Skip to main content
Mathematics LibreTexts

8.3: Solve Systems of Nonlinear Equations

  1. Last updated
  2. Save as PDF
  • Page ID
    66403

    • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Learning Objectives

    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
    Be Prepared

    Before you get started, take this readiness quiz.

    1. Solve the system by graphing: \(\left\{\begin{array}{l}{x-3 y=-3} \\ {x+y=5}\end{array}\right.\).
    2. Solve the system by substitution: \(\left\{\begin{array}{l}{x-4 y=-4} \\ {-3 x+4 y=0}\end{array}\right.\)
    3. Solve the system by elimination: \(\left\{\begin{array}{l}{3 x-4 y=-9} \\ {5 x+3 y=14}\end{array}\right.\)

    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.

    \(\left\{\begin{array}{l}{x^{2}+y^{2}=9} \\ {x^{2}-y=9}\end{array}\right. \)

    \(\left\{\begin{array}{l}{9 x^{2}+y^{2}=9} \\ {y=3 x-3}\end{array}\right.\)

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

    Definition \(\PageIndex{1}\)

    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.

    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.

    Solve a System of Nonlinear Equations by Graphing.

    Example \(\PageIndex{2}\)

    Solve the system by graphing: \(\left\{\begin{array}{l}{x-y=-2} \\ {y=x^{2}}\end{array}\right.\)

    Solution
    Identify each graph. \(\left\{\begin{array}{ll}{x-y=-2} & {\text { line }} \\ {y=x^{2}} & {\text { parabola }}\end{array}\right.\)
    Sketch the possible options for intersection of a parabola and a line. .

    Graph the line, \(x-y=-2\).

    Slope-intercept form \(y=x+2\).

    Graph the parabola, \(y=x^{2}\).

    		 .
    Identify the points of intersection. The points of intersection appear to be \((2,3)\) and \((-1,1)\).

    Check to make sure each solution makes both equations true.

    \((2,4)\)

    \((-1,1)\)

    		  
      The solutions are \((2,4)\) and \((-1,1)\).
    Try It \(\PageIndex{3}\)

    Solve the system by graphing: \(\left\{\begin{array}{l}{x+y=4} \\ {y=x^{2}+2}\end{array}\right.\).

    Answer
    This graph shows the equations of a system, x plus y is equal to 4 and y is equal x squared plus 2, and the x y-coordinate plane. The line has a slope of negative 1 and a y-intercept at 4. The vertex of the parabola is (0, 2) and opens upward. The line and parabola intersect at the points (negative 2, 6) and (1, 3), which are labeled.
     
    Try It \(\PageIndex{4}\)

    Solve the system by graphing: \(\left\{\begin{array}{l}{x-y=-1} \\ {y=-x^{2}+3}\end{array}\right.\)

    Answer
    This graph shows the equations of a system, x minus y is equal to negative 1 and y is equal to negative x squared plus three, and the x y-coordinate plane. The line has a slope of 1 and a y-intercept at 1. The vertex of the parabola is (0, negative 3) and opens upward. The line and parabola intersect at the points (negative 2, negative 1) and (1, 2), which are labeled.
     

    To identify the graph of each equation, keep in mind the characteristics of the \(x^{2}\) and \(y^{2}\) terms of each conic.

    Example \(\PageIndex{5}\)

    Solve the system by graphing: \(\left\{\begin{array}{l}{y=-1} \\ {(x-2)^{2}+(y+3)^{2}=4}\end{array}\right.\).

    Solution
    Identify each graph. \(\left\{\begin{array}{ll}{y=-1} & {\text { line }} \\ {(x-2)^{2}+(y+3)^{2}=4} & {\text { circle }}\end{array}\right.\)
    Sketch the possible options for the intersection of a circle and a line. .

    Graph the circle, \((x-2)^{2}+(y+3)^{2}=4\)

    Center: \((2,-3)\) radius: \(2\)

    Graph the line, \(y=-1\).

