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Unformatted text preview: MAC1105: Quiz #7 Solutions
July 12, 2010 In the topright corner of a clean sheet of paper, write your name, UFID, and section number. Please use a pen with blue or black ink. When you are nished, FOLD your paper in half lengthwise and write your name on the back. 1. State which of the following are linear equations. For those that are linear, solve: (a) 2x  4 = x2  5 Not linear (b) x + 1 = x  3 3 2
x 3  x 2 = 3  1 2x 6  3x 6 = 4 x 6 = 4 x = 24
3.8 0.9 (c) 0.3(x + 5) = 1.2x  2.3
0.3x + 1.5 = 1.2x  2.3 3.8 = 0.9x x = =
38 9 2. Solve: (x + 3)(x  2) = (x  4)(x + 1)
x2  2x + 3x  6 = x2 + x  4x  4 x  6 = 3x  4 4x = 2 x =
1 2 4 x1 3. Does this equation have a solution: +2 2 = 2x ? If so, what is it? x1 The only way to see if there is a solution is to attempt to nd one and then see if it works in the original equation. First we clear the denominator by multiplying both sides by x  1:
4 x1 +2 2 (x  1) = (x  1) 2x x1 4 + 2(x  1) = 2x 2 This reduces to 2 + (x  1) = 2x, from which we obtain:
2 + (x  1) = 2x 1 + x = 2x x = 1. But since x = 1 causes us to divide by 0 in the original problem, this equation does not have a solution. 1 ...
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 Summer '10
 Picklesimer
 Algebra

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