SOLUTION: Hi 4 oranges 3 apples and 2 pears cost $15.50. 2 oranges 6 pears cost $28.60.3 oranges and 4 apples cost $6.80. how much does each fruit cost. thanks

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Question 1105396: Hi
4 oranges 3 apples and 2 pears cost $15.50. 2 oranges 6 pears cost $28.60.3 oranges and 4 apples cost $6.80.
how much does each fruit cost.
thanks
Found 2 solutions by jim_thompson5910, greenestamps:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

x = price of one orange
y = price of one apple
z = price of one pear 

4 oranges 3 apples and 2 pears cost $15.50
2 oranges              6 pears cost $28.60
3 oranges 4 apples             cost $6.80


"4 oranges 3 apples 2 pears cost $15.50" means that 4x%2B3y%2B2z+=+15.50
"2 oranges 0 apples 6 pears cost $28.60" means that 2x%2B0y%2B6z+=+28.60
"3 oranges 4 apples 0 pears cost $6.80" means that 3x%2B4y%2B0z+=+6.80

We have this system
Equation (1): 4x%2B3y%2B2z+=+15.50
Equation (2): 2x%2B0y%2B6z+=+28.60
Equation (3): 3x%2B4y%2B0z+=+6.80

Solve Equation (2) for x
2x%2B0y%2B6z+=+28.60
2x%2B6z+=+28.60
2%28x%2B3z%29+=+28.60
x%2B3z+=+28.60%2F2
x%2B3z+=+14.30
x+=+-3z%2B14.30 <<-- Call this equation (4)

Plug equation (4) into equation (1). Simplify and rearrange terms til you have standard form
4x%2B3y%2B2z+=+15.50
4%28x%29%2B3y%2B2z+=+15.50
4%28-3z%2B14.30%29%2B3y%2B2z+=+15.50 x has been replaced with -3z+14.30
-12z%2B57.20%2B3y%2B2z+=+15.50
-12z%2B3y%2B2z+=+15.50-57.20
3y-10z+=+-41.70 <<-- Call this equation (5)

Plug equation (4) into equation (3). Simplify and rearrange terms til you have standard form
3x%2B4y%2B0z+=+6.80
3%28x%29%2B4y+=+6.80
3%28-3z%2B14.30%29%2B4y+=+6.80 x has been replaced with -3z+14.30
-9z%2B42.9%2B4y+=+6.80
-9z%2B4y+=+6.80-42.9
4y-9z+=+-36.10 <<-- Call this equation (6)

--------------------------------------------------------------------------------------------------------------------------------
Focus on equation (5) and equation (6). We have the new system
3y-10z+=+-41.70
4y-9z+=+-36.10

Multiply the top equation by 4. Multiply the bottom equation by 3. Doing so gives you this new system
12y-40z+=+-166.8
12y-27z+=+-108.3

Subtract the like terms straight down and you get
0y-13z+=+-58.5
which turns into
-13z+=+-58.5

Divide both sides by -13 to isolate z
-13z+=+-58.5
z+=+%28-58.5%29%2F%28-13%29
z+=+4.50

Now use this value of z to find x
x+=+-3z%2B14.30
x+=+-3%284.50%29%2B14.30
x+=+-13.5%2B14.30
x+=+0.80

Then use x = 0.80 and z = 4.50 to find y
4x%2B3y%2B2z+=+15.50
4%280.80%29%2B3y%2B2%284.50%29+=+15.50
3.20%2B3y%2B9+=+15.50
12.20%2B3y+=+15.50
3y+=+15.50-12.20
3y+=+3.30
y+=+3.30%2F3
y+=+1.10
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Summary:
x = 0.80
y = 1.10
z = 4.50

Which means,
Each orange costs $0.80 (ie 80 cents)
Each apple costs $1.10
Each pear costs $4.50

Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


As the other tutor showed, the system of equation we have for this problem is
(1) 4x%2B3y%2B2z+=+15.50
(2) 2x%2B6z+=+28.60
(3) 3x%2B4y+=+6.80

When a system of 2 or more equations has all the equations in this form, I find it is nearly always less work to solve the system using elimination instead of substitution. So here is what I would do with this problem....

Variable x appears in all three equations. So eliminate x between (1) and (2) and again between (1) and (3) to obtain two equations in y and z:

Double equation (2) and subtract equation (1):
4x%2B12z+=+57.20
4x%2B3y%2B2z+=+15.50
(4) -3y%2B10z+=+41.70

Multiply (1) times 3 and (3) times 4 and subtract:
12x%2B9y%2B6z+=+46.50
12x%2B16y+=+27.20
(5) -7y%2B6z+=+19.30

Now eliminate z between equations (4) and (5): multiply (4) times 3 and (5) times -5 and add:
-9y%2B30z+=+125.10
35y-30z+=+-96.50
26y+=+28.60
(6) y+=+1.10

Now substitute (6) in (5) to solve for z:
-7.70%2B6z+=+19.30
6z+=+27
(7) z+=+4.50

And finally substitute (7) in (2) to solve for x:
2x%2B27+=+28.60
2x+=+1.60
(8) x+=+0.80

(6), (7), and (8) tell us that...
an orange costs 0.80;
an apple costs 1.10; and
a pear costs $4.50