    It is a horizontal line.

    		 .
    Identify the points of intersection. The point of intersection appears to be\((2,-1)\).

    Check to make sure the solution makes both equations true.

    \((2,-1)\)

    \(\begin{array} {r r} {(x-2)^{2}+(y+3)^{2}=4} \quad\quad {y=-1} \\ {(2-2)^{2}+(-1+3)^{2}\stackrel{?}{=}4}\quad{-1=-1} \\ {(0)^{2}+(2)^{2}\stackrel{?}{=}4}\quad\quad\quad\quad\quad \\ {4=4}\quad\quad\quad\quad\quad \end{array}\)
      The solution is \((2,-1)\)
    Try It \(\PageIndex{6}\)

    Solve the system by graphing: \(\left\{\begin{array}{l}{x=-6} \\ {(x+3)^{2}+(y-1)^{2}=9}\end{array}\right.\)

    Answer
    This graph shows the equations of a system, x is equal to negative 6 and the quantity x plus 3 squared plus the quantity y minus 1 squared is equal to 9, which is a circle, on the x y-coordinate plane. The line is a vertical line. The center of the circle is (negative 3, 1) and it has a radius of 3 units. The point of intersection between the line and circle is (negative 6, 1).
     
    Try It \(\PageIndex{7}\)

    Solve the system by graphing: \(\left\{\begin{array}{l}{y=4} \\ {(x-2)^{2}+(y+3)^{2}=4}\end{array}\right.\)

    Answer
    This graph shows the equations of a system, y is equal to negative 4 and the quantity x minus 2 squared plus the quantity y plus 3 squared is equal to 4, which is a circle, on the x y-coordinate plane. The line is a horizontal line. The center of the circle is (2, negative 3) and it has a radius of 2 units. There is no point of intersection between the line and circle, so the system has no solution.
     

    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.

    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 \(\PageIndex{8}\)

    Solve the system by using substitution: \(\left\{\begin{array}{l}{9 x^{2}+y^{2}=9} \\ {y=3 x-3}\end{array}\right.\)

    Solution
    Identify each graph. \(\left\{\begin{array}{ll}{9 x^{2}+y^{2}=9} & {\text { ellipse }} \\ {y=3 x-3} & {\text { line }}\end{array}\right.\)
    Sketch the possible options for intersection of an ellipse and a line. .
    The equation \(y=3x-3\) is solved for \(y\). .
      .
    Substitute \(3x-3\) for \(y\) in the first equation. .
    Solve the equation for \(x\). .
      .
    Substitute \(x=0\) and \(x=1\) into \(y=3x-3\) to find \(y\)-. .
      .
      The ordered pairs are \((0,-3), (1,0)\).

    Check both ordered pairs in both equations.

    \((0,-3)\)

    \(\begin{array} {r l}{9 x^{2}+y^{2}=9} &\quad { y=3 x-3} \\ {9\cdot0^{2}+(-3)^{2}\stackrel{?}{=}9}&\quad{-3\stackrel{?}{=}3\cdot0-3} \\ {0+9\stackrel{?}{=}9}&\quad{-3\stackrel{?}{=}0-3} \\ {9=9}&\quad{-3=-3} \end{array}\)

    \((1,0)\)

    \(\begin{array} {r l}{9 x^{2}+y^{2}=9} &\quad { y=3 x-3} \\ {9\cdot 1^{2}+(0)^{2}\stackrel{?}{=}9}&\quad{0\stackrel{?}{=}3\cdot 1-3} \\ {9+0\stackrel{?}{=}9}&\quad{0\stackrel{?}{=}3-3} \\ {9=9}&\quad{0=0} \end{array}\)
      The solutions are \((0,-3), (1,0)\).
    Try It \(\PageIndex{9}\)

    Solve the system by using substitution: \(\left\{\begin{array}{l}{x^{2}+9 y^{2}=9} \\ {y=\dfrac{1}{3} x-3}\end{array}\right.\)

