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

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) (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
"2 oranges 0 apples 6 pears cost $28.60" means that
"3 oranges 4 apples 0 pears cost $6.80" means that

We have this system
Equation (1):
Equation (2):
Equation (3):

Solve Equation (2) for x

<<-- Call this equation (4)

Plug equation (4) into equation (1). Simplify and rearrange terms til you have standard form

x has been replaced with -3z+14.30

<<-- Call this equation (5)

Plug equation (4) into equation (3). Simplify and rearrange terms til you have standard form

x has been replaced with -3z+14.30

<<-- Call this equation (6)

--------------------------------------------------------------------------------------------------------------------------------
Focus on equation (5) and equation (6). We have the new system

Multiply the top equation by 4. Multiply the bottom equation by 3. Doing so gives you this new system

Subtract the like terms straight down and you get

which turns into

Divide both sides by -13 to isolate z

Now use this value of z to find x

Then use x = 0.80 and z = 4.50 to find y

--------------------------------------------------------------------------------------------------------------------------------
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) (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)
(2)
(3)

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):

(4)

Multiply (1) times 3 and (3) times 4 and subtract:

(5)

Now eliminate z between equations (4) and (5): multiply (4) times 3 and (5) times -5 and add:

(6)

Now substitute (6) in (5) to solve for z:

(7)

And finally substitute (7) in (2) to solve for x:

(8)

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