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

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.

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