Checking Answers Math

\(\left\{ \begin{array}{l}{{x}^{2}}+{{y}^{2}}=61\\y-x=1\end{array} \right.\)

 

Check the answers after solving:

\(\begin{align}{{\left( {-6} \right)}^{2}}+{{\left( {-5} \right)}^{2}}&=61\,\,\,\surd \\\left( {-5} \right)-\left( {-6} \right)&=1\,\,\,\,\,\,\surd \\{{\left( 5 \right)}^{2}}+{{\left( 6 \right)}^{2}}&=61\,\,\,\surd \\6-5&=1\,\,\,\,\,\,\surd \end{align}\)

First solve for \(y\) in terms of \(x\) in second equation, and then substitute in first equation:

 

\(\begin{array}{c}y=x+1\\{{x}^{2}}+{{\left( {x+1} \right)}^{2}}=61\\{{x}^{2}}+{{x}^{2}}+2x+1=61\\2{{x}^{2}}+2x-60=0\\{{x}^{2}}+x-30=0\end{array}\)

Now factor, and we have two answers for \(x\). Plug each into easiest equation to get \(y\)’s:

 

\(\begin{array}{c}{{x}^{2}}+x-30=0\\\left( {x+6} \right)\left( {x-5} \right)=0\\x=-6\,\,\,\,\,\,\,\,\,x=5\\y=-6+1=-5\,\,\,\,\,y=5+1=6\end{array}\)

 

Answers are: \(\left( {-6,-5} \right)\) and \(\left( {5,6} \right)\)

 

\(\left\{ \begin{array}{l}{{x}^{2}}+{{y}^{2}}=41\\xy=20\end{array} \right.\)

 

Check the answers after solving:

\(\displaystyle \begin{array}{c}{{\left( 4 \right)}^{2}}+\,\,{{\left( 5 \right)}^{2}}=41\,\,\,\surd \\{{\left( {-4} \right)}^{2}}+\,\,{{\left( {-5} \right)}^{2}}=41\,\,\,\surd \\{{\left( 5 \right)}^{2}}+\,\,{{\left( 4 \right)}^{2}}=41\,\,\,\surd \\{{\left( {-5} \right)}^{2}}+\,\,{{\left( {-4} \right)}^{2}}=41\,\,\,\surd \\\left( 4 \right)\left( 5 \right)=20\,\,\,\surd \\\left( {-4} \right)\left( {-5} \right)=20\,\,\,\surd \\\left( 5 \right)\left( 4 \right)=20\,\,\,\surd \\\left( {-5} \right)\left( {-4} \right)=20\,\,\,\surd \,\,\,\,\,\,\end{array}\)

First solve for \(y\) in terms of \(x\) in the second equation, and substitute. We’ll have to multiply both sides by \({{x}^{2}}\) to get rid of fractions:

 

\(\displaystyle \begin{array}{c}y=\tfrac{{20}}{x}\\\,{{x}^{2}}+{{\left( {\tfrac{{20}}{x}} \right)}^{2}}=41\\{{x}^{2}}\left( {{{x}^{2}}+\tfrac{{400}}{{{{x}^{2}}}}} \right)=\left( {41} \right){{x}^{2}}\\\,{{x}^{4}}+400=41{{x}^{2}}\\\,{{x}^{4}}-41{{x}^{2}}+400=0\end{array}\)

Now factor, and we have four answers for \(x\). Plug each into easiest equation to get \(y\)’s:

 

\(\begin{array}{c}{{x}^{4}}-41{{x}^{2}}+400=0\\\left( {{{x}^{2}}-16} \right)\left( {{{x}^{2}}-25} \right)=0\\{{x}^{2}}-16=0\,\,\,\,\,\,{{x}^{2}}-25=0\\x=\pm 4\,\,\,\,\,\,\,\,\,\,x=\pm 5\end{array}\)

 

For \(x=4\): \(y=5\)      \(x=5\): \(y=4\)

\(x=-4\): \(y=-5\)       \(x=-5\): \(y=-4\)

 

Answers are: \(\left( {4,5} \right),\,\,\left( {-4,-5} \right),\,\,\left( {5,4} \right),\) and \(\left( {-5,-4} \right)\)

\(\left\{ \begin{array}{l}4{{x}^{2}}+{{y}^{2}}=25\\3{{x}^{2}}-5{{y}^{2}}=-33\end{array} \right.\)

 

Check the answers after solving:

