# Bob Miller's Algebra for the Clueless, 2nd edition (Miller, by Bob Miller By Bob Miller

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Extra info for Bob Miller's Algebra for the Clueless, 2nd edition (Miller, Robert, Clueless Series.)

Example text

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 ᎏ՘ᎏ 5 9 Because 5/5 = 9/9, the problem checks. Some checks are nice. 1. In the case of 2 fractions, in order to clear fractions we can cross multiply. Because . . a/b = c/d if ad = bc or bc = ad. We could have written 9(2z − 3) = 5(4z − 7). No step 2. Step 3, multiply out ( ).

Divide each side by the whole coefficient of x, including the sign. We will go down each step for each example. Let us rewrite Example 1. EXAMPLE 1— Solve for x: 4x + 8 = 10x − 16. 1. Multiply each term by the LCD. We can’t do this because there are no fractions. 2. If the x terms are only on the right, switch the sides. We can’t do this because there are x terms on both sides. 3. Multiply out all parentheses, brackets, and braces. None here. 4x + 8 = 10x − 16 First-Degree Equations 4x + 8 = 10x − 16 −10x = −10x 39 4.

X2 = square of the smallest; one more than the largest = x + 4 + 1; product (answer in multiplication) of the smallest and the middle = x(x + 2). x2 + x + 4 + 1 = x(x + 2) x2 + x + 5 = x2 + 2x x2’s cancel x + 5 = 2x x=5 The answers are 5, 7, and 9. As I said, most students like this kind of problem.