College Algebra/Solving Systems of Equations/Solving Systems of Nonlinear Equations

Solving Systems of Equations

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Solving Systems of Nonlinear Equations

Graphs of Nonlinear Systems

Graphs of nonlinear systems of equations can be used to determine the number of solutions and estimate their values.

A nonlinear system of equations includes at least one equation that is not linear. The nonlinear equation(s) may have at least one variable raised to a power greater than 1 or a product of variables. Graphing the equations in the system gives a visual representation of how many solutions the system has. The possible number of solutions in a system depends on the types of equations in the system.

For example, the graph of a system with one linear equation and one quadratic equation includes a line and a parabola. If they do not intersect, there are no solutions. If they intersect in one point, there is one solution. If they intersect in two points, there are two solutions.
A system of one linear equation and one quadratic equation may have no solution, one solution, or two solutions.
In some cases, exact solutions can be identified from the graph. These solutions should be verified by substituting them back into the original equations. In some systems, a graph may only give an approximate solution.
Step-By-Step Example
Graph Nonlinear Systems of Equations
Solve by graphing:
{y−x=4[1]4y=(x+4)2[2]\left\{\begin{matrix}y-x=4&[1]\\4y=(x+4)^2&[2] \end{matrix}\right.
Step 1
Rewrite each equation so that it is solved for yy.
y−x=4y=x+44y=(x+4)2y=(x+4)24\begin{aligned}y-x&=4\\y&=x+4\end{aligned}\hspace{25pt} \begin{aligned}4y&=(x+4)^2\\y&=\frac{(x+4)^2}{4}\end{aligned}
Step 2

Graph each equation.

The first equation is:
y=x+4y=x+4
The graph is a line. The yy-intercept is 4. The slope is 1. The second equation is:
y=(x+4)24y=\frac{(x+4)^2}{4}
The graph is a parabola. The parabola opens upward. So its vertex is its lowest point. The vertex is at (−4,0)(-4,0). The yy-intercept is 4.
Step 3
Points of intersection are solutions of both equations. Use the graph to find the coordinates of points where the graphs intersect. If a graphing utility was used, the trace function may be helpful in finding the coordinates. The intersection points for the given equations are xx- and yy-intercepts of the graphs.
The graphs intersect at (−4,0)(-4,0) and (0,4)(0,4).
Solution

The solutions are (−4,0)(-4,0) and (0,4)(0,4).

To check that the points of intersection are solutions, substitute the values for xx and yy back into both equations and simplify. If they both result in true equations, then the ordered pairs are solutions.

Check (−4,0)(-4,0):
y−x=4(0)−(−4)=40+4=44=44y=(x+4)24(0)=((−4)+4)20=(0)20=0\begin{aligned}y-x&=4\\(0)-(-4)&=4\\0+4&=4\\4&=4\end{aligned}\hspace{25pt}\begin{aligned}4y&=(x+4)^2\\4(0)&=((-4)+4)^2\\0&=(0)^2\\0&=0\end{aligned}
Check (0,4)(0,4):
y−x=4(4)−(0)=44−0=44=44y=(x+4)24(4)=((0)+4)216=4216=16\begin{aligned}y-x&=4\\(4)-(0)&=4\\4-0&=4\\4&=4\end{aligned}\hspace{25pt}\begin{aligned}4y&=(x+4)^2\\4(4)&=((0)+4)^2\\16&=4^2\\16&=16\end{aligned}

Solving Nonlinear Systems Algebraically

Nonlinear systems can be solved by algebraic methods, such as substitution and elimination.
The same methods of substitution and elimination that are used to solve linear systems of equations can also be utilized to solve nonlinear systems.
Step-By-Step Example
Solve a System of Nonlinear Equations by Substitution
Solve the system of equations by substitution.
{5x2+2y2=−05x2−2y2=−46\begin{cases}5x\phantom{^2}+2y\phantom{^2}=\phantom{-}0\\\phantom{5}x^2-2y^2=-46\end{cases}
Step 1

Solve one of the equations for one of the variables.

If possible, choose a variable that is not squared or raised to a higher power.

Solve for yy.
5x+2y=05x+2y−5x=−5x2y=−5x2y2=−5x2y=−5x2\begin{aligned}5x+2y&=0\\5x+2y-5x&=-5x\\2y&=-5x\\\frac{2y}{2}&=\frac{-5x}{2}\\y&=-\frac{5x}{2}\end{aligned}
Step 2

Substitute the expression from Step 1 for the variable in the other equation.

Substitute −5x2-\frac{5x}{2} for yy in the second equation:
x2−2y2=−46x2−2(−5x2)2=−46\begin{aligned}x^2-2y^2&=-46\\x^2-2\left(-\frac{5x}{2}\right)^2&=-46\end{aligned}
Step 3
Solve for the remaining variable.
x2−2(−5x2)2=−46x2−2(25x24)=−46x2−25x22=−462(x2−25x22)=2(−46)2x2−25x2=−92−23x2=−92−23x2−23=−92−23x2=4x=±2\begin{aligned}x^2-2\left(-\frac{5x}{2}\right)^2&=-46\\x^2-2\left(\frac{25x^2}{4}\right)&=-46\\x^2-\frac{25x^2}{2}&=-46\\2\left(x^2-\frac{25x^2}{2}\right)&=2(-46)\\2x^2-25x^2&=-92\\-23x^2&=-92\\\frac{-23x^2}{-23}&=\frac{-92}{-23}\\x^2&=4\\x&=\pm2\end{aligned}
Step 4
Substitute each of the values found in Step 3 back into either of the original equations and solve to find the corresponding values of the other variable.
5x+2y=05(2)+2y=010+2y=010+2y−10=0−102y=−102y2=−102y=−55x+2y=05(−2)+2y=0−10+2y=0−10+2y+10=0+102y=102y2=102y=5\begin{aligned}5x+2y&=0\\5(2)+2y&=0\\10+2y&=0\\10+2y-10&=0-10\\2y&=-10\\\frac{2y}{2}&=\frac{-10}{2}\\y&=-5\end{aligned}\hspace{20pt}\begin{aligned}5x+2y&=0\\5(-2)+2y&=0\\-10+2y&=0\\-10+2y+10&=0+10\\2y&=10\\\frac{2y}{2}&=\frac{10}{2}\\y&=5\end{aligned}
Solution

