Darlene S.

# 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?

Lily G.

Ralph L. answered • 09/30/13 Tutor 4 (1) Algebra I, II, Visual Basic, Beginning C++ tutor See tutors like this 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 45¢ Upvote • 3 Downvote Add comment Brad M. answered • 10/01/13 Tutor 4.9 (487) 15-20 Percentile-point Improvements Within Reach! About this tutor › 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 :) Upvote • 2 Downvote Add comment Meagan V. answered • 09/30/13 Tutor 4.9 (263) Tutor Specializing in Math & English, All SAT & ACT, and SPSS See tutors like this First, set up an algebraic equation: lady finger = x cookie = y 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 there all right just explained differently
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10/26/20

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