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Elementary Algebra - Practice Test III Problems | MATH 017, Exams of Algebra

Community College of PhiladelphiaAlgebra
Prof. Jere C. Brubaker

Material Type: Exam; Professor: Brubaker; Class: Elementary Algebra; Subject: Mathematics; University: Community College of Philadelphia; Term: Unknown 1989;

Typology: Exams

Pre 2010

Uploaded on 08/16/2009

koofers-user-bzr
koofers-user-bzr 🇺🇸

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Math 017 Unit 3 Practice Test III Name________Answers________________
1 Graph the following ordered pairs on the graph below:
(- 5 , 2 ), (0 , - 3), ( 2 , - 4), ( - 3, - 1), (4, 0 )
2 Give the coordinates of the points graphed below.
A = ( - 6, -2 ) B= ( -1, 3 ) C= ( 0, 5 )
D= ( 4, 2 ) E= ( 7, 0 ) F= ( 7, - 5 )
3 Circle all the ordered pairs which are solutions to the following equation.
2x + y = 10 ( 2, 6 ) ( - 2 , 14 ) ( 3 , - 4 ) ( - 2, -6) ( 7, -4)
2(2) + 6 = 10 so ( 2, 6) is a solution
2( - 2 ) + 14 = 10 so ( - 2, 14 ) is a solution
2(3) + (– 4) = 2 ≠ 10 so ( 3 , - 4 ) is not a solution
2( - 2) + ( - 6 ) = - 10 ≠ 10 so ( - 2 , - 6 ) is not a solution
2( 7 ) + ( - 4 ) = 10 so ( 7 , - 4 ) is a solution
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Math 017 Unit 3 Practice Test III Name________Answers________________ 1 Graph the following ordered pairs on the graph below: (- 5 , 2 ), (0 , - 3), ( 2 , - 4), ( - 3, - 1), (4, 0 ) 2 Give the coordinates of the points graphed below. A = ( - 6, -2 ) B= ( -1, 3 ) C= ( 0, 5 ) D= ( 4, 2 ) E= ( 7, 0 ) F= ( 7, - 5 ) 3 Circle all the ordered pairs which are solutions to the following equation. 2x + y = 10 ( 2, 6 ) ( - 2 , 14 ) ( 3 , - 4 ) ( - 2, -6) ( 7, -4) 2(2) + 6 = 10 so ( 2, 6) is a solution 2( - 2 ) + 14 = 10 so ( - 2, 14 ) is a solution 2(3) + (– 4) = 2 ≠ 10 so ( 3 , - 4 ) is not a solution 2( - 2) + ( - 6 ) = - 10 ≠ 10 so ( - 2 , - 6 ) is not a solution 2( 7 ) + ( - 4 ) = 10 so ( 7 , - 4 ) is a solution

4 Find three solutions to the equation x + 2y = 4 ( 0 , 2 ) ( 4 , 0 ) ( -2, 3 ) For problems 5 – 7, find three solutions and graph each of the following equations. 5 y = - 2x + 6 ( 0 , 6 ) ( 2 , 2 ) ( - 1 , 8 ) 6 3x – 2y = 12 ( 0 ,-6 ) ( 4, 0) ( 2, -3)

For problems 10 – 12 solve the following systems. 10 3x + y = 11 y = 2x + 1 Use substitution rule: substitute 2x + 1 for y in first equation 3x + ( 2x + 1 ) = 11 solve this equation 5x + 1 = 11 add like terms -1 -1 subtract 1 from both sides 5x = 10 5x/5 = 10/5 divide both sides by 5 x = 2 solution for x value: substitute this value for x in either equation since it must be true in both equations y = 2 ( 2) + 1 y = 5 so ( 2 , 5 ) is solution for system. Check in first equation: 3( 2 ) + 5 = 11 True 11 x – 3y = 12 x = 6 – 3y Use substitution rule: substitute 6 – 3y for x in first equation (6 – 3y) – 3y = 12 solve this equation 6 – 6y = 12 add like terms

  • 6 - 6 subtract 6 from both sides
    • 6y = 6
  • 6y/-6 = 6/- 6 divide both sides by - 6 y = - 1 solution for y value: substitute this value for y in either equation since it must be true in both equations x = 6 – 3( - 1) solve this equation x = 9 so ( 9, -1 ) is solution to system Check in other equation: 9 – 3 ( -1) = 12 True. 12 3x – 2y = 8 x = 2y Use substitution rule: substitute 2y for x in first equation 3(2y) – 2y = 8 solve this equation for y part of solution 6y – 2y = 8 multiplication 4y = 8 add like terms 4y/4 = 8/4 divide both sides by 4 y = 2 solution for y value: substitute this value for y in either equation since it must be true in both equations. x = 2 ( 2) solve this equation x = 4 so ( 4, 2) is solution to system Check in other equation: 3( 4) – 2 ( 2 ) = 8 True