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Section 5.4 Systems of Nonlinear Equations

Before Starting this Section, Review

1 Solve systems of linear equations (Sections 5.1 and 5.2)

Objectives

1 Solve systems of nonlinear equations by substitution.
2 Solve systems of nonlinear equations by elimination.
3 Solve applied problems using nonlinear systems.

Investing in the Stock Market

A stock is a certificate that shows you own a small fraction of a corporation. When you own stock in a company, you are referred to as a shareholder. When you buy stock, you are paying for a small percentage of everything the company owns, including a slice of its profits (or losses).

Suppose you inherit some money and want to invest $1000 in the stock market. Where do you invest your money? (This is a serious question. You may have to do a great deal of research on the companies that interest you, or you may have to get help from a broker.)

Suppose you get a tip from a friend who heard it from his brother-in-law (a taxi driver who overheard his customer’s conversation) that Microsoft is coming out with a new version of Microsoft Windows that is going to double the company’s business. Where do you buy Microsoft stock? You go to a broker. A brokerage house (such as Fidelity Investments) is a supermarket of stocks. You can tell any broker that you have $1000 and want to buy as much Microsoft stock as you can. The broker might reply, “A share of Microsoft at this time [the price may fluctuate every second] costs $73 [the actual closing price on September 25, 2017, was $73.26], and I am going to charge you $5 for my services. So you can buy $\frac{1000 - 5}{73} \approx 13.63,$ $\frac{1000 - 5}{73} \approx 13.63,$ or 13 shares of Microsoft.” You then give the broker the money, and you become part owner of Microsoft. (Usually, the broker does not give you the paper stock certificate, but rather transfers ownership to you.)

You can make a profit on your investment if you sell your stock for more than the total amount you paid for it. You can also make money while you own certain stocks by receiving dividends. A dividend on a share of stock is a return on your investment and represents the portion of the company’s profits allocated to that share. The dividend can be in the form of cash or additional shares of the company. In Example 3, we discuss such a scenario.

Systems of Nonlinear Equations

In Sections 5.1 and 5.2, we discussed systems of linear equations. We now consider systems of nonlinear equations involving two variables: In these systems, at least one equation is nonlinear.

To determine whether an ordered pair is a solution of a nonlinear system, you check whether it is a solution of each equation in the system. Thus, (2, 1) is a solution of the system

{\begin{array}{rclcrl} x^{2} & + & 3 y & = & 7 & (1) \\ 5 x^{2} & - & 4 y^{2} & = & 16 & (2) \end{array}

${\begin{array}{rclcrl} x^{2} & + & 3 y & = & 7 & (1) \\ 5 x^{2} & - & 4 y^{2} & = & 16 & (2) \end{array}$

because $2^{2} + 3 (1) = 4 + 3 = 7$ $2^{2} + 3 (1) = 4 + 3 = 7$ and $5 {(2)}^{2} - 4 {(1)}^{2} = 20 - 4 = 16$ $5 {(2)}^{2} - 4 {(1)}^{2} = 20 - 4 = 16$ are both true statements.

Solving Systems of Nonlinear Equations by Substitution

1 Solve systems of nonlinear equations by substitution.

We follow the same steps used in the substitution method for solving a system of linear equations in Section 5.1. See page 517. This method is useful when one of the equations in the system is linear in one of the variables.

Example 1 Using Substitution to Solve a Nonlinear System

Solve the system of equations by the substitution method.

{\begin{array}{r} 4 x & + & y & = & - 3 & (1) \\ - x^{2} & + & y & = & 1 & (2) \end{array}

${\begin{array}{r} 4 x & + & y & = & - 3 & (1) \\ - x^{2} & + & y & = & 1 & (2) \end{array}$

Solution

Because both equations are linear in y, we solve either equation for y.

Step 1 Solve for one variable. In equation (2), express y in terms of x.

$y = x^{2} + 1 (3) Solve equation (2) for y .$ $y = x^{2} + 1 (3) Solve equation (2) for y .$
Step 2 Substitute. Substitute $x^{2} + 1$ $x^{2} + 1$ for y in equation (1).

$\begin{array}{rcll} 4 x + y & = & - 3 & Equation (1) \\ 4 x + (x^{2} + 1) & = & - 3 & Replace y with (x^{2} + 1) . \\ 4 x + x^{2} + 4 & = & 0 & Add 3 to both sides . \\ x^{2} + 4 x + 4 & = & 0 & Rewrite in descending powers of x . \end{array}$ $\begin{array}{rcll} 4 x + y & = & - 3 & Equation (1) \\ 4 x + (x^{2} + 1) & = & - 3 & Replace y with (x^{2} + 1) . \\ 4 x + x^{2} + 4 & = & 0 & Add 3 to both sides . \\ x^{2} + 4 x + 4 & = & 0 & Rewrite in descending powers of x . \end{array}$
Step 3 Solve the resulting equation from Step 2.

$\begin{array}{rcll} (x + 2) (x + 2) & = & 0 & Factor . \\ x + 2 & = & 0 & Zero-product property \\ x & = & - 2 & Solve for x . \end{array}$ $\begin{array}{rcll} (x + 2) (x + 2) & = & 0 & Factor . \\ x + 2 & = & 0 & Zero-product property \\ x & = & - 2 & Solve for x . \end{array}$
Step 4 Back-substitute. Substitute $x = - 2$ $x = - 2$ in equation (3) to find the corresponding y -value.

$\begin{array}{l} y = x^{2} + 1 & Equation (3) \\ y = {(- 2)}^{2} + 1 & Replace x with - 2. \\ y = 5 & Simplify . \end{array}$ $\begin{array}{l} y = x^{2} + 1 & Equation (3) \\ y = {(- 2)}^{2} + 1 & Replace x with - 2. \\ y = 5 & Simplify . \end{array}$
Step 5 From Steps 3 and 4, we have $x = - 2$ $x = - 2$ and $y = 5 .$ $y = 5 .$ The apparent solution set of the system is ${(- 2, 5)} .$ ${(- 2, 5)} .$
Step 6 Check: Replace x with $- 2$ $- 2$ and y with 5 in both equations.

