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Solve Systems of Nonlinear Equations

Solve a system of nonlinear equations using graphing

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.

\[\begin{array}{lllllll}\{\begin{array}{l}{x}^{2}+{y}^{2}=9 \\ {x}^{2}-y=9\end{array} & & & \ \{\begin{array}{l}9{x}^{2}+{y}^{2}=9 \\ y=3x-3\end{array} & & & \ \{\begin{array}{l}x+y=4 \\ y={x}^{2}+2\end{array}\end{array}\]

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.

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

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

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.

Example

Try it.

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

Solution

Identify each graph.\(\{\begin{array}{lll}9{x}^{2}+{y}^{2}=9 & & \text{ellipse} \\ y=3x-3 & & \text{line}\end{array}\)
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.
\((0,-3)\)
\(\begin{array}{llllllllll} \\ 9{x}^{2}+{y}^{2} & = & 9 & & & & & y & = & 3x-3 \\ 9\cdot {0}^{2}+{(-3)}^{2} & \overset{?}{=} & 9 & & & & & -3 & \overset{?}{=} & 3\cdot 0-3 \\ 0+9 & \overset{?}{=} & 9 & & & & & -3 & \overset{?}{=} & 0-3 \\ 9 & = & 9✓ & & & & & -3 & = & -3✓\end{array}\)
\((1,0)\)
\(\ \begin{array}{llllllllll}9{x}^{2}+{y}^{2} & = & 9 & & & & & \ y & = & 3x-3 \\ 9\cdot {1}^{2}+{0}^{2} & \overset{?}{=} & 9 & & & & & \ 0 & \overset{?}{=} & 3\cdot 1-3 \\ 9+0 & \overset{?}{=} & 9 & & & & & \ 0 & \overset{?}{=} & 3-3 \\ 9 & = & 9✓ & & & & & \ 0 & = & 0✓\end{array}\)
The solutions are \((0,-3),(1,0).\)

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

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

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.

Example

Try it.

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

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

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

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

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

Try it.

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.
Second sentence.The sum of the numbers is 5.
Solve the system by substitution
Solve the second equation for x.
Substitute x into the first equation.
Expand and simplify.
Solve for y.
Substitute back into the second equation.
The numbers are 1 and 4.

Example

Try it.

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.
The area of the rectangle is 300 square inches.
Solve the system using substitution.
Solve the second equation for x.
Substitute x into the first equation.
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.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

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 (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. Solve the system by graphing: \(\{\begin{array}{l}x-3y=-3 \\ x+y=5\end{array}.\)
    If you missed this problem, review .

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    \(\left(3,2\right)\)

  2. Solve the system by substitution: \(\{\begin{array}{l}x-4y=-4 \\ -3x+4y=0\end{array}.\)
    If you missed this problem, review .

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    \(\left(2,\frac{3}{2}\right)\)

  3. Solve the system by elimination: \(\{\begin{array}{l}3x-4y=-9 \\ 5x+3y=14\end{array}.\)
    If you missed this problem, review .

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    \(\left(1,3\right)\)

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

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    Identify each graph.\(\{\begin{array}{lll}x-y=-2 & & \text{line} \\ y={x}^{2} & & \text{parabola}\end{array}\)
    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 \((2,4)\) and \((-1,1).\)
    Check to make sure each solution makes
    both equations true.
    \((2,4)\)
    \(\ \begin{array}{llllllll}x-y & = & -2 & & & y & = & {x}^{2} \\ 2-4 & \overset{?}{=} & -2 & & & 4 & \overset{?}{=} & {2}^{2} \\ -2 & = & -2✓ & & & 4 & = & 4✓\end{array}\)

    \((-1,1)\)
    \(\ \begin{array}{llllllll}x-y & = & -2 & & & y & = & {x}^{2} \\ -1-1 & \overset{?}{=} & -2 & & & 1 & \overset{?}{=} & {(-1)}^{2} \\ -2 & = & -2✓ & & & 1 & = & 1✓\end{array}\)
    The solutions are \((2,4)\) and \((-1,1).\)

