By Bob Miller
A is for Algebra-and that's the grade you'll pull in case you use Bob Miller's basic advisor to the mathematics path 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 pissed off 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 little thing a you want to be aware of to excel in Algebra I and II.
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Extra info for Bob Miller's Algebra for the Clueless (Clueless Series)
Also do each problem in the book before you do new ones. Be sure to understand each step. It will really help you. Let’s try some more. EXAMPLE 2— Solve for y: 12 − 2(3y − 4) = 4[7 − y]. 12 − 2(3y − 4) = 4[7 − y] There are no fractions and the y terms are on both sides, but there is a step 3. 40 BOB MILLER’S ALGEBRA FOR THE CLUELESS 12 − 6y + 8 = 28 − 4y 3. Multiply out all ( ) and [ ]. We can combine like terms on the left. ) −6y + 20 = −4y + 28 5. Add the opposite of the y term on the right to both sides.
Opposite (additive inverse): a + (−a) = (−a) + a = 0. The opposite of 3 is −3 because 3 + (−3) = 0. The opposite of −6b is 6b because −6b + 6b = 6b + (−6b) = 0. The opposite of 0 is 0. 38 BOB MILLER’S ALGEBRA FOR THE CLUELESS 1. Fractions cause the most problems. So we get rid of them at once. 1. Multiply each term by the least common denominator (LCD). The purpose is to make all equations look the same, so that you get very good very fast. 2. If the x terms are only on the right, switch the sides.