8.5: Solving Nonlinear Systems

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Oct 6, 2021
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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.

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

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

Answer:

$(2,−1), (−2,1)$

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.

Answer:

$Ø$

Exercise $\PageIndex{1}$

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

Answer

$(2,−3)$

www.youtube.com/v/ToIrjw-8SNA

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 $\PageIndex{2}$
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 $(\pm \sqrt{2}, 0)$ .

Answer:

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

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 $\PageIndex{3}$
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 $(3,0)$ .

Answer:

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

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

This leads to two solutions.

Answer:

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

Exercise $\PageIndex{2}$

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

Answer

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

www.youtube.com/v/n8JJ_ybegkM

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.

$\left\{\begin{array}{c}{x^{2}+y^{2}=10} \\ {x+y=4}\end{array}\right.$
$\left\{\begin{array}{c}{x^{2}+y^{2}=5} \\ {x-y=-3}\end{array}\right.$
$\left\{\begin{array}{l}{x^{2}+y^{2}=30} \\ {x-3 y=0}\end{array}\right.$
$\left\{\begin{array}{c}{x^{2}+y^{2}=10} \\ {2 x-y=0}\end{array}\right.$
$\left\{\begin{array}{c}{x^{2}+y^{2}=18} \\ {2 x-2 y=-12}\end{array}\right.$
$\left\{\begin{aligned}(x-4)^{2}+y^{2} &=25 \\ 4 x-3 y &=16 \end{aligned}\right.$
$\left\{\begin{array}{c}{3 x^{2}+2 y^{2}=21} \\ {3 x-y=0}\end{array}\right.$
$\left\{\begin{aligned} x^{2}+5 y^{2} &=36 \\ x-2 y &=0 \end{aligned}\right.$
$\left\{\begin{array}{c}{4 x^{2}+9 y^{2}=36} \\ {2 x+3 y=6}\end{array}\right.$
$\left\{\begin{array}{c}{4 x^{2}+y^{2}=4} \\ {2 x+y=-2}\end{array}\right.$
$\left\{\begin{array}{c}{2 x^{2}+y^{2}=1} \\ {x+y=1}\end{array}\right.$
$\left\{\begin{array}{c}{4 x^{2}+3 y^{2}=12} \\ {2 x-y=2}\end{array}\right.$
$\left\{\begin{aligned} x^{2}-2 y^{2} &=35 \\ x-3 y &=0 \end{aligned}\right.$
$\left\{\begin{array}{c}{5 x^{2}-7 y^{2}=39} \\ {2 x+4 y=0}\end{array}\right.$
$\left\{\begin{array}{c}{9 x^{2}-4 y^{2}=36} \\ {3 x+2 y=0}\end{array}\right.$
$\left\{\begin{array}{l}{x^{2}+y^{2}=25} \\ {x-2 y=-12}\end{array}\right.$
$\left\{\begin{array}{l}{2 x^{2}+3 y=9} \\ {8 x-4 y=12}\end{array}\right.$
$\left\{\begin{array}{l}{2 x-4 y^{2}=3} \\ {3 x-12 y=6}\end{array}\right.$
$\left\{\begin{aligned} 4 x^{2}+3 y^{2} &=12 \\ x-\frac{3}{2} &=0 \end{aligned}\right.$
$\left\{\begin{aligned} 5 x^{2}+4 y^{2} &=40 \\ y-3 &=0 \end{aligned}\right.$
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.
The diagonal of a rectangle measures $\sqrt{5}$ units and has a perimeter equal to $6$ units. Find the dimensions of the rectangle.
For what values of $b$ will the following system have real solutions? $\left\{\begin{array}{c}{x^{2}+y^{2}=1} \\ {y=x+b}\end{array}\right.$
For what values of $m$ will be the following system have real solutions? $\left\{\begin{array}{c}{x^{2}-y^{2}=1} \\ {y=m x}\end{array}\right.$

Answer

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.

$\left\{\begin{array}{c}{x^{2}+y^{2}=4} \\ {y-x^{2}=2}\end{array}\right.$
$\left\{\begin{array}{l}{x^{2}+y^{2}=4} \\ {y-x^{2}=-2}\end{array}\right.$
$\left\{\begin{array}{c}{x^{2}+y^{2}=4} \\ {y-x^{2}=3}\end{array}\right.$
$\left\{\begin{array}{c}{x^{2}+y^{2}=4} \\ {4 y-x^{2}=-4}\end{array}\right.$
$\left\{\begin{array}{c}{x^{2}+3 y^{2}=9} \\ {y^{2}-x=3}\end{array}\right.$
$\left\{\begin{array}{c}{x^{2}+3 y^{2}=9} \\ {x+y^{2}=-4}\end{array}\right.$
$\left\{\begin{aligned} 4 x^{2}-3 y^{2} &=12 \\ x^{2}+y^{2} &=1 \end{aligned}\right.$
$\left\{\begin{array}{l}{x^{2}+y^{2}=1} \\ {x^{2}-y^{2}=1}\end{array}\right.$
$\left\{\begin{aligned} x^{2}+y^{2} &=1 \\ 4 y^{2}-x^{2}-4 y &=0 \end{aligned}\right.$
$\left\{\begin{aligned} x^{2}+y^{2} &=4 \\ 2 x^{2}-y^{2}+4 x &=0 \end{aligned}\right.$
$\left\{\begin{aligned} 2(x-2)^{2}+y^{2} &=6 \\(x-3)^{2}+y^{2} &=4 \end{aligned}\right.$
$\left\{\begin{array}{c}{x^{2}+y^{2}-6 y=0} \\ {4 x^{2}+5 y^{2}+20 y=0}\end{array}\right.$
$\left\{\begin{array}{l}{x^{2}+4 y^{2}=25} \\ {4 x^{2}+y^{2}=40}\end{array}\right.$
$\left\{\begin{array}{c}{x^{2}-2 y^{2}=-10} \\ {4 x^{2}+y^{2}=10}\end{array}\right.$
$\left\{\begin{array}{c}{2 x^{2}+y^{2}=14} \\ {x^{2}-(y-1)^{2}=6}\end{array}\right.$
$\left\{\begin{array}{c}{3 x^{2}-(y-2)^{2}=12} \\ {x^{2}+(y-2)^{2}=1}\end{array}\right.$
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.
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.
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 $q$ .
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 $q$ .

Answer

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.

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

Answer

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.

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

Answer

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 $\PageIndex{7}$

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

Answer: 1. Answer may vary

Footnotes

³²A system of equations where at least one equation is not linear.

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