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

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

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

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    Identify each graph.\(\{\begin{array}{lll}y=-1 & & \text{line} \\ {(x-2)}^{2}+{(y+3)}^{2}=4 & & \text{circle}\end{array}\)
    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.
    \((2,-1)\)
    \(\begin{array}{llllllll}{(x-2)}^{2}+{(y+3)}^{2} & = & 4 & & & y & = & -1 \\ {(2-2)}^{2}+{(-1+3)}^{2} & \overset{?}{=} & 4 & & & -1 & = & -1✓ \\ {(0)}^{2}+{(2)}^{2} & \overset{?}{=} & 4 & & & & & \\ 4 & = & 4✓ & & & & & \end{array}\)
    The solution is \((2,-1).\)

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

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

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

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    Identify each graph.\(\{\begin{array}{lll}9{x}^{2}+{y}^{2}=9 & & \text{ellipse} \\ y=3x-3 & & \text{line}\end{array}\)
    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.
    \((0,-3)\)
    \(\begin{array}{llllllllll} \\ 9{x}^{2}+{y}^{2} & = & 9 & & & & & y & = & 3x-3 \\ 9\cdot {0}^{2}+{(-3)}^{2} & \overset{?}{=} & 9 & & & & & -3 & \overset{?}{=} & 3\cdot 0-3 \\ 0+9 & \overset{?}{=} & 9 & & & & & -3 & \overset{?}{=} & 0-3 \\ 9 & = & 9✓ & & & & & -3 & = & -3✓\end{array}\)
    \((1,0)\)
    \(\ \begin{array}{llllllllll}9{x}^{2}+{y}^{2} & = & 9 & & & & & \ y & = & 3x-3 \\ 9\cdot {1}^{2}+{0}^{2} & \overset{?}{=} & 9 & & & & & \ 0 & \overset{?}{=} & 3\cdot 1-3 \\ 9+0 & \overset{?}{=} & 9 & & & & & \ 0 & \overset{?}{=} & 3-3 \\ 9 & = & 9✓ & & & & & \ 0 & = & 0✓\end{array}\)
    The solutions are \((0,-3),(1,0).\)

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

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    No solution

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

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    \((-\frac{4}{5},\frac{6}{5}),(0,2)\)

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

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    Identify each graph.\(\{\begin{array}{lll}{x}^{2}-y=0 & & \text{parabola} \\ y=x-2 & & \text{line}\end{array}\)
    Sketch the possible options for
    intersection of a parabola and a line
    The equation \(y=x-2\) is solved for y.
    Substitute \(x-2\) for y in the first equation.
    Solve the equation for x.
    This doesn’t factor easily, so we can
    check the discriminant.
    \(\begin{array}{l}{b}^{2}-4ac \\ {(-1)}^{2}-4\cdot 1\cdot 2 \\ \\ -7\end{array}\)The discriminant is negative, so there is no real solution.
    The system has no solution.

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

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    No solution

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

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    \((\frac{4}{9},-\frac{2}{3}),(1,1)\)

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

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

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

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

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

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

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

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    Identify each graph.\(\{\begin{array}{lll}{x}^{2}+{y}^{2}=7 & & \text{circle} \\ {x}^{2}-{y}^{2}=1 & & \text{hyperbola}\end{array}\)
    Sketch the possible options for intersection
    of a circle and hyperbola.
    Both equations are in standard form.\(\{\begin{array}{l}{x}^{2}+{y}^{2}=7 \\ {x}^{2}-{y}^{2}=1\end{array}\)
    The coefficients of \({y}^{2}\) are opposite, so we
    will add the equations.    		\(\begin{array}{l}\underset{__________}{\{\begin{array}{l}{x}^{2}+{y}^{2}=7 \\ {x}^{2}-{y}^{2}=1\end{array}} \\ \\ 2{x}^{2}\ =8\end{array}\)
    Simplify.\(\begin{array}{lll}{x}^{2} & = & 4 \\ x & = & \text{\pm }2\end{array}\)
    \(x=2\ x=-2\)
    Substitute \(x=2\) and \(x=-2\) into one of the
    original equations. Then solve for y.    		\(\begin{array}{lllllllll}{x}^{2}+{y}^{2} & = & 7 & & & & {x}^{2}+{y}^{2} & = & 7 \\ {2}^{2}+{y}^{2} & = & 7 & & & & {(-2)}^{2}+{y}^{2} & = & 7 \\ 4+{y}^{2} & = & 7 & & & & 4+{y}^{2} & = & 7 \\ {y}^{2} & = & 3 & & & & {y}^{2} & = & 3 \\ y & = & \text{\pm }\sqrt{3} & & & & y & = & \text{\pm }\sqrt{3}\end{array}\)
    Write each solution as an ordered pair.The ordered pairs are \((-2,\sqrt{3}),\) \((-2,\text{-}\sqrt{3}),\)
    \((2,\sqrt{3}),\) and \((2,\text{-}\sqrt{3}).\)
    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 \((-2,\sqrt{3}),\) \((-2,\text{-}\sqrt{3}),\) \((2,\sqrt{3}),\)
    and \((2,\text{-}\sqrt{3}).\)