$\begin{array}{l} Equation (1) & Equation (2) \\ \begin{array}{rcl} 4 x + y & = & - 3 \\ 4 (- 2) + 5 & \overset{?}{=} & - 3 \\ - 8 + 5 & = & - 3 ✓ \end{array} & \begin{array}{rcl} - x^{2} + y & = & 1 \\ - {(- 2)}^{2} + 5 & \overset{?}{=} & 1 Replace x with - 2 and y with 5. \\ - 4 + 5 & = & 1 ✓ \end{array} \end{array}$ $\begin{array}{l} Equation (1) & Equation (2) \\ \begin{array}{rcl} 4 x + y & = & - 3 \\ 4 (- 2) + 5 & \overset{?}{=} & - 3 \\ - 8 + 5 & = & - 3 ✓ \end{array} & \begin{array}{rcl} - x^{2} + y & = & 1 \\ - {(- 2)}^{2} + 5 & \overset{?}{=} & 1 Replace x with - 2 and y with 5. \\ - 4 + 5 & = & 1 ✓ \end{array} \end{array}$

The graphs of the line $4 x + y = - 3$ $4 x + y = - 3$ and the parabola $y = x^{2} + 1$ $y = x^{2} + 1$ confirm that the solution set is ${(- 2, 5)} .$ ${(- 2, 5)} .$ See Figure 5.10 .

Figure 5.10

Practice Problem 1

Solve the system of equations by the substitution method.

${\begin{array}{l} x^{2} + y = 2 & (1) \\ 2 x + y = 3 & (2) \end{array}$ ${\begin{array}{l} x^{2} + y = 2 & (1) \\ 2 x + y = 3 & (2) \end{array}$

Solving Systems of Nonlinear Equations by Elimination

2 Solve systems of nonlinear equations by elimination.

We follow the same steps outlined for solving systems of linear equations by the elimination method. See page 519. This method works well when both equations of the system are of the same degree in the same variable.

Example 2 Using Elimination to Solve a Nonlinear System

Solve the system of equations by the elimination method.

{\begin{array}{lcrr} x^{2} + y^{2} & = & 25 & (1) \\ x^{2} - y & = & 5 & (2) \end{array}

${\begin{array}{lcrr} x^{2} + y^{2} & = & 25 & (1) \\ x^{2} - y & = & 5 & (2) \end{array}$

Solution

Step 1 Adjust the coefficients. Because x has the same power in both equations, we choose the variable x for elimination. We multiply equation (2) by $- 1$ $- 1$ and obtain the equivalent system:
Step 2

$\begin{array}{l} \begin{array}{l} {\underline{\begin{array}{rcl} x^{2} + y^{2} & = & 25 \\ - x^{2} + y & = & - 5 \end{array}} & \begin{array}{l} (1) \\ (3) \end{array} \end{array} Multiply equation (2) by - 1. \\ y^{2} + y = 20 (4) Add equations (1) and (3) . \end{array}$ $\begin{array}{l} \begin{array}{l} {\underline{\begin{array}{rcl} x^{2} + y^{2} & = & 25 \\ - x^{2} + y & = & - 5 \end{array}} & \begin{array}{l} (1) \\ (3) \end{array} \end{array} Multiply equation (2) by - 1. \\ y^{2} + y = 20 (4) Add equations (1) and (3) . \end{array}$
Step 3 Solve the equation found in Step 2: $y^{2} + y = 20 .$ $y^{2} + y = 20 .$

$\begin{array}{l} \begin{array}{r} y^{2} + y - 20 = 0 \\ (y + 5) (y - 4) = 0 \end{array} & \begin{array}{l} Subtract 20 from both sides . \\ Factor . \end{array} \\ \begin{array}{l} y + 5 = 0 or y - 4 = 0 \\ y = - 5 or y = 4 \end{array} & \begin{array}{l} Zero-product property \\ Solve for y . \end{array} \end{array}$ $\begin{array}{l} \begin{array}{r} y^{2} + y - 20 = 0 \\ (y + 5) (y - 4) = 0 \end{array} & \begin{array}{l} Subtract 20 from both sides . \\ Factor . \end{array} \\ \begin{array}{l} y + 5 = 0 or y - 4 = 0 \\ y = - 5 or y = 4 \end{array} & \begin{array}{l} Zero-product property \\ Solve for y . \end{array} \end{array}$
Step 4 Back-substitute the values in one of the original equations to solve for the other variable.
1. Substitute $y = - 5$ $y = - 5$ in equation (2) and solve for x.
  
  $\begin{array}{rcll} x^{2} - y & = & 5 & Equation (2) \\ x^{2} - (- 5) & = & 5 & Replace y with - 5. \\ x^{2} & = & 0 & Simplify . \\ x & = & 0 & Solve for x . \end{array}$ $\begin{array}{rcll} x^{2} - y & = & 5 & Equation (2) \\ x^{2} - (- 5) & = & 5 & Replace y with - 5. \\ x^{2} & = & 0 & Simplify . \\ x & = & 0 & Solve for x . \end{array}$
  
  So $(0, - 5)$ $(0, - 5)$ is a solution of the system.
2. Substitute $y = 4$ $y = 4$ in equation (2) and solve for x.
  
  $\begin{array}{rcll} x^{2} - y & = & 5 & Equation (2) \\ x^{2} - 4 & = & 5 & Replace y with 4. \\ x^{2} & = & 9 & Add 4 to both sides . \\ x & = & \pm 3 & Square root property \end{array}$ $\begin{array}{rcll} x^{2} - y & = & 5 & Equation (2) \\ x^{2} - 4 & = & 5 & Replace y with 4. \\ x^{2} & = & 9 & Add 4 to both sides . \\ x & = & \pm 3 & Square root property \end{array}$
  
  So (3, 4) and $(- 3, 4)$ $(- 3, 4)$ are solutions of the system.
  
