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In this lesson we will review
some of the basic rules
for translating verbal phrases into algebraic language.
We will also specifically deal
with translating
verbal inequalities into algebraic
language.
We will NOT be solving the
inequalities in this lesson!
Remember an inequality means just
that...
the quantities we are comparing are NOT equal.
That means that there are 4
possibilities:
One quantity may be LESS THAN
another
or
One quantity may be GREATER THAN
another
or
One quantity may be LESS THAN OR EQUAL TO
another
or
One quantity may be GREATER THAN OR
EQUAL TO another
Each inequality has it's own
symbol:
less than <
greater than >
less than or equal to ≤
greater than or equal to ≥
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Example #1
Sue and David are friends. David is 6 years older than
Sue.
If the sum of Sue and David's age is less than 42 years, what is
the greatest age each can be?
When
translating, you must begin with a "let" statement.
In a let statement you assign a variable to stand for the
quantity (ies) you are trying to solve for.
In this problem there are two quantities we are interested in
finding.
That's right...Sue's age and David's age.....
Let x =
Sue's age
Let x+6 = David's age
Now for the translation into an algebraic inequality:
(x) + (x + 6) < 42
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Example #2
On a farm, the number of cows is 50 more than twice the number
of sheep. If there are at most 250 animals in all, find
the greatest number of cows, and the greatest number of sheep
there could be on this farm.
(Do you understand that that means the
total number
of animals must be less than or equal to 250)
let x = the number of sheep
let 2x+ 50 = the number of cows
(x) + (2x+50) ≤ 250
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Example #3
Bob received grades of 88, 91, 89, and 87 on four science tests.
What is the lowest grade that Bob can receive on the fifth test
in order for his average to be greater than 90?
(Remember to
find an average add up all the terms and divide by the number of
terms)
Let x = the grade he
must receive on the fifth test
(88 + 91 + 89 + 87 + x)
÷ 5 > 90
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Example #4
In Sara's bank, there are twice as many nickels as quarters.
If the value of these coins is at least $8.00, find the smallest
possible number of nickels and quarters in her bank.
Let x =
the number of quarters
Let 2x = the number of
nickels
Then 25x = the value of the
quarters in cents
Then 10x = the value of the
nickels in cents
10x + 25x ≥ 800
(remember $8 = 800 cents!)
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