In This Chapter
* Applying order of operations and algebraic properties
* Using FOIL and other products
* Solving linear and absolute value equations
* Dealing with inequalities
The nice thing about the rules in algebra is that they apply no matter what level of mathematics or what area of math you're studying. Everyone follows the same rules, so you find a nice consistency and orderliness. In this chapter, I discuss and use the basic rules to prepare you for what comes next in Algebra II.
Good Citizenship: Following the Order of Operations and Other Properties
The order of operations in mathematics deals with what comes first (much like the chicken and the egg). When faced with multiple operations, this order tells you the proper course of action.
ALGEBRA RULES
The order of operations states that you use the following sequence of events:
1. Raise to powers or find roots.
2. Multiply or divide.
3. Add or subtract.
Special groupings can override the normal order of operations. For instance, [a.sup.b + c] asks you to add b + c before raising a to a power. If groupings are a part of the expression, first perform whatever's in the grouping symbol. The most common grouping symbols are parentheses, ( ); brackets, ; braces, { }; fraction bars, -; absolute value bars, | |; and radical signs, [square root].
If you find more than one operation from the same level, move from left to right performing those operations.
ALGEBRA RULES
The commutative, associative, and distributive properties allow you to rewrite expressions and not change their value. So what do these properties say? Great question! And here's the answer:
Tip: Rewrite subtraction problems as addition problems so you can use the commutative (and associative) property. In other words, think of a - b as a + -b.
ALGEBRA RULES
The multiplication property of zero states that if the product of a × b = 0, then either a or b (or both) must be equal to 0.
EXAMPLE
Q. Use the order of operations and other properties to simplify the expression 12((5/6 + 3/4) - 1/6)/5[square root of 2[(3).sup.2]+7]
A. 17/25. The big fraction bar is a grouping symbol, so you deal with the numerator and denominator separately. Use the commutative and associative properties to rearrange the fractions in the numerator; square the 3 under the radical in the denominator. Next, in the numerator, combine the fractions that have a common denominator; below the fraction bar, multiply the two numbers under the radical. Reduce the first fraction in the numerator; add the numbers under the radical. Distribute the 12 over the two fractions; take the square root in the denominator. Simplify the numerator and denominator.
Here's what the process looks like: [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.]
1. Simplify -3 [4 - 2([3.sup.2])] =
Solve It
2. Simplify [4.sup.3] ([3.sup.2] + 11)([7.sup.4] - [7.sup.4])([10.sup.10]) =
Solve It
3. Simplify 153 + 187 + 153 + 270 + 471 + 270 + 471 =
Solve It
4. Simplify ([6.sup.2] + 4)[square root of [5.sup.2] - [3.sup.2]]
Solve It
5. Simplify [square root of 4([3.sup.2] + 7)]/[square root of 121 - [square root of 49]
Solve It
6. Simplify 7 + 3|2(8) - [5.sup.2]| (For info on absolute value, see the upcoming "Dealing with Linear Absolute Value Equations" section.)
Solve It
Specializing in Products and FOIL
Multiplying algebraic expressions together can be dandy and nice or downright gruesome. Taking advantage of patterns and processes makes the multiplication quicker, easier, and more accurate.
When multiplying two binomials together, you have to multiply the two terms in the first binomial times the two terms in the second binomial - you're actually distributing the first terms over the second. The FOIL acronym describes a way of multiplying those terms in an organized fashion, saving space and time. FOIL refers to multiplying the two First terms together, then the two Outer terms, then the two Inner terms, and finally the two Last terms. The Outer and Inner terms usually combine. Then you add all the products together. So if you have (ax + b)(cx + d), you can do the multiplication of the terms, or FOIL, like so:
Terms Product
First ax(cx)
Outer ax(d)
Inner b(cx)
Last b(d)
= ac[x.sup.2] + adx + bcx + bd
= ac[x.sup.2] + (ad + bc)x + bd
The following examples show some multiplication patterns to use when multiplying binomials (expressions with two terms).
