SOLUTION: Six apples and 1 pear cost $2.00. At the same store, 3 apples and 6 pears cost $2.10. Find the cost of one apple and the cost of one pear. *** I tried to substitute, but I

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: Six apples and 1 pear cost $2.00. At the same store, 3 apples and 6 pears cost $2.10. Find the cost of one apple and the cost of one pear. *** I tried to substitute, but I       Log On

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Question 32236: Six apples and 1 pear cost $2.00. At the same store, 3 apples and 6 pears cost $2.10. Find the cost of one apple and the cost of one pear.

*** I tried to substitute, but I don't think I did it right.
Here's what I did:
6a+p= 2.00
3a+6p= 2.10
So I tried p=2.00-6a
3a+(2.00-6a)=2.10
3a+2.00-6a=2.10
-3a=2.10-2.00
-3a=.10
and obviously thats not right :(
Found 2 solutions by stanbon, Earlsdon:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
You were good up to the substitution.
You had 3a+6p=2.10
and sustituted p=2.00-6a
You should then have:
3a+6(2.00-6a)=2.10
Can you take it from there?
Cheers,
Stan H.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Well, you started out ok.
6a+p = $2.00 Rewrite as: p = $2.00 - 6a and substitute into 2nd equation.
3a+6p = $2.10
3a+6(2-6a) = 2.10 Simplify and solve for a.
3a+12-36a = 2.10
12-33a = 2.10 Subtract 12 from both sides.
-33a = -9.90 Divide both sides by -33.
a = $0.30
p = 2.00 - 6a
p = 2.00 - 6(0.30)
p = 2.00 - 1.80
p = $0.20
Apples cost $0.30 each.
Pears cost $0.20 each.