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carcleer carcleer
  • 05-08-2016
  • Mathematics
Answered

One number is 7 more than another. The difference between their squares is 161. What are their numbers?

Verified Answer :

0100010
0100010 0100010
  • 05-08-2016

x-first number
y-second number 
Assumption:
x>y
Equations:
x=y+7
x^2-y^2=161

Put x from first equation to second:
(y+7)^2-y^2=161
y^2+14y+49-y^2=161
14y+49=161
14y=161-49
14y=112
y=8, so x=y+7=15

The solution is a pair of values: 8 and 15.

Answer Link
caylus
caylus caylus
  • 05-08-2016

Hello,

Let's assume a the greatest number and b the smallest.

a-b=7
a²-b²=161==>(a-b)(a+b)=161==>a+b=161/7==>a+b=23

a+b=23
a-b=7
==>2a=30==>a=15 and b=15-7=8
Proof:
15²-8²=225-64=161

Answer Link

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