EXAMPLE
Q. Find the square of the binomial: [(2x + 3).sup.2]
A. 4[x.sup.2]+ 12x + 9. When squaring a binomial, you square both terms and put twice the product of the two original terms between the squares: [(a + b).sup.2] = [a.sup.2] + 2ab + [b.sup.2]. So [(2x + 3).sup.2] = [(2x).sup.2] + 2(2x)(3) + [3.sup.2] = 4[x.sup.2] + 12x + 9.
Q. Multiply the two binomials together using FOIL: (3x - 5)(4x + 7)
A. 12[x.sup.2] + x - 35. Find the products: First, 3x(4x) = 12[x.sup.2], plus Outer, 3x(7) = 21x, plus Inner, (-5)(4x) = -20x, plus Last, (-5)(7) = -35. So the product of (3x - 5)(4x + 7) is (3x)(4x) + (3x)(7) + (-5)(4x) + (-5)(7) = 12[x.sup.2] + 21x - 20x - 35 = 12[x.sup.2] + x - 35
Q. Find the product of the binomial and the trinomial: (2x + 7)(3[x.sup.2] + x - 5)
A. 6[x.sup.3] + 23[x.sup.2] - 3x - 35. Distribute the 2x over the terms in the trinomial, and then distribute the 7 over the same terms. Combine like terms to simplify. The product of (2x + 7)(3[x.sup.2] + x - 5) is
= 2x(3[x.sup.2] + x - 5) + 7(3[x.sup.2] + x - 5) = 6[x.sup.3] + 2[x.sup.2] - 10x + 21[x.sup.2] + 7x - 35 = 6[x.sup.3] + 23[x.sup.2] - 3x - 35
7. Square the binomial: [(4x - 5).sup.2]
Solve It
8. Multiply (5y - 6)(5y + 6)
Solve It
9. Multiply (8z - 3)(2z + 5)
Solve It
10. Multiply (2x - 5)(4[x.sup.2] + 10x + 25)
Solve It
Variables on the Side: Solving Linear Equations
A linear equation has the general format ax + b = c, where x is the variable and a, b, and c are constants. When you solve a linear equation, you're looking for the value that x takes on to make the linear equation a true statement. The general game plan for solving linear equations is to isolate the term with the variable on one side of the equation and then multiply or divide to find the solution.
EXAMPLE
Q. Solve for x in the equation 3(x + 7) - 5/4 = 2x+9
A. x = -4. First multiply each side by 4 to get rid of the fraction. Then distribute the 3 over the terms in the parentheses. Combine the like terms on the left. Next, you want all variable terms on one side of the equation, so subtract 8x and 16 from each side. Finally, divide each side by -5.
[??]/1 × 3(x + 7) - 5/[??] = 4 (2x + 9) 3(x + 7)-5 = 8x + 36 3x + 16 = 8x + 36 -5x = 20 x = -4
11. Solve for x: 4x - 5 = 3(3x + 10)
Solve It
12. Solve for x: 5[9 - x(x + 8)] = 5x(1 - x)
Solve It
13. Solve for x: x/4 + x - 6/5 = x + 2/3
Solve It
14. Solve for x: (x - 1) + 2(x - 2) + 3(x - 3) + 4(x - 4) + 5(x - 5) = 0
Solve It
Dealing with Linear Absolute Value Equations
The absolute value of a number is the number's distance from 0. The formal definition of absolute value is, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.]. In other words, the absolute value of a number is exactly that number unless the number is negative; when the number is negative, its absolute value is the opposite, or a positive. The absolute value of a number, then, is the number's value without a sign; it's never negative.
When solving linear absolute value equations, you have two possibilities: one that the quantity inside the absolute value bars is positive, and the other that it's negative. Because you have to consider both situations, you usually get two different answers, one from each scenario. The two answers come from setting the quantity first equal to a positive value and then equal to a negative value.
Before setting the quantity equal to the positive and negative values, first isolate the absolute value term on one side of the equation by adding or subtracting the other terms (if you have any) from each side of the equation.