    Answer

    No solution

    Try It \(\PageIndex{10}\)

    Solve the system by using substitution: \(\left\{\begin{array}{l}{4 x^{2}+y^{2}=4} \\ {y=x+2}\end{array}\right.\)

    Answer

    \(\left(-\dfrac{4}{5}, \dfrac{6}{5}\right),(0,2)\)

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

    Example \(\PageIndex{11}\)

    Solve the system by using substitution: \(\left\{\begin{array}{l}{x^{2}-y=0} \\ {y=x-2}\end{array}\right.\)

    Solution
    Identify each graph. \(\left\{\begin{array}{ll}{x^{2}-y=0} & {\text { parabola }} \\ {y=x-2} & {\text { line }}\end{array}\right.\)
    Sketch the possible options for intersection of a parabola and a line. .
    The equation \(y=x-2\) is solved for \(y\).

    \(y=x-2\)
     

    \(x^2-y=0\)
    Substitute \(x-2\) for \(y\) in the first equation.

    \(x^2-(x-2)=0\)
    Solve the equation for \(x\).

    \(x^2-x+2=0\)
    This doesn't factor easily, so we can check the discriminant.  
     

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

    The system has no solution.
    Try It \(\PageIndex{12}\)

    Solve the system by using substitution: \(\left\{\begin{array}{l}{x^{2}-y=0} \\ {y=2 x-3}\end{array}\right.\)

    Answer

    No solution

    Try It \(\PageIndex{13}\)

    Solve the system by using substitution: \(\left\{\begin{array}{l}{y^{2}-x=0} \\ {y=3 x-2}\end{array}\right.\)

    Answer

    \(\left(\dfrac{4}{9},-\dfrac{2}{3}\right),(1,1)\)

    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.

    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 \(\PageIndex{14}\)

    Solve the system by elimination: \(\left\{\begin{array}{l}{x^{2}+y^{2}=4} \\ {x^{2}-y=4}\end{array}\right.\)

    Solution
    Identify each graph. .
    Sketch the possible options for intersection of a circle and a parabola. .
    Both equations are in standard form. .
    To get opposite coefficients of \(x^{2}\), we will multiply the second equation by \(-1\). .
    Simplify. .
    Add the two equation to eliminate \(x^{2}\)/ .
    Solve for \(y\). .
      .
    Substitute \(y=0\) and \(y=-1\) into one of the original equations. Then solve for \(x\). .
      .
    Write each solution as an ordered pair. The ordered pairs are \((-2,0)(2,0)\). \((\sqrt{3},-1)(-\sqrt{3},-1)\)
    Check that each ordered pair is a solution to both original equations.  
    We will leave the checks for each of the four solutions to you. The solutions are \((-2,0),(2,0),(\sqrt{3},-1)\), and \((-\sqrt{3},-1)\).
     
    Try It \(\PageIndex{15}\)

    Solve the system by elimination: \(\left\{\begin{array}{l}{x^{2}+y^{2}=9} \\ {x^{2}-y=9}\end{array}\right.\)

    Answer

    \((-3,0),(3,0),(-2 \sqrt{2},-1),(2 \sqrt{2},-1)\)

    Try It \(\PageIndex{16}\)

    Solve the system by elimination: \(\left\{\begin{array}{l}{x^{2}+y^{2}=1} \\ {-x+y^{2}=1}\end{array}\right.\)

    Answer

    \((-1,0),(0,1),(0,-1)\)

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

    Example \(\PageIndex{17}\)

    Solve the system by elimination: \(\left\{\begin{array}{l}{x^{2}+y^{2}=7} \\ {x^{2}-y^{2}=1}\end{array}\right.\)

    Solution
    Identify each graph. \(\left\{\begin{array}{ll}{x^{2}+y^{2}=7} & {\text { circle }} \\ {x^{2}-y^{2}=1} & {\text { hyperbola }}\end{array}\right.\)
    Sketch the possible options for intersection of a circle and hyperbola. .
    Both equations are in standard form. \(\left\{\begin{array}{l}{x^{2}+y^{2}=7} \\ {x^{2}-y^{2}=1}\end{array}\right.\)
    The coefficients of \(y^{2}\) are opposite, so we will add the equations.

