Nonlinear Systems of Equations: Examples

Nonlinear systems of equations are systems where at least one equation is not a line, such as a circle or a parabola. The solutions are the points where the graphs cross. You solve them by substitution or elimination, just like linear systems, but you usually end up solving a quadratic equation.

Before this lesson: You should know Linear-Quadratic Systems, Substitution Method, Elimination Method, Equation of a Circle and Quadratic Inequalities.

What Makes a System Nonlinear

A system is a set of equations that share the same variables. In a nonlinear system, at least one equation has a squared variable, a product like $xy$ , or another term that does not make a line.

Each solution is an ordered pair $(x, y)$ that makes every equation true. On a graph, it is a point where the curves meet. A line and a circle can meet at 0, 1 or 2 points. Two curves like circles and parabolas can meet at up to 4 points.

Graph of the circle x^2 + y^2 = 25, centered at the origin with radius 5, and the line y = x + 1. The line crosses the circle at two points, (−4, −3) and (3, 4), which are marked with dots and labeled. — The line y = x + 1 crosses the circle x^2 + y^2 = 25 at two points, so the system has two solutions.

How to Solve a Nonlinear System

Substitution works best when one equation is easy to solve for a variable, such as $y = x + 1$ or $y = x^2$ .

Solve one equation for one variable (it may already be solved).
Substitute that expression into the other equation.
Solve the new equation. It is often a quadratic.
Put each value back into the simpler equation to find the other coordinate.
Check each pair in both original equations.

Elimination works best when both equations have the same squared terms, such as $x^2$ and $y^2$ in each. Add or subtract the equations to cancel one of them.

Worked Examples

Example 1: A line and a circle

Solve the system $x^2 + y^2 = 25$ and $y = x + 1$ .

Substitute $x + 1$ for $y$ in the circle (see substitution method): $x^2 + (x + 1)^2 = 25$
Square the binomial (see special products of binomials): $x^2 + x^2 + 2x + 1 = 25$
Combine like terms: $2x^2 + 2x + 1 = 25$
Subtract 25 from both sides: $2x^2 + 2x - 24 = 0$
Divide every term by 2: $x^2 + x - 12 = 0$
Factor the trinomial (see factoring trinomials): $(x + 4)(x - 3) = 0$
Set each factor equal to zero (see solving quadratic equations by factoring): $x = -4$ or $x = 3$
Find $y$ with the line. For $x = -4$ : $y = -4 + 1 = -3$
For $x = 3$ : $y = 3 + 1 = 4$

Answer: $(-4, -3)$ and $(3, 4)$

Check $(3, 4)$ : $3^2 + 4^2 = 9 + 16 = 25$ , and $4 = 3 + 1$ . Both are true. These are the two points on the graph above.

Example 2: Two parabolas

Solve the system $y = x^2 - 4$ and $y = -x^2 + 2x + 8$ .

Both right sides equal $y$ , so set them equal: $x^2 - 4 = -x^2 + 2x + 8$
Add $x^2$ to both sides: $2x^2 - 4 = 2x + 8$
Subtract $2x$ from both sides: $2x^2 - 2x - 4 = 8$
Subtract 8 from both sides: $2x^2 - 2x - 12 = 0$
Divide every term by 2: $x^2 - x - 6 = 0$
Factor the trinomial: $(x - 3)(x + 2) = 0$
Solve each factor: $x = 3$ or $x = -2$
Find $y$ with the first equation. For $x = 3$ : $y = 3^2 - 4 = 5$
For $x = -2$ : $y = (-2)^2 - 4 = 0$

Answer: $(3, 5)$ and $(-2, 0)$

Check $(3, 5)$ in the second equation: $-9 + 6 + 8 = 5$ . True.

Example 3: Elimination with four solutions

Solve the system $x^2 + y^2 = 13$ and $x^2 - y^2 = 5$ .

Add the equations to eliminate $y^2$ (see elimination method): $2x^2 = 18$
Divide both sides by 2: $x^2 = 9$
Take the square root of both sides (see solving quadratics with square roots): $x = \pm 3$
Substitute $x^2 = 9$ into the first equation: $9 + y^2 = 13$
Subtract 9 from both sides: $y^2 = 4$
Take the square root of both sides: $y = \pm 2$
Pair every $x$ with every $y$ . Each pair works, because only the squares appear.

Answer: $(3, 2)$ , $(3, -2)$ , $(-3, 2)$ and $(-3, -2)$

The graphs are a circle and a hyperbola, and they cross in four places.

Example 4: A circle and a parabola

Solve the system $x^2 + y^2 = 20$ and $y = x^2$ .

