Algebra I Workbook For Dummies
Page 15
When you replace the x with any number smaller than –3, the factor (x – 4) is negative, and the factor (x + 3) is also negative. When you choose a number between –3 and 4 — say 0, for instance — the factor (x – 4) is negative, and the factor (x + 3) is positive. When you replace the x with any number bigger than 4, the factor (x – 4) is positive, and the factor (x + 3) is also positive.
Recall that, when you multiply an even number of negative numbers, the product is positive, and when the number of negative numbers is odd, the product is negative. The inequality asks for when the product is positive; according to the signs on the line, this is when x < –3 (when a negative is multiplied by a negative) or x > 4 (when a positive is multiplied by a positive).
9. Solve for x: (x + 7)(x – 1) > 0.
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10. Solve for x: .
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11. Solve for x: .
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12. Solve for x: .
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Dealing with Polynomial and Rational Inequalities
The same process that gives you solutions to quadratic inequalities is used to solve some other types of inequality problems (see “Solving Quadratic Inequalities” earlier in this chapter for more info). You use the number line process for polynomials (higher degree expressions) and for rational expressions (fractions where a variable ends up in the denominator).
Q. Solve for x: .
A. or . The polynomial is already factored (thank goodness). So set each factor equal to 0 to determine where the factors might change from positive to negative or vice versa (critical numbers). Place the zeros on a number line, test each factor in each interval on the number line, and determine the sign of the expression in that interval. Use that information to solve the problem.
The solution consists of all the numbers that make the expression either negative or equal to 0. The product is negative when you have an odd number of negative factors. Notice that the sign of the product didn’t change when x went from smaller than 1 to larger than 1. That’s because that factor was squared, making that factor either positive or 0 all the time.
Q. Solve for x: .
A. or . Solving an inequality with a fraction works pretty much the same way as solving one with a polynomial. Find all the zeros of the factors, put them on a number line, and check for the sign of the result. The only thing you need to be careful of is not including some number that would result in a 0 factor in the denominator. The sign can change from one side to the other of that number; you just can’t use it in your answer. See this number line for clarification.
Remember the rule for dividing a signed number is the same as it is in multiplication. According to the sign line, the fraction is positive for numbers smaller than –2 or for numbers greater than +3. Also, you get a 0 when x = –2.
13. Solve for x: .
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14. Solve for x: .
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15. Solve for x: .
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16. Solve for x: .
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Solving Absolute Value Inequalities
Put the absolute value function together with an inequality, and you create an absolute value inequality. When solving absolute value equations (refer to Chapter 15, if you need a refresher), you rewrite the equations without the absolute value function in them. To solve absolute value inequalities, you also rewrite the form to create simpler inequality problems — types you already know how to solve. Solve the new problem or problems for the solution to the original.
You need to keep two different situations in mind. (Always assume that the c is a positive number or 0.) In the following list, I show only the rules for > and <, but the same holds if you’re working with ≤ or ≥ :
If you have , change the problem to ax + b > c or ax + b < –c and solve the two inequalities.
If you have , change the problem to –c < ax + b < c and solve the one compound inequality.
Q. Solve for x: .
A. x < , x > 1 or . Change the absolute value inequality to the two separate inequalities 9x – 5 > 4 or 9x – 5 < –4. Solving the two inequalities, you get the two solutions 9x > 9, x > 1 or 9x < 1, x < . So numbers bigger than 1 or smaller than satisfy the original statement.
Q. Solve for x: .
A. < x < 1 or . This is like the first example, except that the sense has been turned around. Rewrite the inequality as –4 < 9x – 5 < 4. Solving it, the solution is 1 < 9x < 9, < x < 1. The inequality statement says that all the numbers between and 1 work. Notice that the interval in this answer is what was left out in writing the solution to the problem when the inequality was reversed.
17. Solve for x: .
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18. Solve for x: .
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19. Solve for x: .
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20. Solve for x: .
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Solving Complex Inequalities
A complex inequality — one with more than two sections (intervals or expressions sandwiched between inequalities) to it — could be just compound and have variables in the middle, or they can have variables in more than one section where adding and subtracting can’t isolate the variable in one place. When this happens, you have to break up the inequality into solvable sections and write the answer in terms of what the different solutions share.