  From (i) and (ii), the apparent solution set for the system is ${(0, - 5), (3, 4), (- 3, 4)} .$ ${(0, - 5), (3, 4), (- 3, 4)} .$
Step 5 Check:

$\begin{array}{c} \begin{array}{l} x^{2} + y^{2} = 25 \\ x^{2} - y = 5 \end{array} & \begin{array}{c} (0, - 5) \\ \begin{array}{rcl} 0^{2} + {(- 5)}^{2} & \overset{?}{=} & 25 \\ 25 & = & 25 ✓ \\ 0^{2} - (- 5) & \overset{?}{=} & 5 \\ 5 & = & 5 ✓ \end{array} \end{array} & \begin{array}{c} (3, 4) \\ \begin{array}{rcl} 3^{2} + 4^{2} & \overset{?}{=} & 25 \\ 9 + 16 & = & 25 \\ 25 & = & 25 ✓ \\ 3^{2} - 4 & \overset{?}{=} & 5 \\ 9 - 4 & = & 5 ✓ \end{array} \end{array} & \begin{array}{c} (- 3, 4) \\ \begin{array}{rcl} {(- 3)}^{2} + 4^{2} & \overset{?}{=} & 25 \\ 9 + 16 & \overset{?}{=} & 25 \\ 25 & = & 25 ✓ \\ {(- 3)}^{2} - 4 & \overset{?}{=} & 5 \\ 9 - 4 & = & 5 ✓ \end{array} \end{array} \end{array}$ $\begin{array}{c} \begin{array}{l} x^{2} + y^{2} = 25 \\ x^{2} - y = 5 \end{array} & \begin{array}{c} (0, - 5) \\ \begin{array}{rcl} 0^{2} + {(- 5)}^{2} & \overset{?}{=} & 25 \\ 25 & = & 25 ✓ \\ 0^{2} - (- 5) & \overset{?}{=} & 5 \\ 5 & = & 5 ✓ \end{array} \end{array} & \begin{array}{c} (3, 4) \\ \begin{array}{rcl} 3^{2} + 4^{2} & \overset{?}{=} & 25 \\ 9 + 16 & = & 25 \\ 25 & = & 25 ✓ \\ 3^{2} - 4 & \overset{?}{=} & 5 \\ 9 - 4 & = & 5 ✓ \end{array} \end{array} & \begin{array}{c} (- 3, 4) \\ \begin{array}{rcl} {(- 3)}^{2} + 4^{2} & \overset{?}{=} & 25 \\ 9 + 16 & \overset{?}{=} & 25 \\ 25 & = & 25 ✓ \\ {(- 3)}^{2} - 4 & \overset{?}{=} & 5 \\ 9 - 4 & = & 5 ✓ \end{array} \end{array} \end{array}$

The graphs of the circle $x^{2} + y^{2} = 25$ $x^{2} + y^{2} = 25$ and the parabola $y = x^{2} - 5$ $y = x^{2} - 5$ sketched in Figure 5.11 confirm the three solutions.

Practice Problem 2

Solve the system of equations.

${\begin{array}{r} x^{2} & + & 2 y^{2} & = & 34 \\ x^{2} & - & y^{2} & = & 7 \end{array}$ ${\begin{array}{r} x^{2} & + & 2 y^{2} & = & 34 \\ x^{2} & - & y^{2} & = & 7 \end{array}$

Applications

3 Solve applied problems using nonlinear systems.

In the next example, we solve an investment problem that involves a nonlinear system of equations.

Example 3 Investing in Stocks

Danielle bought 240 shares of Alpha Airlines stock at $40 per share and paid $100 in broker’s fees. She kept these shares for three years, during which time she received a number of additional shares of the stock as dividends. At the end of three years, her dividends were worth $2400 and she sold all of her stock and made a profit of $7100 (after paying $100 in initial broker’s commission). How many shares of the stock did she receive as dividends? What was the selling price of each share of the stock?

Solution

$\begin{array}{rcl} Let x & = & number of shares she received as dividends \\ p & = & selling price per share \end{array}$ $\begin{array}{rcl} Let x & = & number of shares she received as dividends \\ p & = & selling price per share \end{array}$

Then

x p = 2400 (1) The total dividend is $ 2400.

$x p = 2400 (1) The total dividend is $ 2400.$

The number of shares she sold at p dollars per share is $240 + x .$ $240 + x .$

and

\begin{array}{rcll} Revenue - Cost & = & Profit \\ (240 + x) p - 9700 & = & 7100 \\ (240 + x) p & = & 16,800 & Add 9700 to both sides . \\ 240 p + x p & = & 16,800 (2) & Distributive property \end{array}

$\begin{array}{rcll} Revenue - Cost & = & Profit \\ (240 + x) p - 9700 & = & 7100 \\ (240 + x) p & = & 16,800 & Add 9700 to both sides . \\ 240 p + x p & = & 16,800 (2) & Distributive property \end{array}$

We solve the system of equations:

{\begin{array}{rclll} x p & = & 2400 & (1) & Revenue from the dividends \\ 240 p + x p & = & 16,800 & (2) & Revenue - Cost = profit \end{array}

${\begin{array}{rclll} x p & = & 2400 & (1) & Revenue from the dividends \\ 240 p + x p & = & 16,800 & (2) & Revenue - Cost = profit \end{array}$

Substitute $x p = 2400$ $x p = 2400$ from equation (1) into equation (2).

\begin{array}{rcrcll} 240 p & + & x p & = & 16,800 & (2) \\ 240 p & + & 2400 & = & 16,800 & (3) & Substitute 2400 for x p in (2) . \\ 240 p & = & 14,400 & Subtract 2400 from both sides . \\ p & = & 60 & Solve for p . \end{array}