\(\displaystyle \begin{align}4{{\left( 2 \right)}^{2}}+{{\left( 3 \right)}^{2}}&=25\,\,\surd \,\\\,\,4{{\left( 2 \right)}^{2}}+{{\left( {-3} \right)}^{2}}&=25\,\,\surd \\4{{\left( {-2} \right)}^{2}}+{{\left( 3 \right)}^{2}}&=25\,\,\surd \\4{{\left( {-2} \right)}^{2}}+{{\left( {-3} \right)}^{2}}&=25\,\,\surd \\3{{\left( 2 \right)}^{2}}-5{{\left( 3 \right)}^{2}}&=-33\,\,\surd \\\,\,\,3{{\left( 2 \right)}^{2}}-5{{\left( {-3} \right)}^{2}}&=-33\,\,\surd \\3{{\left( {-2} \right)}^{2}}-5{{\left( 3 \right)}^{2}}&=-33\,\,\surd \,\\3{{\left( {-2} \right)}^{2}}-5{{\left( {-3} \right)}^{2}}&=-33\,\,\surd \end{align}\)

Use linear elimination, and first multiply the first equation by 5:

 

\(\displaystyle \begin{array}{l}5\left( {4{{x}^{2}}+{{y}^{2}}} \right)=5\left( {25} \right)\\\,\,\,20{{x}^{2}}+5{{y}^{2}}=\,125\\\,\,\underline{{\,\,\,3{{x}^{2}}-5{{y}^{2}}=-33}}\\\,\,\,\,23{{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,=92\\\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,=4\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=\pm 2\end{array}\)

For the two answers of \(x\), plug into either equation to get \(y\):

 

\(\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=2:\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=-2:\\4{{\left( 2 \right)}^{2}}+{{y}^{2}}=25\,\,\,\,\,\,\,\,4{{\left( 2 \right)}^{2}}+{{y}^{2}}=25\\{{y}^{2}}=25-16=9\,\,\,\,\,{{y}^{2}}=25-16=9\\\,\,\,\,\,\,\,\,\,y=\pm 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=\pm 3\end{array}\)

 

Answers are: \(\left( {2,3} \right),\,\,\left( {2,-3} \right),\,\,\left( {-2,3} \right),\) and \(\left( {-2,-3} \right)\)

\(\left\{ \begin{array}{l}y={{x}^{3}}-2{{x}^{2}}-3x+8\\y=x\end{array} \right.\)

 

Check the answers after solving:

\(\displaystyle \begin{array}{c}-2={{\left( {-2} \right)}^{3}}-2{{\left( {-2} \right)}^{2}}-3\left( {-2} \right)+8\,\,\surd \\-2=-8-8+6+8\,\,\,\surd \,\end{array}\)

Substitute and solve:

 

\(\begin{array}{c}x={{x}^{3}}-2{{x}^{2}}-3x+8\\{{x}^{3}}-2{{x}^{2}}-4x+8=0\\{{x}^{2}}\left( {x-2} \right)-4\left( {x-2} \right)=0\\\left( {{{x}^{2}}-4} \right)\left( {x-2} \right)=0\\x=\pm 2\end{array}\)

Plug into easiest equation to get \(y\)’s:

 

Answers are: \((-2,-2)\) and \((2,2)\)

\(\left\{ \begin{array}{l}{{x}^{2}}+xy=4\\{{x}^{2}}+2xy=-28\end{array} \right.\)

 

Check the answers after solving:

\(\displaystyle \begin{array}{c}{{\left( 6 \right)}^{2}}+\,\,\left( 6 \right)\left( {-\frac{{16}}{3}} \right)=4\,\,\,\surd \\{{\left( {-6} \right)}^{2}}+\,\,\left( {-6} \right)\left( {\frac{{16}}{3}} \right)=4\,\,\,\surd \\{{6}^{2}}+2\left( 6 \right)\left( {-\frac{{16}}{3}} \right)=-28\,\,\,\surd \\{{\left( {-6} \right)}^{2}}+2\left( {-6} \right)\left( {\frac{{16}}{3}} \right)=-28\,\,\,\surd \end{array}\)

Substitute and solve:

 

\(\require{cancel} \begin{array}{c}y=\frac{{4-{{x}^{2}}}}{x}\\{{x}^{2}}+2\cancel{x}\left( {\frac{{4-{{x}^{2}}}}{{\cancel{x}}}} \right)=-28\\{{x}^{2}}+8-2{{x}^{2}}=-28\\-{{x}^{2}}=-36\\x=\pm 6\end{array}\)

Plug into easiest equation to get \(y\)’s:

 

\(\begin{array}{c}x=6:\,\,\,\,\,\,\,\,\,\,\,\,\,x=-6:\\y=\frac{{4-{{6}^{2}}}}{6}\,\,\,\,\,\,\,\,\,y=\frac{{4-{{{\left( {-6} \right)}}^{2}}}}{{-6}}\\y=-\frac{{16}}{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=\frac{{16}}{3}\end{array}\)

 

Answers are: \(\displaystyle \left( {6,\,\,-\frac{{16}}{3}} \right)\) and \(\displaystyle \left( {-6,\,\,\frac{{16}}{3}} \right)\)

You can also use your graphing calculator:

Non-Linear Problem Math

\(\displaystyle \begin{array}{c}y={{e}^{x}}\\y-4{{x}^{2}}+1=0\end{array}\)

 

Put in graphing calculator:

 

\(\displaystyle \begin{align}{{Y}_{1}}&={{e}^{x}}\\{{Y}_{2}}&=4{{x}^{2}}-1\end{align}\)

 

 

 

Let’s do this one on our graphing calculator. Solve for \(y\) in both cases, graph, and find the intersection. To get my window right, I used ZOOM 6, and then changed the ranges of \(x\) and \(y\) in WINDOW to see the intersections.

