$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 8.5: Solving Nonlinear Systems

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

LEARNING OBJECTIVES

• Identify nonlinear systems.
• Solve nonlinear systems using the substitution method.

## Nonlinear Systems

A system of equations where at least one equation is not linear is called a nonlinear system32. In this section we will use the substitution method to solve nonlinear systems. Recall that solutions to a system with two variables are ordered pairs $$(x,y)$$ that satisfy both equations.

Example $$\PageIndex{1}$$:

Solve: $$\left\{\begin{array}{l}{x+2 y=0} \\ {x^{2}+y^{2}=5}\end{array}\right.$$.

Solution

In this case we begin by solving for x in the first equation.

$$\left\{\begin{array}{c}{x+2 y=0} \\ {x^{2}+y^{2}=5}\end{array}\Longrightarrow x=-2y \right.$$

Substitute $$x=−2y$$ into the second equation and then solve for $$y$$.

\begin{aligned}(\color{Cerulean}{-2y }\color{black}{)}^{2}+y^{2} &=5 \\ 4 y^{2}+y^{2} &=5 \\ 5 y^{2} &=5 \\ y^{2} &=1 \\ y &=\pm 1 \end{aligned}

Here there are two answers for $$y$$; use $$x=−2y$$ to find the corresponding $$x$$-values.

Using $$y=-1$$ Using $$y=1$$
\begin{aligned} x &=-2 y \\ &=-2(-1) \\ &=2 \end{aligned} \begin{aligned} x &=-2 y \\ &=-2(1) \\ &=-2 \end{aligned}

Table 8.5.1

This gives us two ordered pair solutions, $$(2,−1)$$ and $$(−2,1)$$.

$$(2,−1), (−2,1)$$

In the previous example, the given system consisted of a line and a circle. Graphing these equations on the same set of axes, we can see that the two ordered pair solutions correspond to the two points of intersection.

If we are given a system consisting of a circle and a line, then there are $$3$$ possibilities for real solutions—two solutions as pictured above, one solution, or no solution.

Example $$\PageIndex{2}$$

Solve: $$\left\{\begin{array}{c}{x+y=3} \\ {x^{2}+y^{2}=2}\end{array}\right.$$.

Solution

Solve for $$y$$ in the first equation.

$$\left\{\begin{array}{c}{x+y} \\ {x^{2}+y^{2}}\end{array}\right.$$

Next, substitute $$y=3−x$$ into the second equation and then solve for $$x$$.

$$\begin{array}{r}{x^{2}+(\color{Cerulean}{3-x}\color{black}{)}^{2}=2} \\ {x^{2}+9-6 x+x^{2}=2} \\ {2 x^{2}-6 x+9=2} \\ {2 x^{2}-6 x+7=0}\end{array}$$

The resulting equation does not factor. Furthermore, using $$a=2$$,$$b=−6$$, and $$c=7$$ we can see that the discriminant is negative:

\begin{aligned} b^{2}-4 a c &=(-6)^{2}-4(2)(7) \\ &=36-56 \\ &=-20 \end{aligned}

We conclude that there are no real solutions to this equation and thus no solution to the system.

$$Ø$$

Exercise $$\PageIndex{1}$$

Solve: \left\{\begin{aligned} x-y &=5 \\ x^{2}+(y+1)^{2} &=8 \end{aligned}\right.

$$(2,−3)$$

If given a circle and a parabola, then there are $$5$$ possibilities for solutions.

When using the substitution method, we can perform the substitution step using entire algebraic expressions. The goal is to produce a single equation in one variable that can be solved using the techniques learned up to this point in our study of algebra.

Example $$\PageIndex{3}$$:

Solve: $$\left\{\begin{array}{c}{x^{2}+y^{2}=2} \\ {y-x^{2}=-2}\end{array}\right.$$.

