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## Algebra word problem for 5th grade

one lady finger and six cookies cost \$3.05. At the same prices, two lady fingers and four cookies cost \$2.50. How much does one cookie cost?

Let F be the cost of one lady finger
Let C be the cost of on cookie.

so,

F + 6C = 3.05          eq. 1
2F + 4C  = 2.50       eq. 2

use eq. 1 to solve for F:

F = 3.05 - 6C          eq. 3

subsitute eq. 3 to eq. 2

2(3.05 - 6C) + 4C = 2.50

6.10 - 12C + 4C = 2.50

6.10 - 8C = 2.50

6.10 - 2.50 = 8C

3.60 = 8C

so C = \$.45

so, each cookie costs 45ยข

Hey Darlene!
"cut in half" idea: one LF and two C costs \$1.25 ... 4 less cookies saves \$1.80
"halving tool" again => 4C is 1.80 ... 2C is 0.90 ... one cookie is 45 cents ... Regards :)
First, set up an algebraic equation:
lady finger = x

x+6y=3.05
2x+4y=2.50

Then, solve for one variable...
x=3.05-6y

Then, plug in your new equation into the other equation...

2(3.05-6y)+4y=2.50

6.1-12y+4y=2.5

3.6=8y
y= 0.45

Since Y is the price of a cookie, the price of a cookie is 45 cents/\$0.45.