    \(\left\{\begin{array}{l}{x^{2}+y^{2}=7} \\ {x^{2}-y^{2}=1}\end{array}\right.\)

    \(2 x^{2}=8\)
    Simplify. \(x^{2}=4\)
    \(x=\pm 2\)
    \(x=2 \quad x=-2\)
    Substitue \(x=2\) and \(x=-2\) into one of the original equations. Then solve for \(y\). \(\begin{array}{rl}{x^{2}+y^{2} = 7} &\quad { x^{2}+y^{2}=7} \\ {2^{2}+y^{2}=7} & \quad{(-2)^{2}+y^{2}=7} \\ {4+y^{2}=7} &\quad {4+y^{2}=7} \\ {y^{2}=3} &\quad {y^{2}=3} \\ {y=\pm \sqrt{3}} &\quad {y=\pm \sqrt{3}}\end{array}\)
    Write each solution as an ordered pair. The ordered pairs are \((-2, \sqrt{3}),(-2,-\sqrt{3})\), \((2, \sqrt{3}),\) and \((2,-\sqrt{3})\).
    Check that the ordered pair is a solution to both original equations.  
    We will leave the checks for each of the four solutions to you. The solutions are \((-2, \sqrt{3}),(-2,-\sqrt{3}),(2, \sqrt{3})\), and \((2,-\sqrt{3})\).
    Try It \(\PageIndex{18}\)

    Solve the system by elimination: \(\left\{\begin{array}{l}{x^{2}+y^{2}=25} \\ {y^{2}-x^{2}=7}\end{array}\right.\)

    Answer

    \((-3,-4),(-3,4),(3,-4),(3,4)\)

    Try It \(\PageIndex{19}\)

    Solve the system by elimination: \(\left\{\begin{array}{l}{x^{2}+y^{2}=4} \\ {x^{2}-y^{2}=4}\end{array}\right.\)

    Answer

    \((-2,0),(2,0)\)

    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 \(\PageIndex{20}\)

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

    Solution
    Identify what we are looking for. Two different numbers.
    Define the variables.

    \(x\)=first number

    \(y\)=second number
    Translate the information into a system of equations.  
    First sentence. The difference of the squares of two numbers is \(15\).
      \(x^2-y^2=15\)
    Second sentence. The sum of the numbers is \(5\).
      \(x+y=5\)
    Solve the system by substitution. .
    Solve the second equation for \(x\). \(x=5-y\)
    Substitute \(x\) into the first equation. \(x^2-y^2=15\)
      \((5-y)^2-y^2=15\)
    Expand and simplify. \((25-10y+y^2)-y^2=15\)
     

    \(\begin{aligned} 25-10y+y^2-y^2&=15 \\ 25-10y&=15\end{aligned}\)
    Solve for \(y\). \(-10y=-10\)
      \(y=1\)
    Substitute back into the second equation. \(x+y=5\)
     

    \(x+1=5\)

    \(x=4\)
      The numbers are \(1\) and \(4\).
    Try It \(\PageIndex{21}\)

    The difference of the squares of two numbers is \(−20\). The sum of the numbers is \(10\). Find the numbers.

    Answer

    \(4\) and \(6\)

    Try It \(\PageIndex{22}\)

    The difference of the squares of two numbers is \(35\). The sum of the numbers is \(−1\). Find the numbers.

    Answer

    \(-18\) and \(17\)

    Example \(\PageIndex{23}\)

    Myra purchased a small \(25\)” 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
    Identify what we are looking for. The length and width of the rectangle.
    Define the variables.

