Coordinate Algebra Practice EOCT Answers Unit 2

= ax – w = 3 –w = 3 – ax –w 3 – ax –1 w = –3 + ax w = ax – 3 –ax –ax #1 Unit 2 Which equation shows ax – w = 3 solved for w ? ax – w = 3 –ax –ax A. w = ax – 3 B. w = ax + 3 C. w = 3 – ax D. w = 3 + ax –w = 3 – ax = –1 –w 3 – ax w = –3 + ax w = ax – 3

Least Common Denominator #2 Unit 2 Which equation is equivalent to ? Least Common Denominator 8 A. 17x = 88 B. 11x = 88 C. 4x = 44 D. 2x = 44 14x – 3x = 88 11x = 88

Method #1 Method #2 4n = 2(t – 3) 4n = 2(t – 3) 4n = 2t – 6 2 #3 Unit 2 Which equation shows 4n = 2(t – 3) solved for t ? Method #1 Method #2 4n = 2(t – 3) 4n = 2(t – 3) 4n = 2t – 6 2 4n 2(t – 3) = +6 +6 4n + 6 = 2t 2n = t – 3 +3 +3 4n + 6 2t 2 = 2n + 3 = t 2n + 3 = t

6(x + 4) = 2(y + 5) 6x + 24 = 2y + 10 –10 –10 6x + 14 = 2y 6x + 14 2y #4 Unit 2 Which equation shows 6(x + 4) = 2(y + 5) solved for y ? 6(x + 4) = 2(y + 5) A. y = x + 3 B. y = x + 5 C. y = 3x + 7 D. y = 3x + 17 6x + 24 = 2y + 10 –10 –10 6x + 14 = 2y 6x + 14 2y 2 = 3x + 7 = y

Least Common Denominator #5 Unit 2 This equation can be used to find h, the number of hours it takes Flo and Bryan to mow their lawn. How many hours will it take them to mow their lawn? Least Common Denominator 6 A. 6 B. 3 C. 2 D. 1 3h 6 3 = 2h + h = 6 h = 2 3h = 6

Least Common Denominator #6 Unit 2 This equation can be used to determine how many miles apart the two communities are. What is m, the distance between the two communities? A. 0.5 miles B. 5 miles C. 10 miles D. 15 miles Least Common Denominator 20 2m = m + 10 –m –m m = 10

Least Common Denominator #7 Unit 2 For what values of x is the inequality true? Least Common Denominator 3 A. x < 1 B. x > 1 C. x < 5 D. x > 5 2 + x > 3 –2 –2 x > 1

CX = 7b + 50 CY = 9b + 30 CX = CY 7b + 50 = 9b + 30 50 = 2b + 30 Unit 2 A manager is comparing the cost of buying ball caps with the company emblem from two different companies. #8 Company X charges a $50 fee plus $7 per cap. Company Y charges a $30 fee plus $9 per cap. A. 10 caps B. 20 caps C. 40 caps D. 100 caps For what number of ball caps (b) will the manager’s cost be the same for both companies? Cost Formula: Company X CX = 7b + 50 Cost Formula: Company Y CY = 9b + 30 CX = CY 7b + 50 = 9b + 30 (Subtract 7b on both sides) 50 = 2b + 30 (Subtract 30 on both sides) 20 = 2b (Divide 2 on both sides) 10 = b

Method #1 Substitution x + y = 9 2x + 5y = 36 2(9 – y) + 5y = 36 A shop sells one-pound bags of peanuts for $2 and three-pound bags of peanuts for $5. If 9 bags are purchased for a total cost of $36, how many three-pound bags were purchased? Unit 2 #9 Method #1 Substitution Let x = # of one-pound bags Let y = # of three-pound bags (Total number of bags) x + y = 9 Equation #1: (Total value of bags) 2x + 5y = 36 Equation #2: Substitute x = 9 – y into Equation #2 Solve Equation #1 for x x + y = 9 2(9 – y) + 5y = 36 18 – 2y + 5y = 36 –y 18 + 3y = 36 x = 9 – y 3y = 18 6 three-pound bags y = 6

Method #2 Elimination x + y = 9 2x + 5y = 36 –2x – 2y = –18 A shop sells one-pound bags of peanuts for $2 and three-pound bags of peanuts for $5. If 9 bags are purchased for a total cost of $36, how many three-pound bags were purchased? Unit 2 #9 Method #2 Elimination Let x = # of one-pound bags Let y = # of three-pound bags (Total number of bags) x + y = 9 Equation #1: (Total value of bags) 2x + 5y = 36 Equation #2: Add New Equation #1 and Equation #2 Multiply Equation #1 by –2 –2x – 2y = –18 2x + 5y = 36 –2(x + y) = –2(9) –2x – 2y = –18 3y = 18 (New Equation #1) 6 three-pound bags y = 6

#10 Unit 2 Which graph represents a system of linear equations that has multiple common coordinate pairs? A. B. Multiple common coordinate pairs (Two lines overlap)  Has one common coordinate pair D. C. Has no common coordinate pairs Has one common coordinate pair 

x > 3 x < 3 x > 3 x < 3 Which graph represents x > 3 ? #11 Unit 2 Which graph represents x > 3 ? A. x > 3 x < 3 B. C. x > 3 D. x < 3

Which pair of inequalities is shown in the graph? #12 Unit 2 A. y > –x + 1 and y > x – 5 B. y > x + 1 and y > x – 5 Line 1 Line 2 Both given inequalities have slopes equal to positive one. This is a contradiction to the slope of Line 1 being negative. Note Line 1 graph has a negative slope. Line 2 graph has a positive slope.

Which pair of inequalities is shown in the graph? #12 Unit 2 C. y > –x + 1 and y > –x – 5 D. y > x + 1 and y > –x – 5 Both given inequalities have slopes equal to negative one. This is a contradiction to the slope of Line 2 being positive. Line 1 Line 2 Line 2 has a positive slope with a negative y-intercept. However, the line y > x + 1 has a positive slope, but the y-intercept is positive. Note Line 1 graph has a negative slope. Line 2 graph has a positive slope.

Which pair of inequalities is shown in the graph? #12 Unit 2 A. y > –x + 1 and y > x – 5 B. y > x + 1 and y > x – 5 Line 1 Line 2 Both given inequalities have slopes equal to positive one. This is a contradiction to the slope of Line 1 being negative. Note Line 1 graph has a negative slope. Line 2 graph has a positive slope.