Another algebra word problem

[charmstot]

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Since trial and error is not a good idea. How would you set up this problem

Julie has $5.50 in nickel and dimes. She has five more nickels and dimes. How many dimes does she have?

 

[sandman]

set up as follows

x = nickles
y = dimes

.05(x) + .10👍 = 5.50

remove decimals

5x + 10y = 550

2nd equation

x = y + 5

because you have 5 more nickles than dimes

substitute new equation in for (x) of original equation

5 (y + 5) + 10 y = 550

5y + 25 +10y = 550
15 y = 525

y = 35

35 dimes
40 nickels

or

half of 5.50 is roughly 2.75 since you have about as many dimes as nickels
so think somewhere aound 30 dimes. This will be one where it would make more sense to solve the problem backwardss from the 4 choices they give you
especially since you're working in 10s and 5s

 

[Dr. Steve Brule]

yep i would have done it the exact same way..except instead of x and y use d and n for dimes and nickels..just makes it a bit easier on the old noggin