Download presentation

Presentation is loading. Please wait.

Published byJody Hutchinson Modified over 6 years ago

1
Bell Quiz 2-6 1. Horizontal asymptotes = ? 2. Vertical asymptotes = ? 3. Zeroes = ? 4. What is the “end behavior” of

2
2.7 Solving Equations in one variable.

3
What you’ll learn about Solving Rational Equations Extraneous Solutions Applications … and why Applications involving rational functions as models often require that an equation involving fractions be solved.

4
Extraneous Solutions When we divide an equation by an expression containing variables, the resulting equation may have solutions that are not solutions of the original equation. These are extraneous solutions. For this reason we must check each solution of the resulting equation in the original equation.

5
Fraction Review 1. Eliminate the denominator by either: (1) eliminate each denomator one at a time (2) Multiply by the LCD (using the property of equality) (using the property of equality) 3. Solve the resulting equation.

6
Solving by Clearing Fractions 1. Eliminate the denominator. 2. Multiply by the LCD (using the property of equality) (using the property of equality) 3. Solve the resulting quadratic equation. 4.Check for extraneous solutions (if you multiplied or divided by an expression with a variable in it). by an expression with a variable in it).

7
Your turn: solve for ‘x’. 1. 2. 3.

8
Eliminating Extraneous Solutions 1.These are fractions. Get a common denominator.

9
Eliminating Extraneous Solutions Now that you have a common denominator multiple both sides by the common denominator. sides by the common denominator.

10
Eliminating Extraneous Solutions 2. Solve the resulting quadratic equation. 4. Check for extraneous solutions (if you multiplied or divided by an expression with a variable in it). by an expression with a variable in it). If x = 3, the result will be division by zero x = 3 is an extraneous solution and must be rejected. extraneous solution and must be rejected.

11
Your turn: solve for ‘x’, check for extraneous roots. 4. 5.

12
Finding a Minimum Perimeter 1. Find the dimensions of the rectangle with minimum perimeter, if its area is 300 square meters. perimeter, if its area is 300 square meters. 2. Find this least perimeter. Area = 300 x What is the length What is the length of this side ? of this side ? Can you write an expression that uses expression that uses the length of 1 side the length of 1 side and the given area? and the given area?

13
Finding a Minimum Perimeter 1. Find the dimensions of the rectangle with minimum perimeter, if its area is 300 square meters. perimeter, if its area is 300 square meters. 2. Find this least perimeter. Area = 300 x A= L*w 300 = x * w w= 300/x

14
Finding a Minimum Perimeter 1. Find the dimensions of the rectangle with minimum perimeter, if its area is 300 square meters. perimeter, if its area is 300 square meters. 2. Find this least perimeter. Area = 300 x Perimeter = ? Solve the resulting equation graphically. Minimum perimeter = 69.3

15
Finding a Minimum Perimeter 1. Find the dimensions of the rectangle with minimum perimeter, if its area is 300 square meters. perimeter, if its area is 300 square meters. Area = 300 x Solve graphically. 17.7 16.9

16
Your turn: 6. Find the minimum perimeter of this rectangle. Area = 250 5x

17
Finding the Dimensions of a Can A juice company uses 2 liter cans vol = 2000 cubic cm. The can has a surface area of 1000 square cm. Find the radius and height of the can. 1. Find an equation for volume. 2. Find an equation for surface area.

18
Finding the Dimensions of a Can A juice company uses 2 liter cans vol = 2000 cubic cm. The can has a surface area of 1000 square cm. Find the radius and height of the can. 3. Use the substitution method to write a single equation with one variable. single equation with one variable. 4. Solve graphically.

19
Finding the Dimensions of a Can A juice company uses 2 liter cans vol = 2000 cubic cm. The can has a surface area of 1000 square cm. Find the radius and height of the can. 4. Solve graphically. Radius = 4.6 cm or 9.7 cm. If radius = 4.6 cm then height = 29.8 cm If radius = 9.7 cm then height = 6.83 cm

20
HOMEWORK Section 2-7

Similar presentations

© 2021 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google