2.4 – Writing Linear Equations. 2.4 – Writing Linear Equations Forms:
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Transcript of Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 1.5 Linear Equations,...
Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
1.5 Linear Equations, Functions, Zeros,
and Applications
Solve linear equations. Solve applied problems using linear models. Find zeros of linear functions.
Slide 1.5 - 2Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Equations and Solutions
An equation is a statement that two expressions are equal.
To solve an equation in one variable is to find all the values of the variable that make the equation true.
Each of these numbers is a solution of the equation.
The set of all solutions of an equation is its solution set.
Some examples of equations in one variable are
2x 3 5, 3 x 1 4x 5,x 3
x 41,
and x2 3x 2 0.
Slide 1.5 - 3Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Linear Equations
A linear equation in one variable is an equation that can be expressed in the form mx + b = 0, where m and b are real numbers and m ≠ 0.
Slide 1.5 - 4Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Equivalent Equations
Equations that have the same solution set are equivalent equations.
For example, 2x + 3 = 5 and x = 1 are equivalent equations because 1 is the solution of each equation.
On the other hand, x2 – 3x + 2 = 0 and x = 1 are not equivalent equations because 1 and 2 are both solutions of x2 – 3x + 2 = 0 but 2 is not a solution of x = 1.
Slide 1.5 - 5Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Equation-Solving Principles
For any real numbers a, b, and c:
The Addition Principle:
If a = b is true, then a + c = b + c is true.
The Multiplication Principle:
If a = b is true, then ac = bc is true.
Slide 1.5 - 6Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Example
Solve:
Solution: Start by multiplying both sides of the equation by the LCD to clear the equation of fractions.
3
4x 1
7
5
203
4x 1
20
7
5
203
4x 201 28
15x 20 28
15x 20 20 28 20
15x 48
15x
15
48
15
x 48
15
x 16
5
Slide 1.5 - 7Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Example (continued)
Check:
3
416
5 1 ?
7
5
12
5
5
57
5
7
5
3
4x 1
7
5
The solution is 16
5.
TRUE
Slide 1.5 - 8Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Example - Special Case
Solve:
Solution:
24x 7 17 24x
24 7 17 24x x
24x 24x 7 24x 17 24x
7 17
Some equations have no solution.
No matter what number we substitute for x, we get a false sentence.Thus, the equation has no solution.
Slide 1.5 - 9Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Example - Special Case
Solve:
Solution:
3 1
3x
1
3x 3
1
3x 3
1
3x
1
3x
1
3x 3
3 3
There are some equations for which any real number is a solution.
Replacing x with any real number gives a true sentence. Thus any real number is a solution. The equation has infinitely many solutions. The solution set is the set of real numbers, {x | x is a real number}, or (–∞, ∞).
Slide 1.5 - 10Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Applications Using Linear Models
Mathematical techniques are used to answer questions arising from real-world situations.
Linear equations and functions model many of these situations.
Slide 1.5 - 11Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Five Steps for Problem Solving
1. Familiarize yourself with the problem situation.
Make a drawing Find further information
Assign variables Organize into a chart or table
Write a list Guess or estimate the answer
2. Translate to mathematical language or symbolism.
3. Carry out some type of mathematical manipulation.
4. Check to see whether your possible solution actually fits the problem situation.
5. State the answer clearly using a complete sentence.
Slide 1.5 - 12Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
The Motion Formula
The distance d traveled by an object moving at rate r in time t is given by
d = rt.
Slide 1.5 - 13Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Example
America West Airlines’ fleet includes Boeing 737-200’s, each with a cruising speed of 500 mph, and Bombardier deHavilland Dash 8-200’s, each with a cruising speed of 302 mph (Source: America West Airlines). Suppose that a Dash 8-200 takes off and travels at its cruising speed. One hour later, a 737-200 takes off and follows the same route, traveling at its cruising speed. How long will it take the 737-200 to overtake the Dash 8-200?
Slide 1.5 - 14Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Example (continued)
1. Familiarize. Make a drawing showing both the known and unknown information. Let t = the time, in hours, that the 737-200 travels before it overtakes the Dash 8-200. Therefore, the Dash 8-200 will travel t + 1 hours before being overtaken. The planes will travel the same distance, d.
Slide 1.5 - 15Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Example (continued)
We can organize the information in a table as follows.
2. Translate. Using the formula d = rt , we get two expressions for d:
d = 500t and d = 302(t + 1).Since the distance are the same, the equation is:
500t = 302(t + 1)
Distance Rate Time
737-200 d 500 t
Dash 8-200 d 302 t + 1
d = r • t
Slide 1.5 - 16Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Example (continued)
500t = 302(t + 1)500t = 302t + 302198t = 302 t = 302/198 ≈ 1.53
4. Check. If the 737-200 travels about 1.53 hours, it travels about 500(1.53) ≈ 765 mi; and the Dash 8-200 travels about 1.53 + 1, or 2.53 hours and travels about 302(2.53) ≈ 764.06 mi, the answer checks. (Remember that we rounded the value of t).
5. State. About 1.53 hours after the 737-200 has taken off, it will overtake the Dash 8-200.
3. Carry out. We solve the equation.
Slide 1.5 - 17Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Simple-Interest Formula
I = Prt
I = the simple interest ($)
P = the principal ($)
r = the interest rate (%)
t = time (years)
Slide 1.5 - 18Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Example
Jared’s two student loans total $12,000. One loan is at 5% simple interest and the other is at 8%. After 1 year, Jared owes $750 in interest. What is the amount of each loan?
Slide 1.5 - 19Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Example (continued)
Solution: 1. Familiarize. We let x = the amount borrowed at 5%
interest. Then the remainder is $12,000 – x, borrowed at 8% interest.
Amount Borrowed
Interest Rate
Time Amount of Interest
5% loan x 0.05 1 0.05x
8% loan 12,000 – x 0.08 1 0.08(12,000 – x)
Total 12,000 750
Slide 1.5 - 20Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Example (continued)
2. Translate. The total amount of interest on the two loans is $750. Thus we write the following equation.
0.05x + 0.08(12,000 x) = 750
3. Carry out. We solve the equation.
0.05x + 0.08(12,000 x) = 750
0.05x + 960 0.08x = 750
0.03x + 960 = 750
0.03x = 210
x = 7000
If x = 7000, then 12,000 7000 = 5000.
Slide 1.5 - 21Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Example (continued)
4. Check. The interest on $7000 at 5% for 1 yr is $7000(0.05)(1), or $350. The interest on $5000 at 8% for 1 yr is $5000(0.08)(1) or $400. Since $350 + $400 = $750, the answer checks.
5. State. Jared borrowed $7000 at 5% interest and $5000 at 8% interest.
Slide 1.5 - 22Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Zeros of Linear Functions
An input c of a function f is called a zero of the function, if the output for the function is 0 when the input is c. That is, c is a zero of f if f (c) = 0.
A linear function f (x) = mx + b, with m 0, has exactly one zero.
Slide 1.5 - 23Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
ExampleFind the zero of f (x) = 5x 9.
Algebraic Solution: 5x 9 = 0 5x = 9 x = 1.8
Visualizing the Solution:
The intercept of the graph is (9/5, 0) or (1.8, 0).