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Solve Systems of Nonlinear Equations

Solve Systems of Nonlinear Equations

By the end of this section, you will be able to:

  • Solve a system of nonlinear equations using graphing
  • Solve a system of nonlinear equations using substitution
  • Solve a system of nonlinear equations using elimination
  • Use a system of nonlinear equations to solve applications

Solve a System of Nonlinear Equations Using Graphing

We learned how to solve systems of linear equations with two variables by graphing, substitution and elimination. We will be using these same methods as we look at nonlinear systems of equations with two equations and two variables. A system of nonlinear equations is a system where at least one of the equations is not linear.

For example, each of the following systems is a system of nonlinear equations.

{x2+y2=9x2−y=9{9x2+y2=9y=3x−3{x+y=4y=x2+2 \left\{\begin{array}{l}x^2+y^2=9\\x^2-y=9\end{array}\right. \qquad \left\{\begin{array}{l}9x^2+y^2=9\\y=3x-3\end{array}\right. \qquad \left\{\begin{array}{l}x+y=4\\y=x^2+2\end{array}\right.

System of Nonlinear Equations

A system of nonlinear equations is a system where at least one of the equations is not linear.

Just as with systems of linear equations, a solution of a nonlinear system is an ordered pair that makes both equations true. In a nonlinear system, there may be more than one solution. We will see this as we solve a system of nonlinear equations by graphing.

When we solved systems of linear equations, the solution of the system was the point of intersection of the two lines. With systems of nonlinear equations, the graphs may be circles, parabolas or hyperbolas and there may be several points of intersection, and so several solutions. Once you identify the graphs, visualize the different ways the graphs could intersect and so how many solutions there might be.

To solve systems of nonlinear equations by graphing, we use basically the same steps as with systems of linear equations modified slightly for nonlinear equations. The steps are listed below for reference.

How To: Solve a system of nonlinear equations by graphing

  1. Identify the graph of each equation. Sketch the possible options for intersection.
  2. Graph the first equation.
  3. Graph the second equation on the same rectangular coordinate system.
  4. Determine whether the graphs intersect.
  5. Identify the points of intersection.
  6. Check that each ordered pair is a solution to both original equations.

Example 11.33

Solve the system by graphing:

{x−y=−2y=x2 \left\{\begin{array}{l}x-y=-2\\y=x^2\end{array}\right.

Solution. The first equation is a line and the second is a parabola. Write the line in slope-intercept form, y=x+2y=x+2, and graph it with y=x2y=x^2. The points of intersection appear to be (2,4)(2,4) and (−1,1)(-1,1).

Check (2,4)(2,4):

2−4=−2 ✓4=22 ✓ \begin{array}{rcl} 2-4&=&-2\ \checkmark\\[4pt] 4&=&2^2\ \checkmark \end{array}

Check (−1,1)(-1,1):

−1−1=−2 ✓1=(−1)2 ✓ \begin{array}{rcl} -1-1&=&-2\ \checkmark\\[4pt] 1&=&(-1)^2\ \checkmark \end{array}

The solutions are (2,4)(2,4) and (−1,1)(-1,1).

Solve the system {x+y=4y=x2+2\left\{\begin{array}{l}x+y=4\\y=x^2+2\end{array}\right.. Enter both ordered-pair solutions.

To identify the graph of each equation, keep in mind the characteristics of the x2x^2 and y2y^2 terms of each conic.

Example 11.34

Solve the system by graphing:

{y=−1(x−2)2+(y+3)2=4 \left\{\begin{array}{l}y=-1\\(x-2)^2+(y+3)^2=4\end{array}\right.

Solution. The first graph is a line. The second is a circle with center (2,−3)(2,-3) and radius 2. The line y=−1y=-1 is horizontal and touches the circle at (2,−1)(2,-1). Checking gives

(2−2)2+(−1+3)2=4and−1=−1. (2-2)^2+(-1+3)^2=4 \quad\text{and}\quad -1=-1.

The solution is (2,−1)(2,-1).

Solve the system {x=−6(x+3)2+(y−1)2=9\left\{\begin{array}{l}x=-6\\(x+3)^2+(y-1)^2=9\end{array}\right.. Enter the solution as an ordered pair.

Solve a System of Nonlinear Equations Using Substitution

The graphing method works well when the points of intersection are integers and so easy to read off the graph. But more often it is difficult to read the coordinates of the points of intersection. The substitution method is an algebraic method that will work well in many situations. It works especially well when it is easy to solve one of the equations for one of the variables.

