Ch.2 Solution Manual Ed.1 (v5)

# Ch.2 Solution Manual Ed.1 (v5) - Exercise 2.1 Exercise 2.1...

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Exercise 2.1 Exercise 2.1, Solution 1: a. The 2 nd term is 7 x y , and the 3 rd term is 4 y b. The 3 rd term is y , and the 4 th term is 3 c. The 1 st term is 9 x y , and the 3 rd term is 6 y Exercise 2.1, Solution 3: a. The coefficient of the first term 5 x 2 is 5, the coefficient of the second term 3 x y is 3, and the constant term is 5. b. The coefficient of the first term, 2 y 2 is 2, the coefficient of the second term, 3 x , is 3 and the constant term is 1. c. The coefficient of the first term, 2 x y 2 is 2, the coefficient of the second term, 2 x 2 y , is 2 and the constant term is 7. d. The coefficient of the first term, 8 y 2 , is 8, and the constant term is 4. Exercise 2.1, Solution 5: a. 13 x 2 + 8 x – 2 x 2 + 9 x Grouping like terms together, we obtain = 13 x 2 – 2 x 2 + 8 x + 9 x Adding and subtracting like terms, we obtain = 11 x 2 + 17 x Therefore, the simplified given expression is 11 x 2 + 17 x. b. – 18 y – 5 y 2 + 19 y – 2 y 2 Grouping like terms together, we obtain = – 18 y + 19 y – 2 y 2 – 5 y 2 Adding and subtracting like terms, we obtain = y 7 y 2 Therefore, the simplified given expression is y 7 y 2 . c. 6 x – 3 x + 2 y 2 + y 2 Adding and subtracting like terms, we obtain = 3 x + 3 y 2 Therefore, the simplified given expression is 3 x + 3 y 2 . Exercise 2.1, Solution 7: a. Adding and subtracting like terms together, we obtain

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b. Adding and subtracting like terms together, we obtain c. Exercise 2.1, Solution 9: a. 3[5-3(4-x)]-2-5[3(5x-4)+8]-9x = 3(5-12+3x)-2-5(15x-12+8)-9x = 3[(-7)+3x]-2-5(15x-4)-9x = (-21)+9x-2-75x+20-9x = 9x-75x-9x-21-2+20 = -75x-3 b. 6[4(8-y)-5(3+3y)]-21-7[3(7+4y)-4]+198y = 6(32-4y-15-15y)-21-7(21+12y-4)+198y = 192 -24y – 90 – 90y – 21 – 147 – 84y + 28 + 198y = -24y – 90y – 84y + 198y + 192 – 90 – 21 – 147 + 28 = – 38 Exercise 2.1, Solution 11: a. y-{4x-[y-(2y-9)-x]+2} = y-{4x-[y-2y+9-x]+2} = y-{4x-[-y+9-x]+2} 2
= y-(4x+y-9+x+2) = y-(5x+y-7) = y-5x-y+7 = -5x+y-y+7 = -5x+7 b. (x-1)-{[x-(x-3)]-x} = x-1-[(x-x+3)-x] = x-1-(3-x) = x-1-3+x = x+x-1-3 = 2x -4 Exercise 2.1, Solution 13: a. 5{-2y+3[4x-2(3+x)]} = 5[-2y+3(4x-6-2x)] = 5[-2y+3(2x-6)] = 5(-2y+6x-18) = -10y+30x-90 = 10(3x-y-9) b. 2y+{8[3(2y-5)-(8y+9)+6]} =2y+[8(6y-15-8y-9+6)] =2y+[8(6y-8y-15-9+6)] =2y+[8(-2y-18)] =2y-16y-144 =-14y-144 =-2(7y+72) Exercise 2.1, Solution 15: a. (2y-1)(y-4)-(3y+2)(3y-1) =(2y 2 -8y-y+4)-(9y 2 -3y+6y-2) =(2y 2 -9y+4)-(9y 2 +3y-2) =2y 2 -9y+4-9y 2 -3y+2 =2y 2 -9y 2 -9y-3y+4+2 =-7y 2 -12y+6 b. (2x+3)(2x-1)-4(x 2 -7) =(4x 2 -2x+6x-3)-4x 2 +28 =(4x 2 +4x-3)-4x 2 +28 =4x 2 +4x-3-4x 2 +28

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=4x 2 -4x 2 +4x-3+28 =4x+25 c. (5x-6) 2 -(x+5) 2 = (25x 2 -60x+36)-(x 2 +10x+25) = 25x 2 -60x+36-x 2 -10x-25 = 25x 2 -x 2 -60x-10x+36-25 = 24x 2 -70x+11 Exercise 2.1, Solution 17: a. = = -x-y b. = = x-3y+4x+1 = x+4x-3y+1 = 5x-3y+1 c. = = = Exercise 2.1, Solution 19: a. = = = = b. = = = c. = = = Exercise 2.1, Solution 21: a. Substituting
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## This note was uploaded on 04/04/2012 for the course MATH 1052 taught by Professor Kit during the Winter '12 term at Fanshawe.

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Ch.2 Solution Manual Ed.1 (v5) - Exercise 2.1 Exercise 2.1...

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