Q. Solve for x: .
A. or [1, 10). Break it up into the two separate problems: and . Solve the first inequality to get . Solve the second inequality to get x < 10. Putting these two answers together, you get that x must be some number both bigger than or equal to 1 and smaller than 10.
Q. Solve for x: .
A. or . Break the inequality up into two separate problems. The solution to is –1 < x, and the solution to is . The two solutions overlap, with all the common solutions lying to the right of and including .
21. Solve for x: .
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22. Solve for x: .
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Answers to Problems on Working with Inequalities
This section provides the answers (in bold) to the practice problems in this chapter.
1. Starting with 5 > 2, add 4 to each side and then divide by –3. The answer is .
2. Starting with 5 ≥ 1, multiply each side by –4 and then divide each side by –2. The answer is , because you reverse the sense twice.
3. Write , using interval notation. The answer is .
4. Write , using interval notation. The answer is [2, 21].
5. Solve for y: . The answer is or . Subtract 4 and then divide by –5:
6. Solve for x: . The answer is or .
7. Solve for x: . The answer is or (–4,2].
8. Solve for x: . The answer is or (–2,2]. First subtract 7 and then divide by –2 to get the answer:
9. Solve for x: . The answer is or .
when x = –7, and x – 1 = 0 when x = 1, as shown on the following number line.
If , then both factors are positive. If , then is positive, but is negative, making the product negative. If then both factors are negative, and the product is positive. So .
10. Solve for x: . The answer is or [–4, 5]. First subtract 20 from both sides; then factor the quadratic and set the factors equal to 0 to find the values where the factors change signs:
x – 5 = 0 when x = 5, and x + 4 = 0 when x = –4, as shown on the following number line.
If , then and , making the product positive. If then x – 5 < 0, and , resulting in a product that’s negative. If then both factors are negative, and the product is positive. So the product is negative only if . But the solutions of are . Including these two values, the solution is then .
11. Solve for x: . The answer is or . First subtract 9x from each side. Then factor the quadratic and set the factors equal to 0:
, as shown on the following number line.
If , both factors are positive, and . If , then , making the product negative. If , both factors are negative, and . Wh
en . Include those in the solution to get .
12. Solve for x: . The answer is or . First set the expression equal to 0 to help find the factors:
The critical numbers are 5 and –5. Place them on the number line.
From the figure, . When .
13. Solve for x: . The answer is or . First set the factored expression equal to 0. Put these values on the number line:
Assign signs to each of the four regions to get the answer .
14. Solve for x: . The answer is or . First factor the expression by grouping. Then set the factors equal to 0.
The factored form is equal to 0 only if . Put this number on the number line.
Assign signs to each interval to get the answer .
15. Solve for x: . The answer is or .
when the numerator 5 + x = 0, x = –5. When the denominator x = 0, is undefined, so place x = 0 and x = –5 on the number line.
Assign signs to each of the three intervals to get the answer: x > 0 or x < –5.
16. Solve for x: . The answer is or .
The fraction equals 0 when , and it’s undefined when .
Assign signs to each of the four intervals. So when . But only when x = 1 or x = –1, not at x = –3. So the way to write these answers all together with inequality symbols is .
17. Solve for x: . The answer is or . First, rewrite the absolute value as an inequality. Then use the rules for solving inequalities to isolate x in the middle and determine the answer:
Adding 3 to each section gives you . Then divide each section by 4 to get .
18. Solve for x: . The answer is , . First, rewrite the absolute value as two inequalities. The x is then isolated to one side of the inequality by adding and dividing.
Solving for x: , by dividing 2 into each side.
19. Solve for x: . The answer is or [1, 5]. Before rewriting the absolute value, isolate it on one side of the inequality.
Add –6 to each side: . Now rewrite the absolute value as an inequality that can be solved. The x gets isolated in the middle, giving you the answer.
by adding 3 to each section.
20. Solve for x: 5|7x – 4| + 1 > 6. The answer is or . Before rewriting the absolute value, it has to be alone, on the left side. First, add –1 to each side: . Then divide each side by 5 to get . Now you can rewrite the absolute value as two inequality statements. Each statement is solved by performing operations that end up as x greater than or less than some value. Add 4 to each side to get
. And finally divide by 7 to get .