The solutions are (2,−5)(2, -5) and (−2,5)(-2,5).

To check the solutions, substitute the pairs of values for xx and yy back into both equations and simplify. If they both result in true equations, then the solutions are correct.

Check (2,−5)(2,-5):
5x+2y=05(2)+2(−5)=010+(−10)=010−10=00=0x2−2y2=−46(2)2−2(−5)2=−464−2(25)=−464−50=−46−46=−46\begin{aligned}5x+2y&=0\\5(2)+2(-5)&=0\\10+(-10)&=0\\10-10&=0\\0&=0\end{aligned}\hspace{20pt}\begin{aligned}x^2-2y^2&=-46\\(2)^2-2(-5)^2&=-46\\4-2(25)&=-46\\4-50&=-46\\-46&=-46\end{aligned}
Check (−2,5)(-2,5):
5x+2y=05(−2)+2(5)=0−10+10=00=00=0x2−2y2=−46(−2)2−2(5)2=−464−2(25)=−464−50=−46−46=−46\begin{aligned}5x+2y&=0\\5(-2)+2(5)&=0\\-10+10&=0\\0&=0\\\phantom{0=0}\end{aligned}\hspace{20pt}\begin{aligned}x^2-2y^2&=-46\\(-2)^2-2(5)^2&=-46\\4-2(25)&=-46\\4-50&=-46\\-46&=-46\end{aligned}
Step-By-Step Example
Solve a System of Nonlinear Equations by Elimination
Solve the system of equations by elimination.
{3x2+2y2=41[1]3x2+2y2=25[2]\begin{cases}3x\phantom{^2}+2y^2=41& [1]\\\phantom{3}x^2+\phantom{2}y^2=25& [2] \end{cases}
Step 1
Multiply both sides of one equation (or both equations, if necessary) by a constant so that one of its terms is the opposite of a term in the other equation.
{3x2+2y2=41[1]3x2+2y2=25[2]\begin{cases}3x\phantom{^2}+2y^2=41& [1]\\\phantom{3}x^2+\phantom{2}y^2=25& [2] \end{cases}
Multiply equation [2] by –2 so that the y2y^2-term has a coefficient of –2, making it the opposite of 2y22y^2 in equation [1]. Label the resulting equation as [3].
−2(x2+y2)=−2(25)−2x2−2y2=−50[3]\begin{aligned}-2(x^2+y^2)&=-2(25)\\-2x^2-2y^2&=-50\hspace{20pt}[3]\end{aligned}
Step 2
Add equations [1] and [3] to eliminate yy.
{−3x2+2y2=−41[1]−2x2−2y2=−50‾[3]{−2x2+3x2=−9000[3]\begin{aligned}\begin{cases}\phantom{-}3x\phantom{^{2}}+2y^2=\phantom{-}41& [1]\\\underline{-2x^2-2y^2=-50}&[3]\end{cases}\\{\phantom{\lbrace}-2x^2+3x\phantom{^{2}}=-9 \phantom{000[3]}}\end{aligned}
Step 3
Solve the resulting equation for xx.
−2x2+3x=−9−2x2+3x+9=−9+9−2x2+3x+9=0(−2x−3)(x−3)=0\begin{aligned}-2x^2+3x&=-9\\-2x^2+3x+9&=-9+9\\-2x^2+3x+9&=0\\(-2x-3)(x-3)&=0\end{aligned}
Use the zero product property (if pq=0pq=0, then either p=0p=0 or q=0q=0).
−2x−3=0−2x=3x=−1.5x−3=0x=3x=3\begin{aligned}-2x-3&=0\\-2x&=3\\x&=-1.5\end{aligned}\hspace{20pt}\begin{aligned}\\x-3&=0\\x&=3\\\phantom{x=3}\end{aligned}
Step 4

Substitute the values from Step 3 into either of the original equations and then solve for the other variable.

Substitute –1.5 and 3 for xx.
3x+2y2=413(−1.5)+2y2=41−4.5+2y2=412y2=45.5y2=22.75y=±22.753x+2y2=413(3)+2y2=419+2y2=412y2=32y2=16y=±4\begin{aligned}3x+2y^2&=41\\3(-1.5)+2y^2&=41\\-4.5+2y^2&=41\\2y^2&=45.5\\y^2&=22.75\\y&=\pm \sqrt{22.75}\end{aligned}\hspace{20pt}\begin{aligned}3x+2y^2&=41\\3(3) + 2y^2 &= 41\\9+2y^2&=41\\2y^2&=32\\y^2&=16\\y&=\pm4\end{aligned}
Solution

The solutions are (3,±4)(3,\pm 4) and (−1.5,±22.75)(-1.5, \pm \sqrt{22.75}).

The solutions can be checked by substituting the values for xx and yy back into the original equations. If a solution makes both equations true, then the solution is correct.

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