    Let \(x\)= width of the rectangle

    \(y\)=length of the rectangle
    Draw a diagram to help visualize the situation. .
      Area is \(300\) square inches.
    Translate the information into a system of equations. The diagonal of the right triangle is \(25\) inches.
      \(\begin{aligned} x^2+y^2 &= 25^2\\ x^2+y^2 &= 625\end{aligned}\)
      The area of the rectangle is \(300\) square inches.
      .
    Solve the system using substitution. \(xy=300\)
    Solve the second equation for \(x\). \(x=\dfrac{300}{y}\)
    Substitute \(x\) into the first equation. \(x^2+y^2=625\)
      .
    Simplify. .
    Multiply by \(y^{2}\) to clear the fractions. .
    Put in standard form. .
    Solve by factoring. .
      .
      .
    Since \(y\) is a side of the rectangle, we discard the negative values. .
    Substitute back into the second equation. .
      .
      If the length is \(15\) inches, the width is \(20\) inches.
      If the length is \(20\) inches, the width is \(15\) inches.
    Try It \(\PageIndex{24}\)

    Edgar purchased a small \(20\)” TV for his garage. The size of a TV is measured on the diagonal of the screen. The screen also has an area of \(192\) square inches. What are the length and width of the TV screen?

    Answer

    If the length is \(12\) inches, the width is \(16\) inches. If the length is \(16\) inches, the width is \(12\) inches.

    Try It \(\PageIndex{25}\)

    The Harper family purchased a small microwave for their family room. The diagonal of the door measures \(15\) inches. The door also has an area of \(108\) square inches. What are the length and width of the microwave door?

    Answer

    If the length is \(12\) inches, the width is \(9\) inches. If the length is \(9\) inches, the width is \(12\) inches.

    Key Concepts

    • 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.
    • 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.
    • 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.

    Practice Makes Perfect

     

    Note that answers to even exercises are provided.

    Solve a System of Nonlinear Equations Using Graphing

    In the following exercises, solve the system of equations by using graphing.

    1. \(\left\{\begin{array}{l}{y=2 x+2} \\ {y=-x^{2}+2}\end{array}\right.\)
    2. \(\left\{\begin{array}{l}{y=6 x-4} \\ {y=2 x^{2}}\end{array}\right.\)
    3. \(\left\{\begin{array}{l}{x+y=2} \\ {x=y^{2}}\end{array}\right.\)
    4. \(\left\{\begin{array}{l}{x-y=-2} \\ {x=y^{2}}\end{array}\right.\)
    5. \(\left\{\begin{array}{l}{y=\dfrac{3}{2} x+3} \\ {y=-x^{2}+2}\end{array}\right.\)
    6. \(\left\{\begin{array}{l}{y=x-1} \\ {y=x^{2}+1}\end{array}\right.\)
    7. \(\left\{\begin{array}{l}{x=-2} \\ {x^{2}+y^{2}=4}\end{array}\right.\)
    8. \(\left\{\begin{array}{l}{y=-4} \\ {x^{2}+y^{2}=16}\end{array}\right.\)
    9. \(\left\{\begin{array}{l}{x=2} \\ {(x+2)^{2}+(y+3)^{2}=16}\end{array}\right.\)
    10. \(\left\{\begin{array}{l}{y=-1} \\ {(x-2)^{2}+(y-4)^{2}=25}\end{array}\right.\)
    11. \(\left\{\begin{array}{l}{y=-2 x+4} \\ {y=\sqrt{x}+1}\end{array}\right.\)
    12. \(\left\{\begin{array}{l}{y=-\dfrac{1}{2} x+2} \\ {y=\sqrt{x}-2}\end{array}\right.\)
    Answer

    2.