21. Solve for x: . The answer is or [1, 3). First separate into and .
When , subtract 1 from each side to get and then divide each side by 5 to get .
When , subtract 1 from each side and subtract 2x from each side to get and then divide each side by 3 to get . So , which gives the answer . This part of the answer includes the first part, also.
22. Solve for x: . The answer is or . First separate into and . Solving the first inequality, add 3 to each side to get . Then divide each side by 4 to get . Solving the other inequality, you subtract 1 from each side and subtract 4x from each side to get . This second part of the answer is an overlap of the first — all the answers are already covered in the first inequality. For x to satisfy both inequalities, the answer is only .
Part IV
Solving Story Problems and Sketching Graphs
In this part . . .
Now you get to the good part. Not that everything up to this point hasn’t been just wonderful! Applications are what algebra is all about. They’re one of the main reasons to become proficient at algebra. Solving an equation is great, but without some reason for it, what good is that solution? When you solve an equation and get x = 4, you want to be able to say, “See, I told you that the truck had four men in it!” And graphs are wonderful for demonstrating what you found or aiding you in finding it.
Chapter 17
Facing Up to Formulas
In This Chapter
Finding comfort in familiar formulas
Introducing perimeter, area, and volume formulas
Computing interest and percentages
Just as a cook refers to a recipe when preparing delectable concoctions, algebra uses formulas to whip up solutions. In the kitchen, a cook relies on the recipe to turn equal amounts of jalapeños, cream cheese, and beans into a zippy dip. In algebra, a formula is an equation that expresses some relationship you can count on to help you concoct such items as the diagonal distance across a rectangle or the amount of interest paid on a loan.
Formulas such as I = Prt, , and d = rt are much more compact than all the words needed to describe them. And as long as you know what the letters stand for, you can use these formulas to solve problems.
Working with formulas is easy. You can apply them to so many situations in algebra and in real life. Most formulas become old, familiar friends. Plus a certain comfort comes from working with formulas, because you know they never change with time, temperature, or relative humidity.
This chapter provides you several chances to work through some of the more common formulas and to tweak areas where you may need a little extra work. The problems are pretty straightforward, and I tell you which formula to use. In Chapter 18, you get to make decisions as to when and if to use a formula, or whether you get to come up with an equation all on your own.
Working with Formulas
A formula is an equation that expresses some known relationship between given quantities. You use formulas to determine how much a dollar is worth when you go to another country. You use a formula to figure out how much paint to buy when redecorating your home.
Sometimes, solving for one of the variables in a formula is advantageous if you have to repeat the same computation over and over again. In each of the examples and practice exercises, you use the same rules from solving equations, so you see familiar processes in familiar formulas. The examples in this section introduce you to various formulas and then show you how to use them to solve problems.
Q. The formula for changing from degrees Celsius, °C, to degrees Fahrenheit, °F, is . What is the temperature in degrees Fahrenheit when it’s 25 degrees Celsius?
A. 77°F. Replace C with 25, multiply by , and then add 32.
Q. Use the formula for changing Celsius to Fahrenheit to find the average body temperature in degrees Celsius.
A. 37°C. First solve for C by subtracting 32 from each side of the formula and then multiplying by .
Now replace F with 98.6 (average normal body temperature) and compute C.
1. The simple interest formula is I = Prt, where I is the interest earned, P is the principal or what you start with, r is the interest rate written as a decimal, and t is the amount of time in years. Find I if the principal, P, is $10,000, the rate, r is 2%, and the time, t, is 4 years.
Solve It
2. The simple interest formula is I = Prt. Solve for t so you can find out how long it takes for $10,000 to earn $1,000 interest when the rate is 2%.
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3. The formula for the perimeter of a rectangle is P = 2(l + w). Find the perimeter, P, when the length is 7 feet and the width is one yard.
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4. The formula for the area of a trapezoid is . Solve for h so you can determine how high a trapezoid is if the area is 56 square centimeters and the two bases are 6 cm. and 8 cm.