Instead of doing your problems for you, I'll change only the numbers and work
problems exactly like yours step by step, that you can use as models to solve
your problems by. Notice that the numbers are different but they are worked
exactly the same way:
Chris has a change jar that contains $2.51 in quarters and pennies. He
has 17 more pennies than quarters. How many of each type of coin does he have?
q = number of quarters
$0.25q = value of all the quarters
q+17 = number of pennies <-- the more...than... tells us to ADD q and 17.
$0.01(q+17) = value of all the pennies
$0.25q + $0.01(q+17) = $2.51
25q + 1(q+17) = 251
25q + 1q + 17 = 251
26q + 17 = 251
Subtract 17 from both sides
26q = 234
Divide both sides by 26
q = 9 <-- So Chris has 9 quarters
q+17 = number of pennies = 9+17 = 26 <-- So Chris has 26 pennies
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Tiana has a change jar that contains $3.90 in dimes and nickels. She has
11 times as many nickels as dimes. How many of each type of coin does she have?
d = number of dimes
$0.10d = value of all the dimes
11∙d = number of nickels <--the word "times" tells us to multiply 11 by d.
$0.05(11∙d) = value of all the nickels
$0.10d + $0.05(11∙d) = $3.90
10d + 5( 11d) = 390
10d + 55d = 390
65d = 390
Divide both sides by 65
d = 6 <-- So Chris has 6 dimes
11∙d = number of nickels = 11∙6 = 66 <-- So Tiana has 66 nickels
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Kelsey has a change jar that contains $2.40 in nickels and quarters. The number
of nickels is 8 more than three times the number of quarters. How many of each
type of coin does she have?
q = number of quarters
$0.25q = value of all the quarters
3q+8 = number of nickels <--the "more than three times" tells us to ADD 3q and 7.
$0.05(3q+8) = value of all the nickels
$0.25q + $0.05(3q+8) = $2.40
25q + 5(3q+8) = 240
25q + 15q + 40 = 240
40q + 40 = 240
Subtract 40 from both sides
40q = 240
Divide both sides by 35
q = 5 <-- So Kelsey has 5 quarters
3q+8 = number of nickels = 3(5)+8 = 15+8 = 23 <-- So Kelsey has 23 nickels
Edwin
