What is the solution to this system of equations? negative 3 x + 5 y = negative 2. 3 x + 7 y = 26.

Learning Outcomes

• Solve a system that represents the intersection of a parabola and a line using substitution.
• Solve a system that represents the intersection of a circle and a line using substitution.
• Solve a system that represents the intersection of a circle and an ellipse using elimination.

A system of nonlinear equations is a system of two or more equations in two or more variables containing at least one equation that is not linear. Recall that a linear equation can take the form [latex]Ax+By+C=0[/latex]. Any equation that cannot be written in this form in nonlinear. The substitution method we used for linear systems is the same method we will use for nonlinear systems. We solve one equation for one variable and then substitute the result into the second equation to solve for another variable, and so on. There is, however, a variation in the possible outcomes.

Intersection of a Parabola and a Line

There are three possible types of solutions for a system of nonlinear equations involving a parabola and a line.

The graphs below illustrate possible solution sets for a system of equations involving a parabola and a line.

• No solution. The line will never intersect the parabola.
• One solution. The line is tangent to the parabola and intersects the parabola at exactly one point.
• Two solutions. The line crosses on the inside of the parabola and intersects the parabola at two points.

How To: Given a system of equations containing a line and a parabola, find the solution.

1. Solve the linear equation for one of the variables.
2. Substitute the expression obtained in step one into the parabola equation.
3. Solve for the remaining variable.
4. Check your solutions in both equations.

Solve the system of equations.

[latex]\begin{array}{l}x-y=-1\hfill \\ y={x}^{2}+1\hfill \end{array}[/latex]

Yes, but because [latex]x[/latex] is squared in the second equation this could give us extraneous solutions for [latex]x[/latex].

For [latex]y=1[/latex]

[latex]\begin{align}&y={x}^{2}+1 \\ &y={x}^{2}+1 \\ &{x}^{2}=0 \\ &x=\pm \sqrt{0}=0 \end{align}[/latex]

This gives us the same value as in the solution.

For [latex]y=2[/latex]

[latex]\begin{align}&y={x}^{2}+1 \\ &2={x}^{2}+1 \\ &{x}^{2}=1 \\ &x=\pm \sqrt{1}=\pm 1 \end{align}[/latex]

Notice that [latex]-1[/latex] is an extraneous solution.

Solve the given system of equations by substitution.

[latex]\begin{gathered}3x-y=-2 \\ 2{x}^{2}-y=0 \end{gathered}[/latex]

Show Solution

Just as with a parabola and a line, there are three possible outcomes when solving a system of equations representing a circle and a line.

The graph below illustrates possible solution sets for a system of equations involving a circle and a line.

• No solution. The line does not intersect the circle.
• One solution. The line is tangent to the circle and intersects the circle at exactly one point.
• Two solutions. The line crosses the circle and intersects it at two points.

How To: Given a system of equations containing a line and a circle, find the solution.

1. Solve the linear equation for one of the variables.
2. Substitute the expression obtained in step one into the equation for the circle.
3. Solve for the remaining variable.
4. Check your solutions in both equations.

Find the intersection of the given circle and the given line by substitution.

[latex]\begin{gathered}{x}^{2}+{y}^{2}=5 \\ y=3x - 5 \end{gathered}[/latex]

Show Solution

Solve the system of nonlinear equations.

[latex]\begin{array}{l}{x}^{2}+{y}^{2}=10\hfill \\ x - 3y=-10\hfill \end{array}[/latex]

We have seen that substitution is often the preferred method when a system of equations includes a linear equation and a nonlinear equation. However, when both equations in the system have like variables of the second degree, solving them using elimination by addition is often easier than substitution. Generally, elimination is a far simpler method when the system involves only two equations in two variables (a two-by-two system), rather than a three-by-three system, as there are fewer steps. As an example, we will investigate the possible types of solutions when solving a system of equations representing a circle and an ellipse.

The figure below illustrates possible solution sets for a system of equations involving a circle and an ellipse.

• No solution. The circle and ellipse do not intersect. One shape is inside the other or the circle and the ellipse are a distance away from the other.
• One solution. The circle and ellipse are tangent to each other, and intersect at exactly one point.
• Two solutions. The circle and the ellipse intersect at two points.
• Three solutions. The circle and the ellipse intersect at three points.
• Four solutions. The circle and the ellipse intersect at four points.

Solve the system of nonlinear equations.

[latex]\begin{align} {x}^{2}+{y}^{2}=26 \hspace{5mm} \left(1\right)\\ 3{x}^{2}+25{y}^{2}=100 \hspace{5mm} \left(2\right)\end{align}[/latex]

Find the solution set for the given system of nonlinear equations.

[latex]\begin{gathered}4{x}^{2}+{y}^{2}=13\\ {x}^{2}+{y}^{2}=10\end{gathered}[/latex]

In the following video, we present an example of how to solve a system of non-linear equations that represent the intersection of an ellipse and a hyperbola.

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This is a tutorial on how to use the Algebra Calculator, a step-by-step calculator for algebra.

Solving Equations

First go to the Algebra Calculator main page. In the Calculator's text box, you can enter a math problem that you want to calculate.

For example, try entering the equation 3x+2=14 into the text box.

After you enter the expression, Algebra Calculator will print a step-by-step explanation of how to solve 3x+2=14.

Examples

Math Symbols

If you would like to create your own math expressions, here are some symbols that Algebra Calculator understands:

+ (Addition)
- (Subtraction)
* (Multiplication)
/ (Division)
^ (Exponent: "raised to the power")

Graphing

To graph an equation, enter an equation that starts with "y=" or "x=". Here are some examples: y=2x^2+1, y=3x-1, x=5, x=y^2.

To graph a point, enter an ordered pair with the x-coordinate and y-coordinate separated by a comma, e.g., (3,4).

To graph two objects, simply place a semicolon between the two commands, e.g., y=2x^2+1; y=3x-1.

Polynomials

Algebra Calculator can simplify polynomials, but it only supports polynomials containing the variable x.

Here are some examples: x^2 + x + 2 + (2x^2 - 2x), (x+3)^2.

Evaluating Expressions

Algebra Calculator can evaluate expressions that contain the variable x.

To evaluate an expression containing x, enter the expression you want to evaluate, followed by the @ sign and the value you want to plug in for x. For example the command 2x @ 3 evaluates the expression 2x for x=3, which is equal to 2*3 or 6.

Algebra Calculator can also evaluate expressions that contain variables x and y. To evaluate an expression containing x and y, enter the expression you want to evaluate, followed by the @ sign and an ordered pair containing your x-value and y-value. Here is an example evaluating the expression xy at the point (3,4): xy @ (3,4).

Checking Answers For Solving Equations

Just as Algebra Calculator can be used to evaluate expressions, Algebra Calculator can also be used to check answers for solving equations containing x.

As an example, suppose we solved 2x+3=7 and got x=2. If we want to plug 2 back into the original equation to check our work, we can do so: 2x+3=7 @ 2. Since the answer is right, Algebra Calculator shows a green equals sign.

If we instead try a value that doesn't work, say x=3 (try 2x+3=7 @ 3), Algebra Calculator shows a red "not equals" sign instead.

To check an answer to a system of equations containing x and y, enter the two equations separated by a semicolon, followed by the @ sign and an ordered pair containing your x-value and y-value. Example: x+y=7; x+2y=11 @ (3,4).

Tablet Mode

If you are using a tablet such as the iPad, enter Tablet Mode to display a touch keypad.

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