1. Linear Equations, Functions, Zeros, and Applications
Section 1.5
Linear Equations, Functions, Zeros, and Applications

2. Objectives Solve linear equations.
Solve applied problems using linear models.
Find zeros of linear functions.

3. 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

4. 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.

5. 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.

6. 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.

7. Example
Solve:
Start by multiplying both sides of the equation by the LCD to clear the equation of fractions.

8. Example (continued)
Check:
TRUE
The solution is

9. Example - Special Cases
Solve:
No matter what number we substitute for x, we get a false sentence. Thus, the equation has no solution.

10. Example - Special Cases
Solve:
Replacing x with any real number gives a true equation. 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 (–∞, ∞).

11. 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.

12. Five Steps for Problem Solving
1. Familiarize yourself with the problem situation. If the problem is presented in words, this means to read carefully. Some or all of the following can also be helpful.
Make a drawing, if it makes sense to do so.
Make a written list of the known facts and a list of what you wish to find out.
Assign variable to represent unknown quantities.
Organize the information in a chart or a table, if appropriate.
Find further information. Look up a formula, consult a reference book or an expert in the field, or do research on the Internet.
Guess or estimate the answer and check your guess or estimate.

13. Five Steps for Problem Solving
2. Translate the problem situation 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.

14. The Motion Formula
The distance d traveled by an object moving at rate r in time t is given by d = rt.

15. Example
Delta Airlines’ fleet includes B737/800’s, each with a cruising speed of 517 mph, and Saab 340B’s, each with a cruising speed of 290 mph (Source: Delta Airlines). Suppose that a Saab 340B takes off and travels at its cruising speed. One hour later, a B737/800 takes off and follows the same route, traveling at its cruising speed. How long will it take the B737/800 to overtake the Saab 340B?

16. Example (continued)
1. Familiarize. Make a drawing showing both the known and unknown information. Let t = the time, in hours, that the B737/800 travels before it overtakes the Saab 340B. Since the Saab 340B takes off 1 hr before the 737, it will travel t + 1 hr before being overtaken. The planes will travel the same distance, d, when one overtakes the other.

17. Example (continued)
We can organize the information in a table as follows.
d = r • t
Distance
Rate
Time
B737/800 d
517 t
Saab 340B
290
t + 1
2. Translate. Using the formula d = rt , we get two expressions for d:
d = 517t and d = 290(t + 1).
Since the distance are the same, the equation is:
517t = 290(t + 1)

18. Example (continued) 3. Carry out. We solve the equation.
517t = 290(t + 1)
517t = 290t + 290
227t = 290
t ≈ 1.28
4. Check. If the B737/800 travels about 1.28 hr, then the Saab 340B travels for about , or 2.28 hr. In 2.28 hr, the Saab 340B travels 290(2.28), or mi; and in 1.28 hr, the B737/800 travels 517(1.28), or mi. Since mi ≈ mi, the answer checks.

19. Example (continued)
5. State. About 1.28 hr after the B737/800 has taken off, it will overtake the Saab 340B.

20. Simple-Interest Formula
The simple interest I on a principle of P dollars at interest rate r for t years is given by I = Prt I = the simple interest ($) P = the principal ($) r = the interest rate (%) t = time (years)

21. Example
Damarion’s two student loans total $28,000. One loan is at 5% simple interest and the other is at 3% simple interest. After 1 year, Damarion owes $1040 in interest. What is the amount of each loan?

22. Example (continued)
Solution: 1. Familiarize. We let x = the amount borrowed at 5% interest. Then the remainder is $28,000 – x, borrowed at 3% interest.
Amount Borrowed
Interest Rate
Time
Amount of Interest
5% loan x
0.05
1 year
0.05x
3% loan
28,000 – x
0.03
0.03(28,000 – x)
Total
28,000
1040

23. Example (continued)
2. Translate. The total amount of interest on the two loans is $1040. Thus we write the following equation.
0.05x (28,000  x) = 1040
3. Carry out. We solve the equation.
0.05x  0.03x = 1040
0.02x = 1040
0.02x = 200
x = 10,000
If x = 10,000, then 28,000  10,000 = 18,000.

24. Example (continued)
4. Check. The interest on $10,000 at 5% for 1 yr is $10,000(0.05)(1), or $500. The interest on $18,000 at 3% for 1 yr is $18,000(0.03)(1) or $540. Since $500 + $540 = $1040, the answer checks.
5. State. Damarion borrowed $10,000 at 5% interest and $18,000 at 3% interest.

25. 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.

26. Example Find the zero of f (x) = 5x  9. Algebraic Solution:
Using a table in ASK mode we can check the solution. Enter y = 5x – 9 into the equation editor, then enter x = 1.8 into the table and it yields y = 0. That means the zero is 1.8.

27. Example (continued) Find the zero of f (x) = 5x  9.
Graphical Solution or Zero Method:
Graph y = 5x – 9.
Use the ZERO feature from the CALC menu.
The x-intercept is 1.8, so the zero of f (x) = 5x – 9 is 1.8.