The second equation says $x^2 = y$ . Substitute $y$ for $x^2$ in the first: $y + y^2 = 20$
Subtract 20 and write in standard form: $y^2 + y - 20 = 0$
Factor the trinomial: $(y + 5)(y - 4) = 0$
Solve each factor: $y = -5$ or $y = 4$
Reject $y = -5$ , because $x^2 = -5$ has no real solution.
Use $y = 4$ : $x^2 = 4$
Take the square root of both sides: $x = \pm 2$

Answer: $(2, 4)$ and $(-2, 4)$

Check $(2, 4)$ : $2^2 + 4^2 = 4 + 16 = 20$ , and $4 = 2^2$ . True.

Example 5: Word problem with a rectangle

A rectangle has a perimeter of 28 inches and a diagonal of 10 inches. Find its length and width.

Let the sides be $x$ and $y$ . Half the perimeter is the sum of the sides: $x + y = 14$
The diagonal makes a right triangle (see Pythagorean theorem): $x^2 + y^2 = 100$
Solve the first equation for $y$ : $y = 14 - x$
Substitute into the second: $x^2 + (14 - x)^2 = 100$
Square the binomial: $x^2 + 196 - 28x + x^2 = 100$
Combine like terms: $2x^2 - 28x + 196 = 100$
Subtract 100 from both sides: $2x^2 - 28x + 96 = 0$
Divide every term by 2: $x^2 - 14x + 48 = 0$
Factor the trinomial: $(x - 6)(x - 8) = 0$
Solve each factor: $x = 6$ or $x = 8$
Find $y$ : if $x = 6$ , then $y = 14 - 6 = 8$ ; if $x = 8$ , then $y = 6$

Answer: The rectangle is 8 inches long and 6 inches wide.

Check: the perimeter is $2(8) + 2(6) = 28$ , and the diagonal is $\sqrt{64 + 36} = \sqrt{100} = 10$ .

How Many Solutions Can There Be?

After substituting, look at the quadratic you get. Its discriminant tells you how many times a line meets a curve (see the discriminant).

Discriminant of the new quadratic	Line and circle	Number of solutions
Positive	Line cuts through the circle	2
Zero	Line touches the circle once (tangent)	1
Negative	Line misses the circle	0

For example, $x^2 + y^2 = 4$ and $y = x + 5$ give $2x^2 + 10x + 21 = 0$ . Its discriminant is $100 - 168 = -68$ , so the line misses the circle and the system has no real solution.

Common Mistakes

Mistake	Example of the error	Fix
Squaring a binomial wrong	$(x + 1)^2 = x^2 + 1$	$(x + 1)^2 = x^2 + 2x + 1$
Finding $y$ in the circle instead of the line	$x = 3$ in $x^2 + y^2 = 25$ gives $y = \pm 4$ , so $(3, -4)$ is listed	$(3, -4)$ is not on $y = x + 1$ . Use the simpler equation, then check both.
Losing the ±	$x^2 = 9$ , so $x = 3$	$x = \pm 3$ . Both signs can give solutions.
Keeping an impossible value	$y = -5$ with $y = x^2$	$x^2 = -5$ has no real solution, so drop $y = -5$ .
Stopping at the x-values	Answer: $x = -4, 3$	A solution is a point. Give $(-4, -3)$ and $(3, 4)$ .

Practice Problems

Solve $y = 2x$ and $x^2 + y^2 = 20$ .
Solve $x^2 + y^2 = 10$ and $y = x + 2$ .
Solve $y = x^2 + 1$ and $y = -x^2 + 9$ .
Solve $x^2 + y^2 = 25$ and $x^2 - y^2 = 7$ .
Solve $y = x^2$ and $x + y = 6$ .
Solve $x^2 + y^2 = 4$ and $y = x + 4$ .
Solve $x^2 + y^2 = 9$ and $y = 3$ .
Solve $xy = 12$ and $y = x + 1$ .
Solve $x^2 + y^2 = 17$ and $y = x^2 - 5$ .
Two numbers add to 11, and the sum of their squares is 61. Find the numbers.