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

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

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

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

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

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    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.
    Second sentence.The sum of the numbers is 5.
    Solve the system by substitution
    Solve the second equation for x.
    Substitute x into the first equation.
    Expand and simplify.
    Solve for y.
    Substitute back into the second equation.
    The numbers are 1 and 4.

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

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    4 and 6

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

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    \(-18\) and 17

  25. 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?

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    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.
    The area of the rectangle is 300 square inches.
    Solve the system using substitution.
    Solve the second equation for x.
    Substitute x into the first equation.
    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.

  26. 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?

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    If the length is 12 inches, the width is 16 inches. If the length is 16 inches, the width is 12 inches.

  27. 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?

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    If the length is 12 inches, the width is 9 inches. If the length is 9 inches, the width is 12 inches.

  28. \(\{\begin{array}{l}y=2x+2 \\ y=\text{-}{x}^{2}+2\end{array}\)

  29. \(\{\begin{array}{l}y=6x-4 \\ y=2{x}^{2}\end{array}\)

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

  31. \(\{\begin{array}{l}x-y=-2 \\ x={y}^{2}\end{array}\)

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  32. \(\{\begin{array}{l}y=\frac{3}{2}x+3 \\ y=\text{-}{x}^{2}+2\end{array}\)

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

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

  35. \(\{\begin{array}{l}y=-4 \\ {x}^{2}+{y}^{2}=16\end{array}\)

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  36. \(\{\begin{array}{l}x=2 \\ {(x+2)}^{2}+{(y+3)}^{2}=16\end{array}\)

  37. \(\{\begin{array}{l}y=-1 \\ {(x-2)}^{2}+{(y-4)}^{2}=25\end{array}\)

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  38. \(\{\begin{array}{l}y=-2x+4 \\ y=\sqrt[]{x}+1\end{array}\)

  39. \(\{\begin{array}{l}y=-\frac{1}{2}x+2 \\ y=\sqrt[]{x}-2\end{array}\)

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  40. \(\{\begin{array}{l}{x}^{2}+4{y}^{2}=4 \\ y=\frac{1}{2}x-1\end{array}\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}
square root, n-th root
The non-negative number whose square (n-th power) is x.
\pm
plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.
\neq
not equal
The two sides are different.
\leq,\ \geq
less/greater than or equal
Inequalities that allow equality; < and > exclude it.
|x|
absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.
i
imaginary unit
i² = −1.
\log_b x,\ \ln x
logarithm, natural log
The exponent b must be raised to for x; ln uses base e.
\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}
number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Solve Systems of Nonlinear Equations

  1. Solve a system of nonlinear equations using graphing
  2. Solve a system of nonlinear equations using substitution
  3. Solve a system of nonlinear equations using elimination
  4. Use a system of nonlinear equations to solve applications
  5. Identify the graph of each equation. Sketch the possible options for intersection.
  6. Graph the first equation.
  7. Graph the second equation on the same rectangular coordinate system.
  8. Determine whether the graphs intersect.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

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Parts of this page are adapted from OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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