 

Then use the intersect feature on the calculator (2nd trace, 5, enter, enter, enter) to find the intersection. (Use trace and arrow keys to get close to each intersection before using intersect). We can see that there are 3 solutions.

The solutions are \(\left( {-.62,.538} \right)\), \(\left( {.945,2.57} \right)\) and \(\left( {4.281,72.303} \right)\).

Non-Linear Equations Application Problems

Here are a few Non-Linear Systems application problems.

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Problem:

The difference of two numbers is 3, and the sum of their cubes is 407. Find the numbers.

Solution:

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Let’s set up a system of non-linear equations: \(\left\{ \begin{array}{l}x-y=3\\{{x}^{3}}+{{y}^{3}}=407\end{array} \right.\). Substituting the \(y\) from the first equation into the second and solving yields:

\(\begin{align}{{x}^{3}}+{{\left( {x-3} \right)}^{3}}&=407\\{{x}^{3}}+\left( {x-3} \right)\left( {{{x}^{2}}-6x+9} \right)&=407\\{{x}^{3}}+{{x}^{3}}-6{{x}^{2}}+9x-3{{x}^{2}}+18x-27&=407\\2{{x}^{3}}-9{{x}^{2}}+27x-434&=0\end{align}\) We’ll have to use synthetic division (let’s try 7; I found this root on the graphing calculator) 😉

\begin{array}{l}\left. {\underline {\,
{\,\,7\,\,} \,}}\! \right| \,\,\,\,\,2\,\,-9\,\,\,\,\,\,27\,\,-434\\\underline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,14\,\,\,\,\,\,\,35\,\,\,\,\,\,\,\,434\,}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,5\,\,\,\,\,\,\,62\,\,\,\,\,\,\,\,\left| \!{\overline {\,
{\,\,0\,\,} \,}} \right. \end{array}

 \(2{{x}^{2}}+5x+62\) is prime (can’t be factored for real numbers), so the only root is 7.

\(x=7\) works, and to find \(y\), we use \(y=x-3\). When \(x=7,\,\,y=4\). The two numbers are 4 and 7. They work!

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Problem:
Lacy is speeding in her car, and sees a parked police car on the side of the road right next to her at \(t=0\) seconds. She immediately decelerates, but the police car accelerates to catch up with her. (Assume the two cars are going in the same direction in parallel paths).

The distance that Lacy has traveled in feet after \(t\) seconds can be modeled by the equation \(d\left( t\right)=150+75t-1.2{{t}^{2}}\). The distance that the police car travels after \(t\) seconds can be modeled by the equation \(d\left( t \right)=4{{t}^{2}}\).

(a)  How long will it take the police car to catch up to Lacy? 

(b)  How many feet has Lacy traveled from the time she saw the police car (time \(t=0\)) until the police car catches up to Lacy?

Solution:

We need to find the intersection of the two functions, since that is when the distances are the same. Remember that the graphs are not necessarily the paths of the cars, but rather a model of the how far they go given a certain time in seconds.

Note that since we can’t factor, we need to use the Quadratic Formula  to get the values for \(t\).

We could also solve the non-linear systems using a Graphing Calculator, as shown below.

Non-Linear Math Graphing Calculator Solution
(a)  We can solve the systems of equations, using substitution by just setting the \(d\left( t \right)\)’s (\(y\)’s) together; we’ll have to use the Quadratic Formula to solve:

 

\(\left\{ \begin{array}{l}d\left( t \right)=150+75t-1.2{{t}^{2}}\\d\left( t \right)=4{{t}^{2}}\end{array} \right.\)

 

\(\displaystyle \begin{array}{c}150+75t-1.2{{t}^{2}}=4{{t}^{2}}\\5.2{{t}^{2}}-75t-150=0\end{array}\)

\(\displaystyle t=\frac{{-\left( {-75} \right)\pm \sqrt{{{{{\left( {-75} \right)}}^{2}}-4\left( {5.2} \right)\left( {-150} \right)}}}}{{2\left( {5.2} \right)}}\)

\(\displaystyle t\approx 16.20\)

 

Note that we only want the positive value for \(t\), so in 16.2 seconds, the police car will catch up with Lacy.

 

(b)  We can plug the \(x\) value (\(t\)) into either equation to get the \(y\) value (\(d(t)\)); it’s easiest to use the second equation: \(d\left( t \right)=4{{\left( {16.2} \right)}^{2}}\approx 1050\). Lacy will have traveled about 1050 feet when the police car catches up to her.

Learn these rules, and practice, practice, practice!

On to Introduction to Vectors  – you are ready!

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