Question

2) The difference between the squares of the two numbers is
72. Eight times the numerically smaller number is 1 more than
5 times the other number. Find the numerically greater
number

Answer 1

Ответ:

Let the numbers be X & Y and X > Y.

X^2 - Y^2 = 72 -->1

8*Y = 5*X +1 --> 2

Y = (5*X+1)/8

Substituting Y in above equation:

X^2-(5*X+1)/8)^2 = 72

X^2-((5*X+1)^2/64)= 72

64*X^2 - (5*X + 1)^2 = 64*72

64*X^2-(25*X^2 + 10*X + 1) = 64*72

64 X^2 - 25 X^2 - 10*X - 1 = 64*72

39*X^2 - 10*X - 1 = 64*72

39*X^2 - 10*X - 1 = 4608

39*X^2 - 10*X - 4609 = 0

Using factorisation method,

39*X^2 - 429*X+419*X-4609 = 0

39*X(X-11) + 419(X-11) = 0

(39*X+419) (X - 11) = 0

X-110, 39*X + 419 = 0

X = 11, X = -419/31

Discarding X = - 419/31 since it's a decimal number.

Therefore the greater number which is X = 11 -->

Answer.

For information:

11^2 - y^2 = 72

Y^2=12172

Y^2 = 49

Y = +/- 7

Y = -7 has to be discarded, and hence can have only one value Y = 7.