SOLUTION: A shopper buys 3 oranges and 5 lemons for $10.26. A second shopper buys 6 oranges and 4 lemons for $11.16. What is the price of each fruit?

Question 112094: A shopper buys 3 oranges and 5 lemons for $10.26. A second shopper buys 6 oranges and 4 lemons for $11.16. What is the price of each fruit?
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Let the letter "o" represent the cost per orange and the letter "L" represent the cost per
lemon. These costs are in dollars.
.
The first shopper buys 3 oranges and therefore spends 3*o on oranges. This shopper also buys
5 lemons and therefore spends 5*L on lemons. Together these two expenditures add to $10.26
.
In equation form this is:
.
3*o + 5*L = 10.26
.
The second shopper buys 6 oranges and spends 6*o on oranges and buys 4 lemons for a total
of 5*L on lemons. So the total expenditure by this shopper is 6*o plus 4*L and is $11.16
.
In equation form this is:
.
6*o + 4*L = 11.16
.
So the two equations that apply are:
.
3*o + 5*L = 10.26 and
6*o + 4*L = 11.16
.
We can use variable elimination to solve this pair. Multiply the top equation (both sides
and all terms) by 2. When you do the equation pair becomes:
.
6*o + 10*L = 20.52
6*o + 4*L = 11.16
.
You now have 6*o in both equations ... so if you subtract the two equations these terms
drop out and when you subtract the vertical columns of these two equations you get:
.
6*L = 9.36
.
Divide both sides of this equation by 6 and you find that L = $1.56 ... so each lemon
costs $1.56
.
Now you can return to any of the equations and substitute $1.56 for L to find out the
price of the oranges. Let's return to the first equation:
.
3*o + 5*L = 10.26
.
Replacing L by 1.56 results in:
.
3*o + 5*1.56 = 10.26
.
Multiplying the 5 times 1.56 reduces the equation to:
.
3*o + 7.80 = 10.26
.
Subtract 7.80 from both sides to get:
.
3*o = 2.46
.
And divide both sides by 3 to solve for "o":
.
o = 2.46/3 = 0.82
.
So each orange costs $0.82 or 82 cents.
.
Hope this helps you to understand the problem ...

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