Show answers

Substitute: $x^2 + (2x)^2 = 20$
Square: $x^2 + 4x^2 = 20$
Combine: $5x^2 = 20$
Divide by 5: $x^2 = 4$
Square root: $x = \pm 2$
Find $y = 2x$ : $(2, 4)$ and $(-2, -4)$
Substitute: $x^2 + (x + 2)^2 = 10$
Square: $x^2 + x^2 + 4x + 4 = 10$
Combine: $2x^2 + 4x + 4 = 10$
Subtract 10: $2x^2 + 4x - 6 = 0$
Divide by 2: $x^2 + 2x - 3 = 0$
Factor: $(x + 3)(x - 1) = 0$
Solve: $x = -3$ or $x = 1$
Find $y = x + 2$ : $(-3, -1)$ and $(1, 3)$
Set equal: $x^2 + 1 = -x^2 + 9$
Add $x^2$ : $2x^2 + 1 = 9$
Subtract 1: $2x^2 = 8$
Divide by 2: $x^2 = 4$
Square root: $x = \pm 2$
Find $y = x^2 + 1 = 5$ : $(2, 5)$ and $(-2, 5)$
Add the equations: $2x^2 = 32$
Divide by 2: $x^2 = 16$
Square root: $x = \pm 4$
Substitute $x^2 = 16$ : $16 + y^2 = 25$
Subtract 16: $y^2 = 9$
Square root: $y = \pm 3$
Answer: $(4, 3)$ , $(4, -3)$ , $(-4, 3)$ , $(-4, -3)$
Substitute $x^2$ for $y$ : $x + x^2 = 6$
Standard form: $x^2 + x - 6 = 0$
Factor: $(x + 3)(x - 2) = 0$
Solve: $x = -3$ or $x = 2$
Find $y = x^2$ : $(-3, 9)$ and $(2, 4)$
Substitute: $x^2 + (x + 4)^2 = 4$
Square: $x^2 + x^2 + 8x + 16 = 4$
Combine: $2x^2 + 8x + 16 = 4$
Subtract 4: $2x^2 + 8x + 12 = 0$
Divide by 2: $x^2 + 4x + 6 = 0$
Discriminant: $4^2 - 4(1)(6)$
Square and multiply: $16 - 24$
Subtract: $-8$ , which is negative
Answer: no real solution (the line misses the circle)
Substitute: $x^2 + 3^2 = 9$
Square: $x^2 + 9 = 9$
Subtract 9: $x^2 = 0$
Solve: $x = 0$
Answer: $(0, 3)$ only. The line is tangent to the circle.
Substitute: $x(x + 1) = 12$
Distribute: $x^2 + x = 12$
Subtract 12: $x^2 + x - 12 = 0$
Factor: $(x + 4)(x - 3) = 0$
Solve: $x = -4$ or $x = 3$
Find $y = x + 1$ : $(-4, -3)$ and $(3, 4)$
Rewrite the parabola: $x^2 = y + 5$
Substitute: $y + 5 + y^2 = 17$
Standard form: $y^2 + y - 12 = 0$
Factor: $(y + 4)(y - 3) = 0$
Solve: $y = -4$ or $y = 3$
If $y = 3$ : $x^2 = 8$ , so $x = \pm 2\sqrt{2}$ (see simplifying square roots)
If $y = -4$ : $x^2 = 1$ , so $x = \pm 1$
Answer: $(2\sqrt{2}, 3)$ , $(-2\sqrt{2}, 3)$ , $(1, -4)$ , $(-1, -4)$
Write the system: $x + y = 11$ and $x^2 + y^2 = 61$
Solve for $y$ : $y = 11 - x$
Substitute: $x^2 + (11 - x)^2 = 61$
Square: $x^2 + 121 - 22x + x^2 = 61$
Combine: $2x^2 - 22x + 121 = 61$
Subtract 61: $2x^2 - 22x + 60 = 0$
Divide by 2: $x^2 - 11x + 30 = 0$
Factor: $(x - 5)(x - 6) = 0$
Solve: $x = 5$ or $x = 6$
Answer: the numbers are 5 and 6

FAQ

What is a nonlinear system of equations?

It is a system where at least one equation is not linear, such as a circle, parabola or $xy = k$ . Its solutions are the points where the graphs intersect.

How do you solve a nonlinear system of equations?

Use substitution when one equation is solved for a variable, or elimination when both equations have the same squared terms. Then solve the resulting quadratic and find the matching coordinates.

How many solutions can a nonlinear system have?

A line and a circle or parabola can have 0, 1 or 2 solutions. Two circles, parabolas or other conic sections can have up to 4.

Can you solve a nonlinear system by graphing?

Yes. Graph both equations and read the intersection points. Graphs give estimates, so check each point in both equations or solve algebraically for exact answers.

Summary

A nonlinear system has at least one equation that is not a line.
Its solutions are the intersection points of the graphs.
Substitute when one equation is solved for a variable; eliminate when the squared terms match.
Find both coordinates of every solution, and check each one in both equations.
Drop values that are impossible, like $x^2 = -5$ .

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