Solution

We can solve for $$x^{2}$$ in the second equation.

$$\left\{\begin{array}{l}{x^{2}+y^{2}=2} \\ {y-x^{2}=-2 \quad \Rightarrow \quad y+2=x^{2}}\end{array}\right.$$

Substitute $$x^{2}=y+2$$ into the first equation and then solve for $$y$$.

\begin{aligned} \color{Cerulean}{y+2}\color{black}{+}y^{2} &=2 \\ y^{2}+y &=0 \\ y(y+1) &=0 \\ y &=0 \quad \text { or } \quad y=-1 \end{aligned}

Back substitute into $$x^{2}=y+2$$ to find the corresponding $$x$$-values.

Table 8.5.2
Using $$y=-1$$ Using $$y=0$$
\begin{aligned} x^{2} &=y+2 \\ x^{2} &=\color{Cerulean}{-1}\color{black}{+}2 \\ x^{2} &=1 \\ x &=\pm 1 \end{aligned} \begin{aligned} x^{2} &=y+2 \\ x^{2} &=\color{Cerulean}{0}\color{black}{+}2 \\ x^{2} &=2 \\ x &=\pm \sqrt{2} \end{aligned}

This leads us to four solutions, $$(±1,−1)$$ and $$(\pm \sqrt{2}, 0)$$.

$$(\pm 1,-1),(\pm \sqrt{2}, 0)$$

Example $$\PageIndex{4}$$

Solve: \left\{\begin{aligned}(x-1)^{2}-2 y^{2} &=4 \\ x^{2}+y^{2} &=9 \end{aligned}\right.

Solution

We can solve for $$y^{2}$$ in the second equation,

$$\left\{\begin{array}{r}{(x-1)^{2}-2 y^{2}=4} \\ {x^{2}+y^{2}=9}\end{array}\right. \Longrightarrow y^{2}=9-x^{2}$$

Substitute $$y^{2}=9−x^{2}$$ into the first equation and then solve for $$x$$.

\begin{aligned}(x-1)^{2}-2\color{black}{\left(\color{Cerulean}{9-x^{2}}\right) }&=4 \\ x^{2}-2 x+1-18+2 x^{2} &=0 \\ 3 x^{2}-2 x-21 &=0 \\(3 x+7)(x-3) &=0 \\ 3 x+7 &=0 \text { or } x-3=0 \\ x &=-\frac{7}{3} \quad x=3 \end{aligned}

Back substitute into $$y^{2}=9−x^{2}$$ to find the corresponding $$y$$-values.

Table 8.5.3
Using $$x=-\frac{7}{3}$$ Using $$x=3$$
$$\begin{array}{l}{y^{2}=9-\color{black}{\left(\color{Cerulean}{-\frac{7}{3}}\right)^{2}}} \\ {y^{2}=\frac{9}{1}-\frac{49}{9}} \\ {y^{2}=\frac{32}{9}} \\ {y=\pm \frac{\sqrt{32}}{3}=\pm \frac{4 \sqrt{2}}{3}}\end{array}$$ \begin{aligned} y^{2} &=9-(\color{Cerulean}{3}\color{black}{)}^{2} \\ y^{2} &=0 \\ y &=0 \end{aligned}

This leads to three solutions, $$\left(-\frac{7}{3}, \pm \frac{4 \sqrt{2}}{3}\right)$$ and $$(3,0)$$.

$$(3,0),\left(-\frac{7}{3}, \pm \frac{4 \sqrt{2}}{3}\right)$$

Example $$\PageIndex{5}$$

Solve: \left\{\begin{aligned} x^{2}+y^{2} &=2 \\ x y &=1 \end{aligned}\right..

