AnswersFL-1200700-Mathematics for College Algebra MISSolving Linear Equations: Variable on One Side

Solving Linear Equations: Variables on Both Sides Answers

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Which number can each term of the equation be multiplied by to eliminate the fractions before solving?m – = 2 + m

Question illustration
A
2
B
3
C
4
D
5
2
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What is the solution to the linear equation?d – 10 – 2d + 7 = 8 + d – 10 – 3d

A
d = –5
B
d = –1
C
d = 1
D
d = 5
3

Which step could be used to help isolate the variable in the following equation?5.6j – 0.12 = 4 + 1.1j

A
Subtract 0.12 from both sides.
B
Subtract 5.6 from both sides.
C
Subtract 4j from both sides.
D
Subtract 1.1j from both sides.
4

What is the solution to the linear equation?4b + 6 = 2 – b + 4

A
b = –2
B
b = 0
C
b = 4
D
b = 6
5

Two runners are saving money to attend a marathon. The first runner has $112 in savings, received a $45 gift from a friend, and will save $25 each month. The second runner has $50 in savings and will save $60 each month.

A
112 – 25m + 45 = 50 – 60m
B
112 + 25 + 45m = 50m + 60
C
112 + 25 – 45m = –50m + 60
D
112 + 25m + 45 = 50 + 60m
6

Which number can each term of the equation be multiplied by to eliminate the fractions before solving?6 – x + = x + 5

Question illustration
A
2
B
3
C
6
D
12
7

Which number can each term of the equation be multiplied by to eliminate the decimals before solving?–m + 0.02 + 2.1m = –1.45 – 4.81m

A
0.01
B
0.1
C
10
D
100
8

The International Business Club starts the school year with $250.50 in their account. They spend $35 each month on activities. The Future Agricultural Leaders Club starts with $300 and spends $45.25 each month.

A
250.5 + 35m = 300 + 45.25m
B
250.5 – 35m = 300 – 45.25m
C
250.5m – 35 = 300m – 45.25
D
250.5m + 35 = 300m + 45.25
9

What is the solution to the linear equation?2.8y + 6 + 0.2y = 5y – 14

A
y = –10
B
y = –1
C
y = 1
D
y = 10
10

The levels of mercury in two different bodies of water are rising. In one body of water the initial measure of mercury is 0.05 parts per billion (ppb) and is rising at a rate of 0.1 ppb each year. In the second body of water the initial measure is 0.12 ppb and the rate of increase is 0.06 ppb each year.

A
0.05 – 0.1y = 0.12 – 0.06y
B
0.05y + 0.1 = 0.12y + 0.06
C
0.05 + 0.1y = 0.12 + 0.06y
D
0.05y – 0.1 = 0.12y – 0.06

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