University Algebra: Week 5 Homework Problems, Solutions, and Analysis

This document provides comprehensive solutions to a Week 5 algebra assignment. The assignment covers a range of algebra topics, including simplifying algebraic expressions using the order of operations (PEMDAS), evaluating expressions, solving linear equations, inequalities and graphing solutions on a number line, and solving for variables in different equations. The assignment also includes word problems involving the perimeter of a rectangle, temperature conversions, and percentage calculations. The solutions are presented with detailed steps and explanations, and the document references relevant mathematical concepts and formulas to aid understanding. The student demonstrates the ability to solve various types of algebraic problems, providing a valuable resource for students seeking help with their algebra homework or looking to improve their problem-solving skills. The document is a past paper and solved assignment available on Desklib.

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Week 5 Problems

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1. Simplify the following expressions.
a. 52 + 8 – (9- 1) ÷ 4
ANS: 52 + 8 – (9- 1) ÷ 4
= 52 + 8 – 8 ÷ 4 (parenthesis)
= 25 + 8 – 8 ÷ 4 (exponent)
= 25 + 8 – 2 (division)
= 33 – 2 (addition)
= 31 (subtraction) (Lee, Licwinko, & Taylor-Buckner, 2013)
b.
ANS:
=
c.
ANS: =
d. [4(x -2) + 11] – [3(3x – 4) -5]
ANS: [4(x -2) + 11] – [3(3x – 4) -5]
= [4x -8 + 11] – [9x – 12 -5]
= [4x + 3] – [9x – 17]
= 4x + 3 – 9x + 17
= 4x – 9x + 3 + 17
= -5x + 20

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e.
ANS:
= (parenthesis)
= (exponent)
= (multiplication)
= (subtraction)
= 95 (modulus)
2. Evaluate:
a. -(-x) when x = 8
ANS: -(-x) = - (-8) = +8 = 8
b. -x when x = - 43
ANS: -x = - (-43) = + 43 = 43
3. Solve and graph on a number line:
a. 5x + 2(x – 3) + 2x = 6x + 3(x +2) - 4 + x
ANS: 5x + 2(x – 3) + 2x = 6x + 3(x +2) - 4 + x
=> 5x + 2x – 6 + 2x = 6x + 3x +6 - 4 + x
=> 5x + 2x + 2x – 6 = 6x + 3x + x +6 – 4
=> (5 + 2 + 2) x – 6 = (6 + 3 +1) x + 6 - 4
=> 9x – 6 = 10x + 6 – 4
=> 9x – 6 =10x + 2
=> 10x – 9x = - 6 – 2
=> x = - 8 (Earnest, 2015)
x = -8 0
b. 5x + 2 > 1 + 7x

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ANS: 5x + 2 > 1 + 7x
=> 2 – 1 > 7x - 5x
=> 1 > 2x
=> 1/ 2 > x
x = 1/2 0
c. -2 < x < 7- 3
ANS: -2 < x < 7- 3 = -2 < x < 4
x = -2 0 x = 4
4. Solve (do not graph):
a. 0.3 x + 0.9 = 0.7x – 0.6
ANS: 0.3 x + 0.9 = 0.7x – 0.6
 0.9 – 0.6 = 0.7x – 0.3 x
 0.3 = (0.7 – 0.3) x
 0.3 = 0.4 x
 0.4 x = 0.3
 x = 0.3 / 0.4 = 3 / 4 = 0.75
 x = 0.75
b. 5(2 – 4t) + 3t = -5t + 7

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ANS: 5(2 – 4t) + 3t = -5t + 7
=> 10 – 20t + 3t = -5t + 7
=> 10 – 17t = -5t + 7
=> 10 – 7 = -5t + 17t
=> 3 = 12t
=> 12t = 3
=> t = 3 / 12 = 1 / 4 = 0.25
=> t = 0.25
c. 2(2y + 3) + 7 – 3y = 2[-4y + 3(y + 3)]
ANS: 2(2y + 3) + 7 – 3y = 2[-4y + 3(y + 3)]
=> 2(2y + 3) + 7 – 3y = 2[-4y + 3y + 9]
=> 2(2y + 3) + 7 – 3y = -8y + 6y + 18
=> 4y + 6 + 7 – 3y = -8y + 6y + 18
=> 4y -3y +6 + 7 = -2y +18
=> y +13 = -2y +18
=> y + 2y = 18 – 13
=> 3y = 5
=> y = 5 / 3 = 1.67
=> y = 1.67
d.

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ANS:
=>
=>
=>
=>
=>
=>
=> 7t = 6
=>
ANS:
5. Solve for k:
ANS:
=>
=>
=>
=>
6. Solve the following problems:
a. The perimeter of a rectangular yard is 92ft. The length is 6 ft greater than
the width. Find the length and width.

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ANS: Let the width = x ft, and hence the length = (6 + x) ft
Now, area of the rectangle = width * length = x * (6 + x) = 6x + x2
According to the problem, 6x + x2 = 92
So, x2 + 6x – 92 = 0
=>
So, width = 7.05 ft, and length = (7.05 + 6) = 13.05 ft
b. The formula to change oF to oC is:
If the temperature is 14.3 oC what is it in oF?
ANS: Temperature = 14.3o C
We know
Hence,
So, temperature is 57.74o F
c. 620 is what percent of 1230?
ANS: 620 is equivalent to fraction of 1230
Hence, 620 is of 1230
Extra credit: Solve the following for r:

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ANS:
 (taking LCM)
 (taking inverse)
So,
References
Earnest, D. (2015). From number lines to graphs in the coordinate plane: Investigating problem
solving across mathematical representations. Cognition and Instruction, 33(1), 46-87.
Lee, J. K., Licwinko, S., & Taylor-Buckner, N. (2013). Exploring mathematical reasoning of the
order of operations: Rearranging the procedural component PEMDAS. Journal of
Mathematics Education at Teachers College, 4(2)