Question

One number is 7 more than another and its square is 77 more than the square of the smaller number. What are the numbers?

Answer 1

i hope dis helps u..
according to the question ,,,

 the 2 numbers be 'x' and 'y'

given, x=7+y

x^2=77+ y^2 (substtuing x=7+y here)

(7+y)^2=77 + y^2 [(a+b)^2= a^2 + b^2 + 2ab]

49 + 14y + y^2 = 77 + y^2 (y^2 term gets cancelled)

14y=28

y=2

but x=7+y therefore, x=9

answer is x=9 and y=2

i have used 'x' as greater number but u can use 'y' as greater number also.

Answer 2

QUESTION ---- One number is 7 more than another and its square is 77 more than the square of the smaller number. What are the numbers?

ANSWER ----

Let the numbers be x and (x+7)

Then, Equation will be

( x + 7 )^2 - x^2 = 77

Using (a+b) square = a square + 2ab + b square

x square + 49 + 14x - x square = 77

49 + 14x = 77
14x = 77 - 49

14x = 28
x = 2

THEN NUMBERS ARE
x = 2
x + 7 = 9

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ANSWER = 2 & 9
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