$\begin{array}{rcrcll} 240 p & + & x p & = & 16,800 & (2) \\ 240 p & + & 2400 & = & 16,800 & (3) & Substitute 2400 for x p in (2) . \\ 240 p & = & 14,400 & Subtract 2400 from both sides . \\ p & = & 60 & Solve for p . \end{array}$

Back-substitute $p = 60$ $p = 60$ in equation (1).

\begin{array}{rcll} x p & = & 2400 & Equation (1) \\ x (60) & = & 2400 & Replace p with 60. \\ x & = & \frac{2400}{60} = 40 & Solve for x . \end{array}

$\begin{array}{rcll} x p & = & 2400 & Equation (1) \\ x (60) & = & 2400 & Replace p with 60. \\ x & = & \frac{2400}{60} = 40 & Solve for x . \end{array}$

Danielle received $x = 40$ $x = 40$ shares as dividends and sold her stock at $p = $ 60$ $p = $ 60$ per share.

Practice Problem 3

Rework Example 3, assuming that Danielle’s dividends were worth $1950 after three years and her profit was $7850.

Section 5.4 Exercises

Concepts and Vocabulary

In a system of nonlinear equations, equation must be nonlinear.
Both the and methods can also be used to solve systems of nonlinear equations.
The solutions of the system

${\begin{array}{rcll} a x + b y & = & c & (1) \\ {(x - h)}^{2} + {(y - k)}^{2} & = & r^{2} & (2) \end{array} represent the points$ ${\begin{array}{rcll} a x + b y & = & c & (1) \\ {(x - h)}^{2} + {(y - k)}^{2} & = & r^{2} & (2) \end{array} represent the points$

of of the graphs of equations (1) and (2).
The system of equations in Example 1 has exactly one solution because the graph of equation (1) is a(n) line to the graph of equation (2).
True or False. The system of equations

${\begin{array}{rcl} y & = & a x^{2} \\ {(x - h)}^{2} + {(y - k)}^{2} & = & r^{2} \end{array}$ ${\begin{array}{rcl} y & = & a x^{2} \\ {(x - h)}^{2} + {(y - k)}^{2} & = & r^{2} \end{array}$ may have 0, 1, 2, 3, or 4 real solutions.
True or False. If $4 x + 3 y = 8$ $4 x + 3 y = 8$ is one of the equations in a system of equations, then the system is linear.
True or False. The system of equations ${\begin{array}{rcl} y & = & a x^{2} \\ y & = & m x + b \end{array}$ ${\begin{array}{rcl} y & = & a x^{2} \\ y & = & m x + b \end{array}$ may have 0, 1, or 2, real solutions.
True or False. The system of equations ${\begin{cases} x y = \sqrt{3} \\ x^{2} + y^{2} = 4 \end{cases}$ ${\begin{cases} x y = \sqrt{3} \\ x^{2} + y^{2} = 4 \end{cases}$ has 4 real solutions.

Building Skills

In Exercises 9–16, determine which of the given ordered pairs are solutions of each system of equations.

${\begin{array}{rcl} 2 x + 3 y & = & 3 \\ x - y^{2} & = & 2 \end{array} (1, - 3), (3, - 1), (6, 3), (5, \frac{1}{2})$ ${\begin{array}{rcl} 2 x + 3 y & = & 3 \\ x - y^{2} & = & 2 \end{array} (1, - 3), (3, - 1), (6, 3), (5, \frac{1}{2})$
${\begin{array}{rcl} x + 2 y & = & 6 \\ y & = & x^{2} \end{array} (2, 2), (- 2, 4), (0, 3), (1, 2)$ ${\begin{array}{rcl} x + 2 y & = & 6 \\ y & = & x^{2} \end{array} (2, 2), (- 2, 4), (0, 3), (1, 2)$
${\begin{array}{rcl} 5 x - 2 y & = & 7 \\ x^{2} + y^{2} & = & 2 \end{array} (\frac{5}{4}, 1), (0, \frac{11}{4}), (1, - 1), (3, 4)$ ${\begin{array}{rcl} 5 x - 2 y & = & 7 \\ x^{2} + y^{2} & = & 2 \end{array} (\frac{5}{4}, 1), (0, \frac{11}{4}), (1, - 1), (3, 4)$
${\begin{array}{lcrcl} x & - & 2 y & = & - 5 \\ x^{2} & + & y^{2} & = & 25 \end{array} (1, 3), (- 5, 0), (3, 4), (3, - 4)$ ${\begin{array}{lcrcl} x & - & 2 y & = & - 5 \\ x^{2} & + & y^{2} & = & 25 \end{array} (1, 3), (- 5, 0), (3, 4), (3, - 4)$
${\begin{array}{rcrcl} 4 x^{2} & + & 5 y^{2} & = & 180 \\ x^{2} & - & y^{2} & = & 9 \end{array} (5, 4), (- 5, 4), (3, 0), (- 5, - 4)$ ${\begin{array}{rcrcl} 4 x^{2} & + & 5 y^{2} & = & 180 \\ x^{2} & - & y^{2} & = & 9 \end{array} (5, 4), (- 5, 4), (3, 0), (- 5, - 4)$
${\begin{array}{rcl} x^{2} - y^{2} & = & 3 \\ x^{2} - 4 x + y^{2} & = & - 3 \end{array} (2, 1), (2, - 1), (- 2, 1), (- 2, - 1)$ ${\begin{array}{rcl} x^{2} - y^{2} & = & 3 \\ x^{2} - 4 x + y^{2} & = & - 3 \end{array} (2, 1), (2, - 1), (- 2, 1), (- 2, - 1)$
${\begin{cases} y = e^{x - 1} \\ y = 2 x - 1 \end{cases} (1, 1), (0, - 1), (2, e), (- 1, - 3)$ ${\begin{cases} y = e^{x - 1} \\ y = 2 x - 1 \end{cases} (1, 1), (0, - 1), (2, e), (- 1, - 3)$
${\begin{cases} y = \ln (x + 1) \\ y = x \end{cases} (1, 1), (0, 0), (- 1, - 1), (e - 1, 1)$ ${\begin{cases} y = \ln (x + 1) \\ y = x \end{cases} (1, 1), (0, 0), (- 1, - 1), (e - 1, 1)$

In Exercises 17–32, solve each system of equations by the substitution method. Check your solutions.

