SOLUTION: One number is 4 more than another. The difference between their squares is 104. What are the numbers?

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Question 1049827: One number is 4 more than another. The difference between their squares is 104. What are the numbers?
Found 3 solutions by josgarithmetic, MathLover1, addingup:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Assign variables and translate the description literally.

Numbers are x and x+4.
%28x%2B4%29%5E2-x%5E2=104
x%5E2%2B8x%2B16-x%5E2=104
8x%2B16=104
8x=88
system%28x=11%2Cx%2B4=15%29

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

let one number be x and another y
if one number is 4 more than another, means
x=y%2B4 ........eq.1
if the difference between their squares is 104, means
x%5E2-y%5E2=104 ........eq.2
substitute x from eq.1 in eq.2
%28y%2B4%29%5E2-y%5E2=104 ........eq.2......solve for y
y%5E2%2B8y%2B16-y%5E2=104
cross%28y%5E2%29%2B8y%2B16-cross%28y%5E2%29=104
8y=104-16
8y=88
y=88%2F8
y=11
now find x

x=y%2B4 ........eq.1
x=11%2B4
x=15
so, one number is 15 and another 11

Answer by addingup(3677) About Me  (Show Source):
You can put this solution on YOUR website!
Let the numbers be x and y
x = y+4
x^2-y^2 = 104 Substitute for x:
(y+4)^2-y^2 = 104
Expand the left side with FOIL (First-Outer-Inner-Last)
(y^2+8y+16)-y^2 = 104
8y+(y^2-y^2)+16 = 104
8y+16 = 104
8y = 88
y = 11
Since x = y+4 = x = 11+4= 15
Now we have:
y = 11
x = 15
>>>>>>>>>>>>>>>>
Check:
x^2-y^2 = 104
15^2-11^2 = 104
225-121 = 104 Correct
:
John