    This graph shows the equations of a system, y is equal to 6 x minus 4 which is a line and y is equal to 2 x squared which is a parabola, on the x y-coordinate plane. The vertex of the parabola is (0, 0) and the parabola opens upward. The line has a slope of 6. The line and parabola intersect at the points (1, 2) and (2, 8), which are labeled. The solutions are (1, 2) and (2, 8).
     

    4.

    This graph shows the equations of a system, x minus y is equal to negative 2 which is a line and x is equal to y squared which is a rightward-opening parabola, on the x y-coordinate plane. The vertex of the parabola is (0, 0) and it passes through the points (1, 1) and (1, negative 1). The line has a slope of 1 and a y-intercept at 2. The line and parabola do not intersect, so the system has no solution.
     

    6.

    This graph shows the equations of a system, y is x minus 1 which is a line and y is equal to x squared plus 1 which is an upward-opening parabola, on the x y-coordinate plane. The vertex of the parabola is (0, 1) and it passes through the points (negative 1, 2) and (1, 2). The line has a slope of 1 and a y-intercept at negative 1. The line and parabola do not intersect, so the system has no solution.
     

    8.

    This graph shows the equations of a system, x is equal to negative 2 which is a line and x squared plus y squared is equal to 16 which is a circle, on the x y-coordinate plane. The line is horizontal. The center of the circle is (0, 0) and the radius of the circle is 4. The line and circle intersect at (negative 2, 0), so the solution of the system is (negative 2, 0).
     

    10.

    This graph shows the equations of a system, x is equal to 2 which is a line and the quantity x minus 2 end quantity squared plus the quantity y minus 4 end quantity squared is equal to 25 which is a circle, on the x y-coordinate plane. The line is horizontal. The center of the circle is (2, 4) and the radius of the circle is 5. The line and circle intersect at (2, negative 1), so the solution of the system is (2, negative 1).
     

    12.

    This graph shows the equations of a system, y is equal to negative one-half x plus 2 which is a line and the y is equal to the square root of x minus 2, on the x y-coordinate plane. The curve for y is equal to the square root of x minus 2 The curve for y is equal to the square root of x plus 1 where x is greater than or equal to 0 and y is greater than or equal to negative 2. The line and square root curve intersect at (4, 0), so the solution is (4, 0).
     
    Solve a System of Nonlinear Equations Using Substitution

    In the following exercises, solve the system of equations by using substitution.

    1. \(\left\{\begin{array}{l}{x^{2}+4 y^{2}=4} \\ {y=\dfrac{1}{2} x-1}\end{array}\right.\)
    2. \(\left\{\begin{array}{l}{9 x^{2}+y^{2}=9} \\ {y=3 x+3}\end{array}\right.\)
    3. \(\left\{\begin{array}{l}{9 x^{2}+y^{2}=9} \\ {y=x+3}\end{array}\right.\)
    4. \(\left\{\begin{array}{l}{9 x^{2}+4 y^{2}=36} \\ {x=2}\end{array}\right.\)
    5. \(\left\{\begin{array}{l}{4 x^{2}+y^{2}=4} \\ {y=4}\end{array}\right.\)
    6. \(\left\{\begin{array}{l}{x^{2}+y^{2}=169} \\ {x=12}\end{array}\right.\)
    7. \(\left\{\begin{array}{l}{3 x^{2}-y=0} \\ {y=2 x-1}\end{array}\right.\)
    8. \(\left\{\begin{array}{l}{2 y^{2}-x=0} \\ {y=x+1}\end{array}\right.\)
    9. \(\left\{\begin{array}{l}{y=x^{2}+3} \\ {y=x+3}\end{array}\right.\)
    10. \(\left\{\begin{array}{l}{y=x^{2}-4} \\ {y=x-4}\end{array}\right.\)
    11. \(\left\{\begin{array}{l}{x^{2}+y^{2}=25} \\ {x-y=1}\end{array}\right.\)
    12. \(\left\{\begin{array}{l}{x^{2}+y^{2}=25} \\ {2 x+y=10}\end{array}\right.\)
    Answer

    14. \((-1,0),(0,3)\)

    16. \((2,0)\)

    18. \((12,-5),(12,5)\)

    20. No solution

    22. \((0,-4),(1,-3)\)

    24. \((3,4),(5,0)\)

    Solve a System of Nonlinear Equations Using Elimination

    In the following exercises, solve the system of equations by using elimination.