${\begin{cases} y = x^{2} \\ y = x + 2 \end{cases}$ ${\begin{cases} y = x^{2} \\ y = x + 2 \end{cases}$
${\begin{array}{rcl} y & = & x^{2} \\ x + y & = & 6 \end{array}$ ${\begin{array}{rcl} y & = & x^{2} \\ x + y & = & 6 \end{array}$
${\begin{array}{l} x^{2} & - & y & = & 6 \\ x & - & y & = & 0 \end{array}$ ${\begin{array}{l} x^{2} & - & y & = & 6 \\ x & - & y & = & 0 \end{array}$
${\begin{array}{rcl} x^{2} & - & y & = & 6 \\ 5 x & - & y & = & 0 \end{array}$ ${\begin{array}{rcl} x^{2} & - & y & = & 6 \\ 5 x & - & y & = & 0 \end{array}$
${\begin{array}{r} x^{2} & + & y^{2} & = & 9 \\ x & = & 3 \end{array}$ ${\begin{array}{r} x^{2} & + & y^{2} & = & 9 \\ x & = & 3 \end{array}$
${\begin{array}{r} x^{2} & + & y^{2} & = & 9 \\ y & = & 3 \end{array}$ ${\begin{array}{r} x^{2} & + & y^{2} & = & 9 \\ y & = & 3 \end{array}$
${\begin{array}{lclcr} x^{2} & + & y^{2} & = & 5 \\ x & - & y & = & - 3 \end{array}$ ${\begin{array}{lclcr} x^{2} & + & y^{2} & = & 5 \\ x & - & y & = & - 3 \end{array}$
${\begin{array}{r} x^{2} & + & y^{2} & = & 13 \\ 2 x & - & 3 y & = & 0 \end{array}$ ${\begin{array}{r} x^{2} & + & y^{2} & = & 13 \\ 2 x & - & 3 y & = & 0 \end{array}$
${\begin{array}{r} x^{2} - 4 x + y^{2} & = & - 2 \\ x - y & = & 2 \end{array}$ ${\begin{array}{r} x^{2} - 4 x + y^{2} & = & - 2 \\ x - y & = & 2 \end{array}$
${\begin{array}{r} x^{2} - 8 x + y^{2} & = & - 6 \\ 2 x - y & = & 1 \end{array}$ ${\begin{array}{r} x^{2} - 8 x + y^{2} & = & - 6 \\ 2 x - y & = & 1 \end{array}$
${\begin{array}{r} x - y & = & - 2 \\ x y & = & 3 \end{array}$ ${\begin{array}{r} x - y & = & - 2 \\ x y & = & 3 \end{array}$
${\begin{array}{r} x - 2 y & = & 4 \\ x y & = & 6 \end{array}$ ${\begin{array}{r} x - 2 y & = & 4 \\ x y & = & 6 \end{array}$
${\begin{array}{r} 4 x^{2} + y^{2} & = & 25 \\ x + y & = & 5 \end{array}$ ${\begin{array}{r} 4 x^{2} + y^{2} & = & 25 \\ x + y & = & 5 \end{array}$
${\begin{array}{r} x^{2} + 4 y^{2} & = & 16 \\ x + 2 y & = & 4 \end{array}$ ${\begin{array}{r} x^{2} + 4 y^{2} & = & 16 \\ x + 2 y & = & 4 \end{array}$
${\begin{array}{r} x^{2} - y^{2} & = & 24 \\ 5 x - 7 y & = & 0 \end{array}$ ${\begin{array}{r} x^{2} - y^{2} & = & 24 \\ 5 x - 7 y & = & 0 \end{array}$
${\begin{array}{r} x^{2} - 7 y^{2} & = & 9 \\ x - y & = & 3 \end{array}$ ${\begin{array}{r} x^{2} - 7 y^{2} & = & 9 \\ x - y & = & 3 \end{array}$

In Exercises 33–42, solve each system of equations (only real solutions) by the elimination method. Check your solutions.

${\begin{array}{r} x^{2} + y^{2} & = & 20 \\ x^{2} - y^{2} & = & 12 \end{array}$ ${\begin{array}{r} x^{2} + y^{2} & = & 20 \\ x^{2} - y^{2} & = & 12 \end{array}$
${\begin{array}{r} x^{2} + 8 y^{2} & = & 9 \\ 3 x^{2} + y^{2} & = & 4 \end{array}$ ${\begin{array}{r} x^{2} + 8 y^{2} & = & 9 \\ 3 x^{2} + y^{2} & = & 4 \end{array}$
${\begin{array}{r} x^{2} & + & 2 y^{2} & = & 12 \\ 7 y^{2} & - & 5 x^{2} & = & 8 \end{array}$ ${\begin{array}{r} x^{2} & + & 2 y^{2} & = & 12 \\ 7 y^{2} & - & 5 x^{2} & = & 8 \end{array}$
${\begin{array}{r} 3 x^{2} & + & 2 y^{2} & = & 77 \\ x^{2} & - & 6 y^{2} & = & 19 \end{array}$ ${\begin{array}{r} 3 x^{2} & + & 2 y^{2} & = & 77 \\ x^{2} & - & 6 y^{2} & = & 19 \end{array}$
${\begin{array}{rcl} x^{2} - y & = & 2 \\ 2 x - y & = & 4 \end{array}$ ${\begin{array}{rcl} x^{2} - y & = & 2 \\ 2 x - y & = & 4 \end{array}$
${\begin{array}{lcrcl} x^{2} & + & 3 y & = & 0 \\ x & - & y & = & - 12 \end{array}$ ${\begin{array}{lcrcl} x^{2} & + & 3 y & = & 0 \\ x & - & y & = & - 12 \end{array}$
${\begin{array}{r} x^{2} & + & y^{2} & = & 5 \\ 3 x^{2} & - & 2 y^{2} & = & - 5 \end{array}$ ${\begin{array}{r} x^{2} & + & y^{2} & = & 5 \\ 3 x^{2} & - & 2 y^{2} & = & - 5 \end{array}$
${\begin{array}{r} x^{2} + 4 y^{2} & = & 5 \\ 9 x^{2} - y^{2} & = & 8 \end{array}$ ${\begin{array}{r} x^{2} + 4 y^{2} & = & 5 \\ 9 x^{2} - y^{2} & = & 8 \end{array}$
${\begin{array}{r} x^{2} & + & y^{2} & + & 2 x & = & 9 \\ x^{2} & + & 4 y^{2} & + & 3 x & = & 14 \end{array}$ ${\begin{array}{r} x^{2} & + & y^{2} & + & 2 x & = & 9 \\ x^{2} & + & 4 y^{2} & + & 3 x & = & 14 \end{array}$
${\begin{array}{r} x^{2} & + & y^{2} & = & 9 \\ x^{2} & + & y^{2} & - & 18 x & = & 0 \end{array}$ ${\begin{array}{r} x^{2} & + & y^{2} & = & 9 \\ x^{2} & + & y^{2} & - & 18 x & = & 0 \end{array}$

