section 1.2 solving equations using a graphing utility; linear and rational equations copyright ©...

Section 1.2

Solving Equations Using a Graphing Utility; Linear and Rational

Equations

Copyright © 2013 Pearson Education, Inc. All rights reserved

An equation in one variable is a statement in which two expressions, at least one containing the variable, are equal.The expressions are called the sides of the equation.The admissible values of the variable, if any, that result in a true statement are called solution, or roots, of the equation.To solve an equation means to find all the solutions of the equation.

Equations in One Variable


Examples of an equation in one variable:Equations in One Variable

x 5 9 x2 5x 2x 2 x2 4x 1

0 x2 9 5

x + 5 = 9 is true when x = 4. 4 is a solution of the equation or we say that 4 satisfies the equation.We write the solution in set notation, this is called the solution set of the equation.

An equation that is satisfied for every choice of the variable for which both sides are defined is called an identity.


Solve Equations Using a Graphing Utility



Using ZERO (or ROOT) to Approximate Solutions of an Equation

Find the solution(s) of the equation x3 – x + 1 = 0. Round answers to two decimal places.The solutions of the equation x3 – x + 1 = 0 are the same as the x-intercepts of the graph of Y1 = x3 – x + 1.

There appears to be one x-intercept (solution) between –2 and –1.


Using ZERO (or ROOT) to Approximate Solutions of an Equation

Using the ZERO (or ROOT) feature of a graphing utility, we determine that the x-intercept, and thus the solution to the equation, is x = –1.32 rounded to two decimal places.


Using INTERSECT to Approximate Solutions of an Equation

Find the solution(s) to the equation 4x4 – 3 = 2x + 1. Round answers to two decimal places.Begin by graphing each side of the equation:

Y1 = 4x4 – 3 and Y2 = 2x + 1.



At a point of intersection of the graphs, the value of the y-coordinate is the same for Y1 and Y2. Thus, the x-coordinate of the point of intersection represents a solution to the equation.



The INTERSECT feature on a graphing utility determines a point of intersection of the graphs. Using this feature, we find that the graphs intersect at (–0.87, –0.73) and (1.12, 3.23) rounded to two decimal places.

One method for solving equations algebraically requires that a series of equivalent equations be developed from the original equation until an obvious solution results.

Solving Equations Algebraically

Solve the equation: 4 7 13x 77 13 74x

4 20x

4 04 4

2x

5x


1 1Solve the equation: 1 3 3 23 4x x

1 11 3 3 23

12 124

x x

4 1 3 12 3 3 2x x

4 4 36 9 6x x

4 40 9 6x x 40 9 40 94 40 9 6x xx x

5 46x 5 465 5x

465

x


Solve the equation: 3 1 2 3 3 2x x x x 2 23 5 2 3 7 6x x x x

5 2 7 6x x

5 7 4x x

2 4x

2x


Solve Rational Equations


A rational equation is an equation that contains a rational expression.

To solve a rational equation, multiply both sides of the equation by the least common multiple of the denominators of the rational expressions that make up the rational equation.

3x 1

2x 1

7x 5x 4

3x 2


2 3 1Solve the equation:

1 3 3 1x x x x

2 3 13 1 3 1

1 3 3 1x x x xx x x x

2 6 3 1 1x x 2 6 3 3 1x x

8x 8x


Solving a Rational Equation

3 6Solve the equation: 12 2x

x x

2 2 ,2

23 6 12x x x x

x x

3 2 6x x

4 2 6x 4 8x 2x no solution


Sometimes the process of creating equivalent equations leads to apparent solutions that are not solutions of the original equation. These are called extraneous solutions.

Extraneous Solutions


A total of $16,000 is invested, some in stocks and some in bonds. If the amount invested in bonds is one fourth that invested in stocks, how much is invested in each category?

We are being asked to find the amount of two investments. These amounts total $16,000.



If x equals the amount invested in stocks, then the rest of the money, 16,000 – x, is the amount invested in bonds.



We also know that: Total amount invest in bonds is one fourth that in stocks

16,000 x 14x



1160004

x x 1160004

x x 5160004x

4 4 5160005 5 4

x

3200 x

So $3200 is invested in stocks and $16000 – $3200 = $12,800 is invested in bonds.



The total invested is $3200 + $12,800 = $16000 and the amount of bonds, $3200 is one fourth of that of stocks, $12,800.


Andy grossed $440 one week by working 50 hours. His employer pays time-and-a-half for all hours worked in excess of 40 hours. What is Andy’s hourly wage?

We are looking for an hourly wage. Our answer will be expressed in dollars per hour.



Let x represent the regularly wage; x is measured in dollars per hour.



The sum of regular salary plus overtime salary wil equal $440. From the table, 40x + 15x = 440.



40x + 15x = 440

55x = 440

x = 8

Andy’s regularly wage is $8.00 per hour.



Forty hours yields a salary of 40(8.00) = $320.00 and 10 hours overtime yields a salary of 10(1.5)(8.00) = $120.00 for a total of $440.00.


section 1.2 solving equations using a graphing utility; linear and rational equations copyright ©...

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