Solution

Solve for $$y$$ in the second equation.

$$\left\{\begin{array}{r}{x^{2}+y^{2}=2} \\ {x y=1}\end{array}\right.\Longrightarrow y=\frac{1}{x}$$

Substitute $$y=\frac{1}{x}$$ into the first equation and then solve for $$x$$.

$$x^{2}+\left(\frac{1}{x}\right)^{2}=2$$
$$x^{2}+\frac{1}{x^{2}}=2$$

This leaves us with a rational equation. Make a note that $$x≠0$$ and multiply both sides by $$x^{2}$$.

\begin{aligned} \color{Cerulean}{x^{2}}\color{black}{\left(x^{2}+\frac{1}{x^{2}}\right)} &=2 \cdot \color{Cerulean}{x^{2}} \\ x^{4}+1 &=2 x^{2} \\ x^{4}-2 x^{2}+1 &=0 \\\left(x^{2}-1\right)\left(x^{2}-1\right) &=0 \end{aligned}

At this point we can see that both factors are the same. Apply the zero product property.

\begin{aligned} x^{2}-1 &=0 \\ x^{2} &=1 \\ x &=\pm 1 \end{aligned}

Back substitute into $$y=\frac{1}{x}$$ to find the corresponding $$y$$-values.

Using $$x=-1$$ Using $$x=1$$
\begin{aligned} y &=\frac{1}{x} \\ &=\frac{1}{\color{Cerulean}{-1}} \\ &=-1 \end{aligned} \begin{aligned} y &=\frac{1}{x} \\ &=\frac{1}{\color{Cerulean}{1}} \\ &=1 \end{aligned}

$$(1,1),(-1,-1)$$

Exercise $$\PageIndex{2}$$

Solve: $$\left\{\begin{array}{r}{\frac{1}{x}+\frac{1}{y}=4} \\ {\frac{1}{x^{2}}+\frac{1}{y^{2}}=40}\end{array}\right.$$

$$\left(-\frac{1}{2}, \frac{1}{6}\right)\left(\frac{1}{6},-\frac{1}{2}\right)$$

## Key Takeaways

• Use the substitution method to solve nonlinear systems.
• Streamline the solving process by using entire algebraic expressions in the substitution step to obtain a single equation with one variable.
• Understanding the geometric interpretation of the system can help in finding real solutions.

Exercise $$\PageIndex{3}$$

Solve.

1. $$\left\{\begin{array}{c}{x^{2}+y^{2}=10} \\ {x+y=4}\end{array}\right.$$
2. $$\left\{\begin{array}{c}{x^{2}+y^{2}=5} \\ {x-y=-3}\end{array}\right.$$
3. $$\left\{\begin{array}{l}{x^{2}+y^{2}=30} \\ {x-3 y=0}\end{array}\right.$$
4. $$\left\{\begin{array}{c}{x^{2}+y^{2}=10} \\ {2 x-y=0}\end{array}\right.$$
5. $$\left\{\begin{array}{c}{x^{2}+y^{2}=18} \\ {2 x-2 y=-12}\end{array}\right.$$
6. \left\{\begin{aligned}(x-4)^{2}+y^{2} &=25 \\ 4 x-3 y &=16 \end{aligned}\right.
7. $$\left\{\begin{array}{c}{3 x^{2}+2 y^{2}=21} \\ {3 x-y=0}\end{array}\right.$$
8. \left\{\begin{aligned} x^{2}+5 y^{2} &=36 \\ x-2 y &=0 \end{aligned}\right.
9. $$\left\{\begin{array}{c}{4 x^{2}+9 y^{2}=36} \\ {2 x+3 y=6}\end{array}\right.$$
10. $$\left\{\begin{array}{c}{4 x^{2}+y^{2}=4} \\ {2 x+y=-2}\end{array}\right.$$
11. $$\left\{\begin{array}{c}{2 x^{2}+y^{2}=1} \\ {x+y=1}\end{array}\right.$$
12. $$\left\{\begin{array}{c}{4 x^{2}+3 y^{2}=12} \\ {2 x-y=2}\end{array}\right.$$
13. \left\{\begin{aligned} x^{2}-2 y^{2} &=35 \\ x-3 y &=0 \end{aligned}\right.
14. $$\left\{\begin{array}{c}{5 x^{2}-7 y^{2}=39} \\ {2 x+4 y=0}\end{array}\right.$$
15. $$\left\{\begin{array}{c}{9 x^{2}-4 y^{2}=36} \\ {3 x+2 y=0}\end{array}\right.$$
16. $$\left\{\begin{array}{l}{x^{2}+y^{2}=25} \\ {x-2 y=-12}\end{array}\right.$$
17. $$\left\{\begin{array}{l}{2 x^{2}+3 y=9} \\ {8 x-4 y=12}\end{array}\right.$$
18. $$\left\{\begin{array}{l}{2 x-4 y^{2}=3} \\ {3 x-12 y=6}\end{array}\right.$$
19. \left\{\begin{aligned} 4 x^{2}+3 y^{2} &=12 \\ x-\frac{3}{2} &=0 \end{aligned}\right.
20. \left\{\begin{aligned} 5 x^{2}+4 y^{2} &=40 \\ y-3 &=0 \end{aligned}\right.
21. The sum of the squares of two positive integers is $$10$$. If the first integer is added to twice the second integer, the sum is $$7$$. Find the integers.
22. The diagonal of a rectangle measures $$\sqrt{5}$$ units and has a perimeter equal to $$6$$ units. Find the dimensions of the rectangle.
23. For what values of $$b$$ will the following system have real solutions? $$\left\{\begin{array}{c}{x^{2}+y^{2}=1} \\ {y=x+b}\end{array}\right.$$
24. For what values of $$m$$ will be the following system have real solutions? $$\left\{\begin{array}{c}{x^{2}-y^{2}=1} \\ {y=m x}\end{array}\right.$$