In Exercises 43–56, use any method to solve each system of equations.

${\begin{array}{r} x + y & = & 8 \\ x y & = & 15 \end{array}$ ${\begin{array}{r} x + y & = & 8 \\ x y & = & 15 \end{array}$
${\begin{array}{r} 2 x + y & = & 8 \\ x^{2} - y^{2} & = & 5 \end{array}$ ${\begin{array}{r} 2 x + y & = & 8 \\ x^{2} - y^{2} & = & 5 \end{array}$
${\begin{array}{r} x^{2} & + & y^{2} & = & 2 \\ 3 x^{2} & + & 3 y^{2} & = & 9 \end{array}$ ${\begin{array}{r} x^{2} & + & y^{2} & = & 2 \\ 3 x^{2} & + & 3 y^{2} & = & 9 \end{array}$
${\begin{array}{r} x^{2} & + & y^{2} & = & 5 \\ 3 x^{2} & - & y^{2} & = & - 11 \end{array}$ ${\begin{array}{r} x^{2} & + & y^{2} & = & 5 \\ 3 x^{2} & - & y^{2} & = & - 11 \end{array}$
${\begin{array}{l} y^{2} & = & 4 x + 4 \\ y & = & 2 x - 2 \end{array}$ ${\begin{array}{l} y^{2} & = & 4 x + 4 \\ y & = & 2 x - 2 \end{array}$
${\begin{array}{rcl} x y & = & 125 \\ y & = & x^{2} \end{array}$ ${\begin{array}{rcl} x y & = & 125 \\ y & = & x^{2} \end{array}$
${\begin{array}{r} x^{2} + 4 y^{2} & = & 25 \\ x - 2 y + 1 & = & 0 \end{array}$ ${\begin{array}{r} x^{2} + 4 y^{2} & = & 25 \\ x - 2 y + 1 & = & 0 \end{array}$
${\begin{array}{r} y = x^{2} - 5 x + 4 \\ 3 x + y = 3 \end{array}$ ${\begin{array}{r} y = x^{2} - 5 x + 4 \\ 3 x + y = 3 \end{array}$
${\begin{array}{rcl} x^{2} - 3 y^{2} & = & 1 \\ x^{2} + 4 y^{2} & = & 8 \end{array}$ ${\begin{array}{rcl} x^{2} - 3 y^{2} & = & 1 \\ x^{2} + 4 y^{2} & = & 8 \end{array}$
${\begin{array}{rcl} 4 x^{2} - y^{2} & = & 12 \\ 4 y^{2} - x^{2} & = & 12 \end{array}$ ${\begin{array}{rcl} 4 x^{2} - y^{2} & = & 12 \\ 4 y^{2} - x^{2} & = & 12 \end{array}$
${\begin{array}{r} x^{2} & - & x y & + & 5 x & = & 4 \\ 2 x^{2} & - & 3 x y & + & 10 x & = & - 2 \end{array}$ ${\begin{array}{r} x^{2} & - & x y & + & 5 x & = & 4 \\ 2 x^{2} & - & 3 x y & + & 10 x & = & - 2 \end{array}$
${\begin{array}{r} x^{2} & - & x y & + & x & = & - 4 \\ 3 x^{2} & - & 2 x y & - & 2 x & = & 4 \end{array}$ ${\begin{array}{r} x^{2} & - & x y & + & x & = & - 4 \\ 3 x^{2} & - & 2 x y & - & 2 x & = & 4 \end{array}$
${\begin{array}{r} x^{2} & + & y^{2} & - & 8 x & = & - 8 \\ x^{2} & - & 4 y^{2} & + & 6 x & = & 0 \end{array}$ ${\begin{array}{r} x^{2} & + & y^{2} & - & 8 x & = & - 8 \\ x^{2} & - & 4 y^{2} & + & 6 x & = & 0 \end{array}$
${\begin{array}{r} 4 x^{2} & + & y^{2} & - & 9 y & = & - 4 \\ 4 x^{2} & - & y^{2} & - & 3 y & = & 0 \end{array}$ ${\begin{array}{r} 4 x^{2} & + & y^{2} & - & 9 y & = & - 4 \\ 4 x^{2} & - & y^{2} & - & 3 y & = & 0 \end{array}$

Applying the Concepts

Exercises 57 and 58 show the demand and supply functions of a product. In each case, p represents the price per unit and x represents the number of units, in hundreds. Determine the price that gives market equilibrium and the number of units that are demanded and supplied at that price.

