SOLUTION: Six apples and three oranges cost $1.77. Two apples and five oranges cost $1.27. Find the cost of each apple and the cost of an orange.

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Six apples and three oranges cost $1.77. Two apples and five oranges cost $1.27. Find the cost of each apple and the cost of an orange.      Log On


   

Question 37669This question is from textbook
: Six apples and three oranges cost $1.77. Two apples and five oranges cost $1.27. Find the cost of each apple and the cost of an orange. This question is from textbook

Found 3 solutions by jessicaherndon, jackytavares, smartguy101:
Answer by jessicaherndon(6) About Me  (Show Source):
You can put this solution on YOUR website!
first, give each a variable. apples = x oranges = y.
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6x+3y=1.77
2x+5y=1.27
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get one variable to cancel out
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6x+3y=1.77
(3)6x+15y=3.81
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you are left with
15y - 3y = 2.04
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12y = 2.04
divide 2.04 by 12
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oranges are .17$
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then, pick one of the ORIGINAL equations and substitute back in.
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6x + 3(.17)=1.77
6x + .51 = 1.77
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6x= 1.26
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x= .21
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apples are .21$ and oranges are .17$

Answer by jackytavares(5) About Me  (Show Source):
You can put this solution on YOUR website!
apples = x and oranges = y
The point for this is to see how much oranges and apples cost. [each]
original equation:
6x+3y=1.77
2x+5y=1.27
(~second equation~ multiply (3) to everything [2x,5y and 1.27]) just like showed below:
(3)2x+(3)5y=(3)1.27
because we are trying to get the same beggining number with the same variable [6x] which would give you:
6x+15y=3.81 [yay! we got [6x] for the second equation like we planned]
now put the 1st equation that we didn't change (6x+3y=1.77) down below the one we change.
It would look like this:
6x+15y=3.81
6x+3y=1.77 [this was the one we didn't changed]
cross out the two 6x [from the first and second equation] and subtract 3.81 and 1.77
you would be left with:
15y-3y=2.04
subtract the 15y and 3y........
12y=2.04
divide 12 to each side.....
y=.17 [so now we know oranges are $.17]
Now we have to find out how much do apples cost.
Now we go way way way back. our original equations

remember...........
6x+3y=1.77
2x+5y=1.27
now that we know how much oranges cost
substute the cost [y=.17] to the original equation [it doesn't matter which one but it has to be only ONE]
lets do the first one....yea?
you have to replace it where the "y" is at. just as shown below.....
[first equation] ->6x+3(.17)=1.77 [replace .17 with the variable [y]
6x+.51=1.77
now subtract 1.77 and .51
so you will get:
6x=1.26
divide [6] to each side......

your answer x=.21
now you know oranges are $.17
and apples are $.21
hope you enjoyed it and learned.
thanx for your attention if u have questions about it cantact me at
tavares261@msn.com
ask for Jacky Tavares

Answer by smartguy101(5) About Me  (Show Source):
You can put this solution on YOUR website!
first, give each a variable. apples = x oranges = y.
6x+3y=1.77
2x+5y=1.27
get one variable to cancel out
6x+3y=1.77
(3)6x+15y=3.81
you are left with
15y - 3y = 2.04
12y = 2.04
divide 2.04 by 12
oranges are .17$
then, pick one of the ORIGINAL equations and substitute back in.
6x + 3(.17)=1.77
6x + .51 = 1.77
6x= 1.26
x= .21
apples are .21$ and oranges are .17$