1. $$(1,3),(3,1)$$

3. $$(-3 \sqrt{3},-\sqrt{3}),(3 \sqrt{3}, \sqrt{3})$$

5. $$(-3,3)$$

7. $$(-1,-3),(1,3)$$

9. $$(0,2),(3,0)$$

11. $$(0,1),\left(\frac{2}{3}, \frac{1}{3}\right)$$

13. $$(-3 \sqrt{5},-\sqrt{5}),(3 \sqrt{5}, \sqrt{5})$$

15. $$\emptyset$$

17. $$\left(\frac{-3+3 \sqrt{5}}{2},-6+3 \sqrt{5}\right) ,\left(\frac{-3-3 \sqrt{5}}{2},-6-3 \sqrt{5}\right)$$

19. $$\left(\frac{3}{2},-1\right),\left(\frac{3}{2}, 1\right)$$

21. $$1,3$$

23. $$b \in[-\sqrt{2}, \sqrt{2}]$$

Exercise $$\PageIndex{4}$$

Solve.

1. $$\left\{\begin{array}{c}{x^{2}+y^{2}=4} \\ {y-x^{2}=2}\end{array}\right.$$
2. $$\left\{\begin{array}{l}{x^{2}+y^{2}=4} \\ {y-x^{2}=-2}\end{array}\right.$$
3. $$\left\{\begin{array}{c}{x^{2}+y^{2}=4} \\ {y-x^{2}=3}\end{array}\right.$$
4. $$\left\{\begin{array}{c}{x^{2}+y^{2}=4} \\ {4 y-x^{2}=-4}\end{array}\right.$$
5. $$\left\{\begin{array}{c}{x^{2}+3 y^{2}=9} \\ {y^{2}-x=3}\end{array}\right.$$
6. $$\left\{\begin{array}{c}{x^{2}+3 y^{2}=9} \\ {x+y^{2}=-4}\end{array}\right.$$
7. \left\{\begin{aligned} 4 x^{2}-3 y^{2} &=12 \\ x^{2}+y^{2} &=1 \end{aligned}\right.
8. $$\left\{\begin{array}{l}{x^{2}+y^{2}=1} \\ {x^{2}-y^{2}=1}\end{array}\right.$$
9. \left\{\begin{aligned} x^{2}+y^{2} &=1 \\ 4 y^{2}-x^{2}-4 y &=0 \end{aligned}\right.
10. \left\{\begin{aligned} x^{2}+y^{2} &=4 \\ 2 x^{2}-y^{2}+4 x &=0 \end{aligned}\right.
11. \left\{\begin{aligned} 2(x-2)^{2}+y^{2} &=6 \\(x-3)^{2}+y^{2} &=4 \end{aligned}\right.
12. $$\left\{\begin{array}{c}{x^{2}+y^{2}-6 y=0} \\ {4 x^{2}+5 y^{2}+20 y=0}\end{array}\right.$$
13. $$\left\{\begin{array}{l}{x^{2}+4 y^{2}=25} \\ {4 x^{2}+y^{2}=40}\end{array}\right.$$
14. $$\left\{\begin{array}{c}{x^{2}-2 y^{2}=-10} \\ {4 x^{2}+y^{2}=10}\end{array}\right.$$
15. $$\left\{\begin{array}{c}{2 x^{2}+y^{2}=14} \\ {x^{2}-(y-1)^{2}=6}\end{array}\right.$$
16. $$\left\{\begin{array}{c}{3 x^{2}-(y-2)^{2}=12} \\ {x^{2}+(y-2)^{2}=1}\end{array}\right.$$
17. The difference of the squares of two positive integers is $$12$$. The sum of the larger integer and the square of the smaller is equal to $$8$$. Find the integers.
18. The difference between the length and width of a rectangle is $$4$$ units and the diagonal measures $$8$$ units. Find the dimensions of the rectangle. Round off to the nearest tenth.