The substitution method is very similar to the substitution method that we used for systems of linear equations. The steps are listed below for reference.

How To: Solve a system of nonlinear equations by substitution

  1. Identify the graph of each equation. Sketch the possible options for intersection.
  2. Solve one of the equations for either variable.
  3. Substitute the expression from Step 2 into the other equation.
  4. Solve the resulting equation.
  5. Substitute each solution in Step 4 into one of the original equations to find the other variable.
  6. Write each solution as an ordered pair.
  7. Check that each ordered pair is a solution to both original equations.

Example 11.35

Solve the system by using substitution:

{9x2+y2=9y=3x−3 \left\{\begin{array}{l}9x^2+y^2=9\\y=3x-3\end{array}\right.

Solution. The first graph is an ellipse and the second is a line. The second equation is already solved for yy. Substitute 3x−33x-3 for yy in the first equation.

9x2+(3x−3)2=99x2+9x2−18x+9=918x2−18x=018x(x−1)=0, \begin{aligned} 9x^2+(3x-3)^2&=9\\ 9x^2+9x^2-18x+9&=9\\ 18x^2-18x&=0\\ 18x(x-1)&=0, \end{aligned}

so x=0x=0 or x=1x=1. Substitute into y=3x−3y=3x-3:

x=0⟹y=−3,x=1⟹y=0. x=0\Longrightarrow y=-3, \qquad x=1\Longrightarrow y=0.

The ordered pairs are (0,−3)(0,-3) and (1,0)(1,0). Substitution in both original equations verifies both solutions.

Solve {4x2+y2=4y=x+2\left\{\begin{array}{l}4x^2+y^2=4\\y=x+2\end{array}\right.. Enter the solution with the larger xx-coordinate.

So far, each system of nonlinear equations has had at least one solution. The next example will show another option.

Example 11.36

Solve the system by using substitution:

{x2−y=0y=x−2 \left\{\begin{array}{l}x^2-y=0\\y=x-2\end{array}\right.

Solution. The first graph is a parabola and the second is a line. Since y=x−2y=x-2, substitute x−2x-2 for yy in the first equation:

x2−(x−2)=0⟹x2−x+2=0. x^2-(x-2)=0 \quad\Longrightarrow\quad x^2-x+2=0.

This does not factor easily, so check the discriminant:

b2−4ac=(−1)2−4(1)(2)=−7. b^2-4ac=(-1)^2-4(1)(2)=-7.

The discriminant is negative, so there is no real solution. The system has no solution.

Solve {x2−y=0y=2x−3\left\{\begin{array}{l}x^2-y=0\\y=2x-3\end{array}\right.. How many real solutions are there?

Solve a System of Nonlinear Equations Using Elimination

When we studied systems of linear equations, we used the method of elimination to solve the system. We can also use elimination to solve systems of nonlinear equations. It works well when the equations have both variables squared. When using elimination, we try to make the coefficients of one variable to be opposites, so when we add the equations together, that variable is eliminated.

The elimination method is very similar to the elimination method that we used for systems of linear equations. The steps are listed for reference.

How To: Solve a system of equations by elimination

  1. Identify the graph of each equation. Sketch the possible options for intersection.
  2. Write both equations in standard form.
  3. Make the coefficients of one variable opposites. Decide which variable you will eliminate. Multiply one or both equations so that the coefficients of that variable are opposites.
  4. Add the equations resulting from Step 3 to eliminate one variable.
  5. Solve for the remaining variable.
  6. Substitute each solution from Step 5 into one of the original equations. Then solve for the other variable.
  7. Write each solution as an ordered pair.
  8. Check that each ordered pair is a solution to both original equations.

Example 11.37

Solve the system by elimination:

{x2+y2=4x2−y=4 \left\{\begin{array}{l}x^2+y^2=4\\x^2-y=4\end{array}\right.

Solution. The graphs are a circle and a parabola. Both equations are in standard form. Multiply the second equation by −1-1 and add:

x2+y2=4−x2+y=−4y2+y=0. \begin{array}{rcl} x^2+y^2&=&4\\ -x^2+y&=&-4\\ \hline y^2+y&=&0. \end{array}

Thus y(y+1)=0y(y+1)=0, so y=0y=0 or y=−1y=-1. Substitute into x2−y=4x^2-y=4:

y=0:x2=4,x=±2,y=−1:x2=3,x=±3. \begin{aligned} y=0&:\quad x^2=4,\quad x=\pm2,\\ y=-1&:\quad x^2=3,\quad x=\pm\sqrt3. \end{aligned}

The solutions are (−2,0)(-2,0), (2,0)(2,0), (3,−1)(\sqrt3,-1), and (−3,−1)(-\sqrt3,-1). We leave the checks for each of the four solutions to you.

