two positive numbers differ by 6,and their squares differ by 96, find their numbers?

Question

please show me how you did this

Answer 1

x-y=6 (1)
x^2-y^2=96
(x-y)*(x+y)=96
x+y=96/6=16 (2)
sum (1) and (2)
2*x=22
x=11
y=11-6=5

Answer 2

Let the numbers be x and y

x - y = 6 and x^2 - y^2 = 96

from the first equation x = y+6 and substituting this in the second equation we get, (y+6)^2 - y^2 = 96
y^2+12y+36 -y^2= 96
12y + 36 = 96
12y = 96-36 = 60
hence y = 5

Putting y = 5 in the first equation we get x = 11 and hence the two numbers are 5 and 11.

Answer 3

This is simple. You know how to square the sum of two numbers? (a+b)^2 = a^2 + 2ab + b^2

Assume your integers are x and y. You know that x+6 = y already. So we solve for y by knowing that y^2 - x^2 = 96.

Applying our rule from above, (x + 6)^2 must be x^2 + 12x + 36 and it is equal to y^2.

y^2 - x^2 = 96 = 12x + 36.

Now subtract 36 from both sides: 96-36=60 and it must equal 12x.

Since 12x = 60 we can see that 60/12 = x.

Therefore, x=5, y=11. x^2 = 25, y^2 = 121. 121-96=25.

Answer 4

11 and 5. trial and error. Started with 10 and 4 and the difference was less than 96, so I went up 1 and that was it.

Answer 5

x - y = 6
x^2 - y^2 = 96

x = y + 6
(y + 6)^2 - y^2 = 96
((y + 6)(y + 6)) - y^2 = 96
(y^2 + 6y + 6y + 36) - y^2 = 96
y^2 + 12y + 36 - y^2 = 96
12y + 36 = 96
12y = 60
y = 5

x = y + 6
x = 5 + 6
x = 11

The numbers are 5 and 11

Answer 6

The difference of the two squares equals the product of their sum and difference. Use that fact to write an equation and make substitutions. That should get you started, others gave you the answer.

Answer 7

11-5=6
5*5=25
11*11=121
121-25=96
Hope this helps

Answer 8

Can't you do your own homework? Does your mom have to wipe your butt too?