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Solving Equations & Inequalities
Word Problems
Objectives:
•To solve word problems involving linear equations and
inequalities.
VOCAB OF WORD PROBLEMS
ADDITION
SUBTRACTION
SUM
MORE THAN
PLUS
INCREASED BY
DIFFERENCE
LESS THAN
MINUS
DECREASED BY
MULTIPLICATION
PRODUCT
TIMES
MULTIPLIED BY
PER
DIVISION
QUOTIENT
DIVIDED BY
VOCAB OF WORD PROBLEMS
WORD
IS/ARE
IS LESS THAN
OPERATION
=
<
IS GREATER THAN
IS LESS THAN OR
EQUAL TO
IS GREATER THAN OR
EQUAL TO
QUANTITY
A NUMBER
>
≤
OPPOSITE OF A
NUMBER
≥
( )
X
-X
Write & solve an inequality. Label each step (DCMAM)
1) Three times a number n decreased by two is
less than or equal to five times a number n.
3n  2  5n
 3n
 3n
 2  2n
2 2
1  n
n  1
Move the variable to one
side
M – Divide by the
coefficient
Write & solve an inequality. Label each step (DCMAM)
2) Three times the sum of a number x and two is
less than twice a number x increased by 6.
3 x  2   2x  6
Distribute
3x  6  2x  6
Move the variable to  2 x
 2x
one side
x6 6
A – Subtract the

6

6
constant
x0
Write & solve an inequality. Label each step (DCMAM)
3) Four times a number n decreased by 7
is greater than or equal to five times
the sum of twice a number n and seven.
Distribute
Move the variable to
one side
A – Subtract the
constant
4n  7  5( 2n  7)
4n  7  10n  35
 4n
 4n
 7  6n  35
 35
 35
 42  6n M – Divide by the
coefficient
6
6
n  7
7  n
Write & solve an inequality.
4) A manufacturer of remote-control race cars sells them for
$18 each. The cost to run the company is $2,000 per day plus
$13 to build each car. What is the minimum number of cars
that must be sold each day for the company to make a profit?
Variable
Key info
Word
Let C = minimum # of cars that must be sold
to make a profit
Income > expenses
equation
Numeric 18c  2000  13c
equation
Solve
18c  2000 13c
 13c
 13c
5c  2000
5
5
c  400
They must sell more than 400 cars to make a
profit. That means, at least 401 cars.
Write & solve an inequality.
5) A store charges $4 per game to rent a video game. A club
charges only $2 per game, but has a membership fee of $50
per year. At least how many games must you rent each year
so that renting from the club costs less than renting from
the store?
Variable
Key info
Word
Let g = minimum # of games you need to rent
each year to make renting from club less
than the store
Club cost < store cost
equation
Numeric
equation
2 g  50  4 g
Solve
2 g  50  4 g
 2g
 2g
50  2 g
2
2
25  g
g  25
You must rent more than 25 games to make
the club costs less than the store cost. That
means, at least 26 games.