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Math 097 - Midterm 1 Solutions 1. Compute −32 + 27 5 ( 12 − 3)2 + −32 + 27 5 ( 21 − 3)2 + 1 2 1 2 = −9 + 27 5 (− 52 )2 + 21 = − 18 5 25 1 + 4 2 = = = −18 5 27 4 4 − 18 5 · 27 8 − 15 . 2. Explain when two terms in an expression can be combined and why. (Saying that they are “like terms” is not suffiecient.) Two terms can be combined when you factor out a common factor and only arithmetic is left in the parentheses. For example, we can factor an x2 out of 2x2 + 4x2 and get just arithmetic: 2x2 + 4x2 = (2 + 4)x2 = 6x2 . If we factor the common y out of 3y 2 + 7y, we get 3y 2 + 7y = (3y + 7)y. Since the parentheses don’t just contain arithmetic with numbers, we can do the computation in the parentheses. So we’re stuck with uncombined terms. Math 097 2 3. Simplify the following expressions. (a) −(3 + 2y) + x(y − x + 2) We’ll distribute everything we can distribute: −(3 + 2y) + x(y − x + 2) = −3 − 2y + xy − x2 + 2x. There aren’t any terms we can combine, so we’re done. (b) (a − b) + 12 (a + b) − 7(2b + 4a) (a − b) + 12 (a + b) − 7(2b + 4a) = a − b + 12 a + 12 b − 14b − 28a = −26 12 a − 14 12 b = − 53 2 a− 29 2 b. Math 097 3 4. A pizza shop has two pricing options for pizza delivery. For option A, the price per pizza is $15 with a $5 delivery fee for the order, and for option B, the price per pizza is $13.50 with a $20 delivery fee for the order. (a) Write an equation that expresses the question “For what number of pizzas will the two options cost the same?” If we let x be the number of pizzas, we get cost for option A = cost for option B $ $ $5 + 15 · x pizzas = $20 + 13.50 pizza · x pizzas pizza 5 + 15x = 20 + 13.50x. (b) For what number of pizzas will the two options cost the same? We just need to solve the equation from part a. 5 + 15x = 20 + 13.50x −13.5x − 5 − 13.5x − 5 1.5x = 15 x = 10. So if we order 10 pizzas, the two pricing options will cost the same. Math 097 4 5. Solve the equations below. Describe the collection of solutions in words. (a) 2x − 3 = 12 2x − 3 = 12 +3 +3 2x = 15 /2 /2 x= We can put 15 2 15 2 . back in the original equation to check that it works: 2( 15 2 ) − 3 = 15 − 3 = 12. (b) −(2 + a) − 2a = 12 − 3(a + 1) −(2 + a) − 2a = 12 − 3(a + 1) −2 − a − 2a = 12 − 3a − 3 −2 − 3a = 9 − 3a +3a + 3a −2 = 9 This equation is always false, so there is no solution. (c) 1 100 y + 1500 1000 = 3 500 − 17 1000 y We can clear the denominators first by multiplying both sides by 1000: 1 100 y + 1500 1000 = 3 500 − 17 1000 y 10y + 1500 = 6 − 17y +17y − 1500 + 17y − 1500 27y = −1494 /27 /27 y = − 1494 27 y = − 166 3 Math 097 5 6. (a) Solve the inequality and draw a picture of the collection of solutions: 7z − 3 ≤ 4(3z − 4) + 18 7z − 3 ≤ 4(3z − 4) + 18 7z − 3 ≤ 12z − 16 + 18 7z − 3 ≤ 12z + 2 −12z + 3 − 12z + 3 −5z ≤ 5 /(−5) /(−5) z ≥ −1 (b) Write an inequality whose collection of solutions is pictured below: 14 We want w-values to the right of 14 3 , that is, bigger than 3 . Also, the solid dot indicates that we want to include 14 3 itself as a solution to our inequality. So we want w≥ 14 3 .