For {x2+y2=9x2−y=9\left\{\begin{array}{l}x^2+y^2=9\\x^2-y=9\end{array}\right., enter the yy-coordinate shared by the solutions (3,y)(3,y) and (−3,y)(-3,y).

There are also four options when we consider a circle and a hyperbola.

Example 11.38

Solve the system by elimination:

{x2+y2=7x2−y2=1 \left\{\begin{array}{l}x^2+y^2=7\\x^2-y^2=1\end{array}\right.

Solution. The graphs are a circle and a hyperbola. Both equations are in standard form. The coefficients of y2y^2 are opposite, so add the equations:

2x2=8,x2=4,x=±2. 2x^2=8,\qquad x^2=4,\qquad x=\pm2.

Substitute x=2x=2 and x=−2x=-2 into either original equation:

4+y2=7,y2=3,y=±3. 4+y^2=7,\qquad y^2=3,\qquad y=\pm\sqrt3.

The solutions are (−2,3)(-2,\sqrt3), (−2,−3)(-2,-\sqrt3), (2,3)(2,\sqrt3), and (2,−3)(2,-\sqrt3). We leave the checks for each of the four solutions to you.

For {x2+y2=25y2−x2=7\left\{\begin{array}{l}x^2+y^2=25\\y^2-x^2=7\end{array}\right., enter the positive yy-coordinate of a solution.

Use a System of Nonlinear Equations to Solve Applications

Systems of nonlinear equations can be used to model and solve many applications. We will look at an everyday geometric situation as our example.

Example 11.39

The difference of the squares of two numbers is 15. The sum of the numbers is 5. Find the numbers.

Solution. Let xx be the first number and yy the second number. Translate the information into a system:

{x2−y2=15x+y=5 \left\{\begin{array}{l}x^2-y^2=15\\x+y=5\end{array}\right.

Solve the second equation for xx, x=5−yx=5-y, and substitute:

(5−y)2−y2=1525−10y+y2−y2=1525−10y=15−10y=−10y=1. \begin{aligned} (5-y)^2-y^2&=15\\ 25-10y+y^2-y^2&=15\\ 25-10y&=15\\ -10y&=-10\\ y&=1. \end{aligned}

Then x+1=5x+1=5, so x=4x=4. The numbers are 1 and 4.

The difference of the squares of two numbers is −20-20. The sum of the numbers is 1010. Enter the smaller number.

Example 11.40

Myra purchased a small 25-inch TV for her kitchen. The size of a TV is measured on the diagonal of the screen. The screen also has an area of 300 square inches. What are the length and width of the TV screen?

Solution. Let xx be the width of the rectangle and yy its length. The diagonal of the right triangle is 25 inches, and the area is 300 square inches:

{x2+y2=625xy=300 \left\{\begin{array}{l}x^2+y^2=625\\xy=300\end{array}\right.

Solve the second equation for xx, x=300yx=\frac{300}{y}, and substitute:

(300y)2+y2=62590,000y2+y2=62590,000+y4=625y2y4−625y2+90,000=0(y2−225)(y2−400)=0. \begin{aligned} \left(\frac{300}{y}\right)^2+y^2&=625\\ \frac{90{,}000}{y^2}+y^2&=625\\ 90{,}000+y^4&=625y^2\\ y^4-625y^2+90{,}000&=0\\ (y^2-225)(y^2-400)&=0. \end{aligned}

Thus y=±15y=\pm15 or y=±20y=\pm20. Since yy is a side of the rectangle, discard the negative values. If the length is 15 inches, the width is 20 inches. If the length is 20 inches, the width is 15 inches.

Edgar purchased a small 20-inch TV. Its screen has an area of 192192 square inches. Enter the shorter side length.

Key terms. A system of nonlinear equations is a system where at least one of the equations is not linear.

This section is adapted from Intermediate Algebra 2e, Section 11.5 by Lynn Marecek and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at OpenStax. Changes: omitted readiness quizzes, practice sets, self-checks, media links, and complex source figures; converted selected Try It problems to interactive questions and described graphing constructions in words.

HyperbolasSequences, Series and Binomial Theorem