19. The diagonal of a rectangle measures $$p$$ units and has a perimeter equal to $$2q$$ units. Find the dimensions of the rectangle in terms of $$p$$ and $$q$$.
20. The area of a rectangle is $$p$$ square units and its perimeter is $$2q$$ units. Find the dimensions of the rectangle in terms of $$p$$ and $$q$$.

1. $$(0,2)$$

3. $$\emptyset$$

5. $$(-3,0),(0,-\sqrt{3}),(0, \sqrt{3})$$

7. $$\emptyset$$

9. $$(0,1),\left(-\frac{2 \sqrt{5}}{5},-\frac{1}{5}\right),\left(\frac{2 \sqrt{5}}{5},-\frac{1}{5}\right)$$

11. $$(3,-2),(3,2)$$

13. $$(-3,-2),(-3,2),(3,-2),(3,2)$$

15. $$(-\sqrt{7}, 0),(\sqrt{7}, 0),\left(-\frac{\sqrt{55}}{3}, \frac{4}{3}\right),\left(\frac{\sqrt{55}}{3}, \frac{4}{3}\right)$$

17. $$2,4$$

19. $$\frac{q+\sqrt{2 p^{2}-q^{2}}}{2}$$ units by $$\frac{q-\sqrt{2 p^{2}-q^{2}}}{2}$$ units

Exercise $$\PageIndex{5}$$

Solve.

1. \left\{\begin{aligned} x^{2}+y^{2} &=26 \\ x y &=5 \end{aligned}\right.
2. \left\{\begin{aligned} x^{2}+y^{2} &=10 \\ x y &=3 \end{aligned}\right.
3. \left\{\begin{aligned} 2 x^{2}-3 y^{2} &=5 \\ x y &=1 \end{aligned}\right.
4. $$\left\{\begin{array}{c}{3 x^{2}-4 y^{2}=-11} \\ {x y=1}\end{array}\right.$$
5. $$\left\{\begin{array}{c}{x^{2}+y^{2}=2} \\ {x y-2=0}\end{array}\right.$$
6. $$\left\{\begin{array}{l}{x^{2}+y^{2}=1} \\ {2 x y-1=0}\end{array}\right.$$
7. \left\{\begin{aligned} 4 x-y^{2} &=0 \\ x y &=2 \end{aligned}\right.
8. $$\left\{\begin{array}{c}{3 y-x^{2}=0} \\ {x y-9=0}\end{array}\right.$$
9. \left\{\begin{aligned} 2 y-x^{2} &=0 \\ x y-1 &=0 \end{aligned}\right.
10. \left\{\begin{aligned} x-y^{2} &=0 \\ x y &=3 \end{aligned}\right.
11. The diagonal of a rectangle measures $$2\sqrt{10}$$ units. If the area of the rectangle is $$12$$ square units, find its dimensions.
12. The area of a rectangle is $$48$$ square meters and the perimeter measures $$32$$ meters. Find the dimensions of the rectangle.
13. The product of two positive integers is $$72$$ and their sum is $$18$$. Find the integers.
14. The sum of the squares of two positive integers is $$52$$ and their product is $$24$$. Find the integers.