    1. \(\left\{\begin{array}{l}{x^{2}+y^{2}=16} \\ {x^{2}-2 y=8}\end{array}\right.\)
    2. \(\left\{\begin{array}{l}{x^{2}+y^{2}=16} \\ {x^{2}-y=4}\end{array}\right.\)
    3. \(\left\{\begin{array}{l}{x^{2}+y^{2}=4} \\ {x^{2}+2 y=1}\end{array}\right.\)
    4. \(\left\{\begin{array}{l}{x^{2}+y^{2}=4} \\ {x^{2}-y=2}\end{array}\right.\)
    5. \(\left\{\begin{array}{l}{x^{2}+y^{2}=9} \\ {x^{2}-y=3}\end{array}\right.\)
    6. \(\left\{\begin{array}{l}{x^{2}+y^{2}=4} \\ {y^{2}-x=2}\end{array}\right.\)
    7. \(\left\{\begin{array}{l}{x^{2}+y^{2}=25} \\ {2 x^{2}-3 y^{2}=5}\end{array}\right.\)
    8. \(\left\{\begin{array}{l}{x^{2}+y^{2}=20} \\ {x^{2}-y^{2}=-12}\end{array}\right.\)
    9. \(\left\{\begin{array}{l}{x^{2}+y^{2}=13} \\ {x^{2}-y^{2}=5}\end{array}\right.\)
    10. \(\left\{\begin{array}{l}{x^{2}+y^{2}=16} \\ {x^{2}-y^{2}=16}\end{array}\right.\)
    11. \(\left\{\begin{array}{l}{4 x^{2}+9 y^{2}=36} \\ {2 x^{2}-9 y^{2}=18}\end{array}\right.\)
    12. \(\left\{\begin{array}{l}{x^{2}-y^{2}=3} \\ {2 x^{2}+y^{2}=6}\end{array}\right.\)
    13. \(\left\{\begin{array}{l}{4 x^{2}-y^{2}=4} \\ {4 x^{2}+y^{2}=4}\end{array}\right.\)
    14. \(\left\{\begin{array}{l}{x^{2}-y^{2}=-5} \\ {3 x^{2}+2 y^{2}=30}\end{array}\right.\)
    15. \(\left\{\begin{array}{l}{x^{2}-y^{2}=1} \\ {x^{2}-2 y=4}\end{array}\right.\)
    16. \(\left\{\begin{array}{l}{2 x^{2}+y^{2}=11} \\ {x^{2}+3 y^{2}=28}\end{array}\right.\)
    Answer

    26. \((0,-4),(-\sqrt{7}, 3),(\sqrt{7}, 3)\)

    28. \((0,-2),(-\sqrt{3}, 1),(\sqrt{3}, 1)\)

    30. \((-2,0),(1,-\sqrt{3}),(1, \sqrt{3})\)

    32. \((-2,-4),(-2,4),(2,-4),(2,4)\)

    34. \((-4,0),(4,0)\)

    36. \((-\sqrt{3}, 0),(\sqrt{3}, 0)\)

    38. \((-2,-3),(-2,3),(2,-3),(2,3)\)

    40. \((-1,-3),(-1,3),(1,-3),(1,3)\)

    Use a System of Nonlinear Equations to Solve Applications

    In the following exercises, solve the problem using a system of equations.

