SOLUTION: find two numbers whose sum is 12 and whose difference between their squares is 48.

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Question 73655: find two numbers whose sum is 12 and whose difference between their squares is 48.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Let the two numbers be a and b. From the problem description, you can write:
1) a%2Bb+=+12 and
2) a%5E2-b%5E2+=+48
Rewrite equation 1) as: a = 12-b and substitute for the a in equation 2).
2a) %2812-b%29%5E2-b%5E2+=+48 Simplify and solve for b.
%28144-24b%2Bb%5E2%29-b%5E2+=+48
144-24b+=+48 Add 24b to both sides.
144+=+48%2B24b Subtract 48 from both sides.
96+=+24b Divide both sides by 24.
4+=+b and...
a+=+12-b
a+=+12-4
a+=+8
The two numbers are 4 and 8
Check:
a%2Bb+=+8%2B4 = 12 Their sum is 12.
a%5E2-b%5E2+=+8%5E2-4%5E2
64-16+=+48 The difference of their squares is 48.