By Bob Miller

ISBN-10: 0071473661

ISBN-13: 9780071473668

**A is for Algebra-and that's the grade you'll pull for those who use Bob Miller's basic advisor to the maths direction each college-bound child needs to take**

With 8 books and greater than 30 years of hard-core school room adventure, Bob Miller is the annoyed student's ally. He breaks down the complexities of each challenge into easy-to-understand items that any math-phobe can understand-and this totally up to date moment variation of *Bob Miller's Algebra for the Clueless* covers every thing a you must recognize to excel in Algebra I and II.

**Read Online or Download Bob Miller's Algebra for the Clueless (Clueless Series) PDF**

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**Extra resources for Bob Miller's Algebra for the Clueless (Clueless Series)**

**Sample text**

6 –4 –2 0 2 4 6 8 Now that we know what an integer is, we would like to add, subtract, multiply, and divide them. ADDITION For adding, you should think about money: + means gain and − means loss. EXAMPLE 3— 7+5 Think (don’t write) (+7) + (+5) Gain 7; gain 5 more. ANSWER 12 or +12 EXAMPLE 4— N OT E You should think of this as an adding problem. −3 − 4 Think (−3) + (−4) Lose 3; lose 4 more. ANSWER −7 Integers Plus More EXAMPLE 5— −7 + 3 Think (−7) + (+3) Lose seven; gain three: lose four. ANSWER −4 NOTE −7 + 3 is the same problem as 3 − 7 (+3 + −7).

Multiply both sides by 2. 2. Switch sides. A = 1⁄2h(b1 + b2) 2A = h(b1 + b2) 3. Distributive law. h(b1 + b2) = 2A 6. Get hb2 to the right. hb1 + hb2 = 2A −hb2 = 8. Divide both sides by h. −hb2 hb1 = 2A − hb2 2A− hb2 2A b1 = ᎏᎏ or ᎏ − b2 h h CHAPTER 4 PROBLEMS WITH WORDS WHY SO MANY STUDENTS H AV E P R O B L E M S O N T H E S AT One of the reasons so many of you have trouble in math, especially on the SAT, is the lack of problems with words. These problems help your brain translate algebra into English.

42 BOB MILLER’S ALGEBRA FOR THE CLUELESS Sometimes the check is worse than doing the problem. This is definitely the case here. ᎏᎏ 7ᎏᎏ − 5 2 17 17 ᎏ ᎏ + ᎏᎏ 3 18 18 4 6 18 17 ᎏᎏ × ᎏᎏ 17 1 ᎏᎏ 4 17 + ᎏᎏ × ᎏᎏ 1 1 7 18 5 17 ᎏᎏ ᎏᎏ − ᎏᎏ ᎏᎏ 1 17 1 17 ᎏᎏ 6 ᎏᎏ 1 126 85 ᎏᎏ − ᎏᎏ 17 17 18 ᎏ + ᎏᎏ 6 ᎏᎏ 68 1 41 17 ᎏᎏ ᎏᎏ 17 1 9 ᎏ+ᎏ 6 17 34 ᎏᎏ ᎏᎏ 1 1 41 9 3 ᎏᎏ+ᎏ 34 3 102 41 27 ᎏ+ᎏ 102 102 68 ᎏ 102 2 2 ᎏ=ᎏ 3 3 Whew! ALGEBRA is more fun than arithmetic, at least for me!!! First-Degree Equations 43 Solve for z: EXAMPLE 4— 2z − 3 4z − 7 ᎏ=ᎏ 5 9 (2z − 3) (4z − 7) ᎏ=ᎏ 5 9 5(4z − 7) = 9(2z − 3) 20z − 35 = 18z − 27 −18z = −18z 2z − 35 = −27 +35 = +35 2z 8 ᎏ=ᎏ 2 2 z=4 To check: 2(4) − 3 4(4) − 7 ᎏᎏ 5 9 Because 5/5 = 9/9, the problem checks.

### Bob Miller's Algebra for the Clueless (Clueless Series) by Bob Miller

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