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Solving Equations and Word Problems in Mathematics - Prof. Robert L. Calico, Assignments of Algebra

Homework assignments for sections 7-2 through 7-3 of a dspm course, focusing on solving equations and word problems in mathematics. It includes instructions for solving equations using algebraic methods and strategies for solving word problems.

Typology: Assignments

Pre 2010

Uploaded on 08/13/2009

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Name___________________
DSPM 0800 HOMEWORK (ASSIGNMENT 2, SECTIONS 7-2 THROUGH 7-3)
Solve the following equations:
1.
7
8
x
8
7
8
8 8(7)
56
x
x
x
Check:
56
8
7
8
7
7 7
x
2.
2 6 13t t
2 6 13
2 6 6 13 6 (Move the numbers to the right--
to move a +6, add -6 to each side.
Note that there are other wa
t t
t t
ys to
solve the equation.)
2 7 (Combine like terms)
2 7 (Once you move numbers to the
t t
t t t t
right, you must move variables to
the left)
7t
Check:
2 6 13
2(7) 6 7 13
14 6 20
20 20
t t
pf3
pf4
pf5

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Name___________________ DSPM 0800 HOMEWORK (ASSIGNMENT 2, SECTIONS 7-2 THROUGH 7-3) Solve the following equations:

  1. 7 8 x

x x x

Check: (^56) 8

x

  1. 2 t^ ^6   t^13 2 6 13 2 6 6 13 6 (Move the numbers to the right-- to move a +6, add -6 to each side. Note that there are other wa

t t t t ys to solve the equation.) 2 7 (Combine like terms) 2 7 (Once you move numbers to the

t t t t t t right, you must move variables to the left) t  7 Check:

t t

  1. 5 t  18  12 5 30 5 5

t t t t t

Check:

t

  1. 5 y^ ^7 ^ y^ ^7 y^ ^21 ^5 y 5 7 7 21 5 6 7 2 21 (It is easier if you combine like terms right away so you don't overlook terms) 6 7 2 2 21 2 4 7 21 4

y y y y y y y y y y y (^4 ) 4 4

y y y y Check:

y y y y

  1. Three parts totaling 54 pounds are packaged for shipping. The second of the three parts weighs 3 pounds more than the first part and the third part weighs 9 pounds less than the first part. Find the weight of each part. There are three unknowns in this problem. However, the weight of the other two parts is given in terms of the first part, so we should let the variable represent the weight of the first part. Let x = the weight of the first part. Then x^ ^3 = the weight of the second part. x  9 = the weight of the third part. (9 pounds less than the first part) 3 60 3 3 ( 3) ( 9) 54 (Three parts weigh 54 pounds) 3 6 54 3 6 6 54 6 3 60 20 3 23 (Weight of second part) 9 11 (Weight of third part)

x x x x x x x x x x Check:

xx   x            The parts weigh __ 20 ___, __ 23 ___, and __ 11 __ pounds.

  1. Two containers of cleaning solution contain a total of 345 gallons. One of them contains four times as much as the other. How much solution is in each container? Let x = the amount of solution in the first container. Then 4 x = the amount of solution in the second container. 5 345 5 5

x x x x x x

Check:

xx       The containers hold ___ 69 ___ and _ 276 gallons.

  1. Robert is responsible for assembling six less than three times as many booklets as Jane. If Robert collates 36 booklets, how many booklets will Jane assemble? Let x = the number of books Jane assembles. 3 6 36 (The problem states that Robert collates 6 less than 3 times as many booklets as Jane. It also states that Robert collates 36 x   3 42 3 3 booklets) 3 6 6 36 6 3 42 14

x x x x Check:

x        Jane assembles ____ 14 _____ booklets.

  1. The formula for the perimeter of a rectangle is P  2 l  2 w. A lot is five feet longer than it is wide. If the perimeter of the lot is 220 feet, what is the length and width of the lot? Since this is a geometric problem, you can draw a rectangle and label the sides. Let w = the width of the rectangle. (Since length is given in terms of width, let width be the variable in the problem) Then w^ ^5 = the length of the rectangle. w w + 5