TIP
If you find more than one absolute value expression in your problem, you have to get down and dirty - consider all the possibilities. A value inside absolute value bars can be either positive or negative, so you look at all the different combinations: Both values within the bars are positive, or the first is positive and the second is negative, or the first is negative and the second positive, or both are negative. Whew!
EXAMPLE
Q. Solve for x in |3x + 7| = 11
A. x = 4/3, -6. First rewrite the absolute value equation as two separate linear equations. In the first equation, assume that the 3x + 7 is positive and set it equal to 11. In the second equation, also equal to 11, assume that the (3x + 7) is negative. For that one, negate (multiply by -1) the whole binomial, and then solve the equation.
3x + 7 = 11; 3x = 4; x = 4/3. -(3x + 7) = 11; -3x - 7 = 11; -3x = 18; x = -6
15. Solve for x in |2x + 1| = 83
Solve It
16. Solve for x in 5|4 - 3x| + 6 = 11
Solve It
17. Solve for x in |x + 3| + |2x - 1| = 6
Solve It
18. Solve for x in 3|4x - 5| + 10 = 7
Solve It
Greater Math Skills: Equalizing Linear Inequalities
A linear inequality resembles a linear equation - except for the relationship. The basic form for a linear inequality is ax + b > c. The > can be replaced by [greater than or equal to] or < or [less than or equal to], but the methods used to solve the inequality stay the same. When the extra bar appears under the inequality symbol, it means or equal to, so you read [less than or equal to] as "is less than or equal to."
The main time to watch out when solving inequalities is when you multiply or divide each side of the inequality by a negative number. When you do that - and yes, you're allowed - you have to reverse the sense or the relationship. The inequality > becomes <, and vice versa.
ALGEBRA RULES
When solving absolute value inequalities (see the preceding section for more on absolute values), you first drop the absolute value bars. You then apply one of two separate rules for absolute value inequalities, depending on which way the inequality symbol faces:
REMEMBER
Two ways of writing your answers are inequality notation and interval notation:
EXAMPLE
Q. Solve the inequality 8x - 15 [greater than or equal to] 10x + 7. Write the answer in both inequality and interval notation.
A. x [less than or equal to] -11. First subtract 10x from each side; then add 15 to each side. This step moves the variable terms to the left and the constants to the right: -2x [greater than or equal to] 22. Now divide each side by -2. Because you're dividing by a negative number, you need to reverse the inequality sign: x [less than or equal to] -11. That's the answer in inequality notation. The solution is that x can be any number either equal to or smaller than -11. In interval notation, you write this (-[infinity], -11].
Q. Solve the inequality |5 - 6x| < 7. Write the answer in both inequality and interval notation.
A. 1/3 < x < 2 and (-1/3, 2. Rewrite the absolute value inequality as the inequality -7 < 5 - 6x < 7.
Subtract 5 from each of the three sections, or intervals, of the inequality to put the variable term by itself: -12 < -6x < 2. Now divide each section by -6, reversing the inequality symbols. Then, after you've gone to all the trouble of reversing the inequalities, rewrite the statement again with the smaller number on the left to correspond to numbers on the number line. This step requires reversing the inequalities again. Here are the details:
-12 < - 6x < 2
2 > x > - 1/3
-1/3 < x < 2
In interval notation, this answer is (-1/3, 2).
This answer looks very much like the coordinates of a point. In instances like this, be very clear about what you're trying to convey with the interval notation.
19. Solve the inequality 5x + 7 [less than or equal to] 22
Solve It
20. Solve the inequality 8(y - 4) [greater than or equal to] 5(3y - 12)
Solve It
21. Solve the inequality |3x + 7| > 4
Solve It
22. Solve the inequality 4|5 - 2x| + 3 < 7
Solve It
(Continues...)
Excerpted from Algebra II Workbook For Dummies by Mary Jane Sterling Copyright © 2007 by Mary Jane Sterling. Excerpted by permission.
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