${\begin{cases} p + 2 x^{2} = 96 \\ p - 13 x = 39 \end{cases}$ ${\begin{cases} p + 2 x^{2} = 96 \\ p - 13 x = 39 \end{cases}$
${\begin{array}{r} p^{2} + 6 p & + & 3 x & = & 75 \\ - p & + & x & = & 13 \end{array}$ ${\begin{array}{r} p^{2} + 6 p & + & 3 x & = & 75 \\ - p & + & x & = & 13 \end{array}$

In Exercises 59–66, use systems of equations to solve the problem.

Numbers. Find two positive numbers whose sum is 24 and whose product is 143.
Numbers. Find two positive numbers whose difference is 13 and whose product is 114.
Commercial land. A commercial parcel of land is in the form of a trapezoid with two right angles, as shown in the figure. The oblique side is 100 meters in length, and two sides that meet at the vertex of one of the right angles are equal. The perimeter of the land is 360 meters. Find
1. the lengths of the remaining sides.
2. the area of the land.
Dividing a pasture. The area of a rectangular pasture is 8750 square feet. The pasture is divided into three smaller pastures by two fences parallel to the shorter sides. The widths of two of the larger pastures are the same, and the width of the third is one-half that of the others. Find the possible dimensions of the original pasture if the perimeter of the smaller of the subdivisions is 190 feet.
Chartering a fishing boat. A group of students of a fraternity chartered a fishing boat that would cost $960. Before the date of the fishing trip, eight more students joined the group and agreed to pay their share of the cost of the charter. The cost per student was reduced by $6. Find the original number of students in the group and the original cost per student.
Carpet sale. A salesperson sold a square carpet and a rectangular carpet whose length was 6 feet more than its width. The combined area of the two carpets was 540 square feet. The price of the square carpet was $12 per square yard, and the price of the rectangular carpet was $10 per square yard. Assuming that the rectangular piece was $50 more than the square piece, find the dimensions of each carpet.
Buying and selling stocks. Flat Fee Brokerage, Inc., charges $100 commission for every transaction (purchase or sale) you make. Anju bought a block of stock containing over 400 shares that cost her $10,000, including the commission. After one year, she received a stock dividend of 30 shares. She then sold all of her stock for $3 more per share than it cost and made a profit of $1900 (after paying the commission). How many shares did she buy, and what was the original price of each share?
Trading stocks. Repeat Exercise 65 with the following details: Anju does not receive dividends and decides to keep 100 shares of the stock. She sells the remaining stock for $10 more per share than it cost and makes a profit of $3900 after paying the commission.

Beyond the Basics

In Exercises 67–70, the given circle intersects the given line at the points A and B. Find the coordinates of A and B and compute the distance d(A, B).

Circle: $x^{2} + y^{2} - 7 x + 5 y + 6 = 0;$ $x^{2} + y^{2} - 7 x + 5 y + 6 = 0;$ line: x-axis
The circle of Exercise 67 and the y-axis
Circle: $x^{2} + y^{2} + 2 x - 4 y - 5 = 0;$ $x^{2} + y^{2} + 2 x - 4 y - 5 = 0;$ line: $x - y + 1 = 0$ $x - y + 1 = 0$
Circle: $x^{2} + y^{2} - 6 x - 8 y - 50 = 0;$ $x^{2} + y^{2} - 6 x - 8 y - 50 = 0;$ line: $2 x + y - 5 = 0$ $2 x + y - 5 = 0$
Show that the line with equation $x + 2 y + 6 = 0$ $x + 2 y + 6 = 0$ is a tangent line to the circle with equation $x^{2} + y^{2} - 2 x + 2 y - 3 = 0 .$ $x^{2} + y^{2} - 2 x + 2 y - 3 = 0 .$ (A line that intersects the circle at exactly one point is a tangent line to the circle.)
Repeat Exercise 71 for the line with equation $3 x - 4 y = 0$ $3 x - 4 y = 0$ and the circle with equation $x^{2} + y^{2} - 2 x - 4 y + 4 = 0 .$ $x^{2} + y^{2} - 2 x - 4 y + 4 = 0 .$

In Exercises 73–76, find the value(s) of c so that the graphs of the equations in the system are tangent.

${\begin{array}{l} x & + & y & = & 2 \\ x^{2} & + & y^{2} & = & c^{2} \end{array}$ ${\begin{array}{l} x & + & y & = & 2 \\ x^{2} & + & y^{2} & = & c^{2} \end{array}$
${\begin{array}{l} c x & - & y & = & 5 \\ x^{2} & + & y^{2} & = & 16 \end{array}$ ${\begin{array}{l} c x & - & y & = & 5 \\ x^{2} & + & y^{2} & = & 16 \end{array}$
${\begin{array}{l} 4 x^{2} & + & y^{2} & = & 25 \\ 8 x & + & 3 y & = & c \end{array}$ ${\begin{array}{l} 4 x^{2} & + & y^{2} & = & 25 \\ 8 x & + & 3 y & = & c \end{array}$
${\begin{array}{l} x^{2} & + & 4 y^{2} & = & 9 \\ c x & + & 2 y & = & - 4 \end{array}$ ${\begin{array}{l} x^{2} & + & 4 y^{2} & = & 9 \\ c x & + & 2 y & = & - 4 \end{array}$

In Exercises 77–80, solve by letting $u = \frac{1}{x}$ $u = \frac{1}{x}$ and $v = \frac{1}{y} .$ $v = \frac{1}{y} .$