1. $$(-5,-1),(5,1),(-1,-5),(1,5)$$

3. $$\left(-\sqrt{3},-\frac{\sqrt{3}}{3}\right),\left(\sqrt{3}, \frac{\sqrt{3}}{3}\right)$$

5. $$\emptyset$$

7. $$(1,2)$$

9. $$\left(\sqrt[3]{2}, \frac{\sqrt[3]{4}}{2}\right)$$

11. $$2$$ units by $$6$$ units

13. $$6,12$$

Exercise $$\PageIndex{6}$$

Solve.

1. $$\left\{\begin{array}{l}{\frac{1}{x}+\frac{1}{y}=4} \\ {\frac{1}{x}-\frac{1}{y}=2}\end{array}\right.$$
2. $$\left\{\begin{array}{l}{\frac{2}{x}-\frac{1}{y}=5} \\ {\frac{1}{x}+\frac{1}{y}=2}\end{array}\right.$$
3. $$\left\{\begin{array}{l}{\frac{1}{x}+\frac{2}{y}=1} \\ {\frac{3}{x}-\frac{1}{y}=2}\end{array}\right.$$
4. $$\left\{\begin{array}{l}{\frac{1}{x}+\frac{1}{y}=6} \\ {\frac{1}{x^{2}}+\frac{1}{y^{2}}=20}\end{array}\right.$$
5. $$\left\{\begin{array}{l}{\frac{1}{x}+\frac{1}{y}=2} \\ {\frac{1}{x^{2}}+\frac{1}{y^{2}}=34}\end{array}\right.$$
6. $$\left\{\begin{array}{l}{x y-16=0} \\ {2 x^{2}-y=0}\end{array}\right.$$
7. $$\left\{\begin{array}{l}{x+y^{2}=4} \\ {y=\sqrt{x}}\end{array}\right.$$
8. $$\left\{\begin{array}{c}{y^{2}-(x-1)^{2}=1} \\ {y=\sqrt{x}}\end{array}\right.$$
9. $$\left\{\begin{array}{l}{y=2^{x}} \\ {y=2^{2 x}-56}\end{array}\right.$$
10. $$\left\{\begin{array}{l}{y=3^{2 x}-72} \\ {y-3^{x}=0}\end{array}\right.$$
11. $$\left\{\begin{array}{l}{y=e^{4 x}} \\ {y=e^{2 x}+6}\end{array}\right.$$
12. $$\left\{\begin{array}{l}{y-e^{2 x}=0} \\ {y-e^{x}=0}\end{array}\right.$$

1. $$\left(\frac{1}{3}, 1\right)$$

3. $$\left(\frac{7}{5}, 7\right)$$

5. $$\left(-\frac{1}{3}, \frac{1}{5}\right),\left(\frac{1}{5},-\frac{1}{3}\right)$$

7. $$(2, \sqrt{2})$$

9. $$(3,8)$$

11. $$\left(\frac{\ln 3}{2}, 9\right)$$

Exercise $$\PageIndex{7}$$

1. How many real solutions can be obtained from a system that consists of a circle and a hyperbola? Explain.
2. Make up your own nonlinear system, solve it, and provide the answer. Also, provide a graph and discuss the geometric interpretation of the solutions.