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1.) find the two numbers which has a sum of 12 and a difference of 48 between their squares.​

Sagot :

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.