${\begin{array}{rcl} \frac{1}{x} - \frac{1}{y} & = & 5 \\ \frac{1}{x y} - 24 & = & 0 \end{array}$ ${\begin{array}{rcl} \frac{1}{x} - \frac{1}{y} & = & 5 \\ \frac{1}{x y} - 24 & = & 0 \end{array}$
${\begin{array}{lcr} \frac{1}{x} + \frac{1}{y} & = & 3 \\ \frac{1}{x^{2}} - \frac{1}{y^{2}} & = & - 3 \end{array}$ ${\begin{array}{lcr} \frac{1}{x} + \frac{1}{y} & = & 3 \\ \frac{1}{x^{2}} - \frac{1}{y^{2}} & = & - 3 \end{array}$
${\begin{array}{r} \frac{5}{x^{2}} - \frac{3}{y^{2}} & = & 2 \\ \frac{6}{x^{2}} + \frac{1}{y^{2}} & = & 7 \end{array}$ ${\begin{array}{r} \frac{5}{x^{2}} - \frac{3}{y^{2}} & = & 2 \\ \frac{6}{x^{2}} + \frac{1}{y^{2}} & = & 7 \end{array}$
${\begin{array}{rcl} \frac{3}{x^{2}} - \frac{2}{y^{2}} & = & 1 \\ \frac{6}{x^{2}} + \frac{5}{y^{2}} & = & 11 \end{array}$ ${\begin{array}{rcl} \frac{3}{x^{2}} - \frac{2}{y^{2}} & = & 1 \\ \frac{6}{x^{2}} + \frac{5}{y^{2}} & = & 11 \end{array}$

In Exercises 81–86, find square roots of each complex number w. [ $z = a + b i$ $z = a + b i$ is a square root of w if $z^{2} = w$ $z^{2} = w$ .]

$w = - 5 + 12 i$ $w = - 5 + 12 i$
$w = - 16 - 30 i$ $w = - 16 - 30 i$
$w = 7 - 24 i$ $w = 7 - 24 i$
$w = 3 - 4 i$ $w = 3 - 4 i$
$w = 13 + 8 \sqrt{3} i$ $w = 13 + 8 \sqrt{3} i$
$w = - 1 - 4 \sqrt{5} i$ $w = - 1 - 4 \sqrt{5} i$

In Exercises 87–92, solve each system of equations and verify the solutions graphically.

${\begin{array}{l} y = 3^{x} + 4^{2} \\ y = 3^{2 x} - 2 \end{array}$ ${\begin{array}{l} y = 3^{x} + 4^{2} \\ y = 3^{2 x} - 2 \end{array}$
${\begin{array}{l} y = \ln x + 1 \\ y = \ln x - 2 \end{array}$ ${\begin{array}{l} y = \ln x + 1 \\ y = \ln x - 2 \end{array}$
${\begin{array}{l} y = 2^{x} + 3^{2} \\ y = 2^{2 x} + 1 \end{array}$ ${\begin{array}{l} y = 2^{x} + 3^{2} \\ y = 2^{2 x} + 1 \end{array}$
${\begin{array}{l} y = 2^{x} + 3 \\ y = 4^{x} + 1 \end{array}$ ${\begin{array}{l} y = 2^{x} + 3 \\ y = 4^{x} + 1 \end{array}$
${\begin{array}{rcl} x & = & 3^{y} \\ 3^{2 y} & = & 3 x - 2 \end{array}$ ${\begin{array}{rcl} x & = & 3^{y} \\ 3^{2 y} & = & 3 x - 2 \end{array}$
${\begin{array}{rcl} x & = & 3^{y} \\ x^{2} & = & 3^{y + 1} - 2 \end{array}$ ${\begin{array}{rcl} x & = & 3^{y} \\ x^{2} & = & 3^{y + 1} - 2 \end{array}$

Critical Thinking / Discussion / Writing

Consider a system of two equations in two variables.

${\begin{cases} y = m x + b \\ y = A x^{3} + B x^{2} + C x + D, A \neq 0 \end{cases}$ ${\begin{cases} y = m x + b \\ y = A x^{3} + B x^{2} + C x + D, A \neq 0 \end{cases}$

Explain whether it is possible for this system to have
1. no real solutions.
2. one real solution.
3. two real solutions.
4. three real solutions.
5. four real solutions.
6. more than four real solutions.
Repeat Exercise 93 for the system.

${\begin{cases} y = m x + b \\ y = A x^{4} + B x^{3} + C x^{2} + D x + E, A \neq 0 \end{cases}$ ${\begin{cases} y = m x + b \\ y = A x^{4} + B x^{3} + C x^{2} + D x + E, A \neq 0 \end{cases}$

Getting Ready for the Next Section

In Exercises 95–102, sketch the graph of each equation.

$2 x + y = 0$ $2 x + y = 0$
$2 y - x = 0$ $2 y - x = 0$
$3 x + 4 y = 12$ $3 x + 4 y = 12$
$4 x - 3 y = 24$ $4 x - 3 y = 24$
$y - 2 x^{2} = 0$ $y - 2 x^{2} = 0$
$y - x^{2} = 3$ $y - x^{2} = 3$
$y + x^{2} - 2 = 0$ $y + x^{2} - 2 = 0$
$x^{2} + y^{2} = 16$ $x^{2} + y^{2} = 16$

In Exercises 103–106, determine which of the given points are on the graph of the equation.

Equation	Points
$5 x + 3 y = 0$ $5 x + 3 y = 0$	$(0, 0), (- 3, 5), (1, - 1), (- 1, 1)$ $(0, 0), (- 3, 5), (1, - 1), (- 1, 1)$
$3 x + 4 y = 12$ $3 x + 4 y = 12$	$(0, 0), (4, 0), (0, 3), (3, 4)$ $(0, 0), (4, 0), (0, 3), (3, 4)$
$y = x^{2} + 3$ $y = x^{2} + 3$	$(0, 0), (1, 4), (- 1, 4), (2, 5)$ $(0, 0), (1, 4), (- 1, 4), (2, 5)$
$y + x^{2} = 2$ $y + x^{2} = 2$	$(0, 0), (0, 2), (2, - 2), (1, 1)$ $(0, 0), (0, 2), (2, - 2), (1, 1)$

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Table of Contents for Section 5.4 Systems of Nonlinear Equations

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Table of Contents for
Section 5.4 Systems of Nonlinear Equations