    1. The sum of two numbers is \(−6\) and the product is \(8\). Find the numbers.
    2. The sum of two numbers is \(11\) and the product is \(−42\). Find the numbers.
    3. The sum of the squares of two numbers is \(65\). The difference of the number is \(3\). Find the numbers.
    4. The sum of the squares of two numbers is \(113\). The difference of the number is \(1\). Find the numbers.
    5. The difference of the squares of two numbers is \(15\). The difference of twice the square of the first number and the square of the second number is \(30\). Find the numbers.
    6. The difference of the squares of two numbers is \(20\). The difference of the square of the first number and twice the square of the second number is \(4\). Find the numbers.
    7. The perimeter of a rectangle is \(32\) inches and its area is \(63\) square inches. Find the length and width of the rectangle.
    8. The perimeter of a rectangle is \(52\) cm and its area is \(165\) \(\mathrm{cm}^{2}\). Find the length and width of the rectangle.
    9. Dion purchased a new microwave. The diagonal of the door measures \(17\) inches. The door also has an area of \(120\) square inches. What are the length and width of the microwave door?
    10. Jules purchased a microwave for his kitchen. The diagonal of the front of the microwave measures \(26\) inches. The front also has an area of \(240\) square inches. What are the length and width of the microwave?
    11. Roman found a widescreen TV on sale, but isn’t sure if it will fit his entertainment center. The TV is \(60\)”. The size of a TV is measured on the diagonal of the screen and a widescreen has a length that is larger than the width. The screen also has an area of \(1728\) square inches. His entertainment center has an insert for the TV with a length of \(50\) inches and width of \(40\) inches. What are the length and width of the TV screen and will it fit Roman’s entertainment center?
    12. Donnette found a widescreen TV at a garage sale, but isn’t sure if it will fit her entertainment center. The TV is \(50\)”. The size of a TV is measured on the diagonal of the screen and a widescreen has a length that is larger than the width. The screen also has an area of \(1200\) square inches. Her entertainment center has an insert for the TV with a length of \(38\) inches and width of \(27\) inches. What are the length and width of the TV screen and will it fit Donnette’s entertainment center?
    Answer

    42. \(-3\) and \(14\)

    44. \(-7\) and \(-8\) or \(8\) and \(7\)

    46. \(-6\) and \(-4\) or \(-6\) and \(4\) or \(6\) and \(-4\) or \(6\) and \(4\)

    48. If the length is \(11\) cm, the width is \(15\) cm. If the length is \(15\) cm, the width is \(11\) cm.

    50. If the length is \(10\) inches, the width is \(24\) inches. If the length is \(24\) inches, the width is \(10\) inches.

    52. The length is \(40\) inches and the width is \(30\) inches. The TV will not fit Donnette’s entertainment center.

    Writing Exercises
    1. In your own words, explain the advantages and disadvantages of solving a system of equations by graphing.
    2. Explain in your own words how to solve a system of equations using substitution.
    3. Explain in your own words how to solve a system of equations using elimination.
    4. A circle and a parabola can intersect in ways that would result in \(0, 1, 2, 3,\) or \(4\) solutions. Draw a sketch of each of the possibilities.
    Answer

    54. Answers may vary

    56. Answers may vary 

     

    Additional Exercise

    57.  Solve for \(x\):

    \(\left\{\begin{array}{l} {x^2-y^2=-4}\\{y=2\sqrt{x}}\end{array}\right.\)

     

     

    Self Check

    a. After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

    This table has four columns and five rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve a system of nonlinear equations using graphing. In row 3, the I can solve a system of nonlinear equations using substitution. In row 4, the I can was solve a system of a nonlinear equations using the elimination. In row 5, the I can was use a system of nonlinear equations to solve applications.
     

    b. After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

     

    This page titled 8.3: Solve Systems of Nonlinear Equations is shared under a CC BY license and was authored, remixed, and/or curated by OpenStax.

    1. Back to top
    • Was this article helpful?

    Recommended articles

    1. Article type
      Section or Page
      Author
      OpenStax
      License
      CC BY
      Show Page TOC
      no
    2. Tags
      1. source[1]-math-5192
      2. source[2]-math-5192