SOLUTION: Find two numbers such that the smaller subtracted from the larger is 3 and the difference of the square of the larger subtracted from square of the smaller is 45.

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Question 1118976: Find two numbers such that the smaller subtracted from the larger is 3 and the difference of the square of the larger subtracted from square of the smaller is 45.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39616) About Me  (Show Source):
Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
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Find two numbers such that the smaller subtracted from the larger is 3 and
the difference of the square of the larger subtracted from square of the smaller is 45.
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You are given


x - y = 3           (1)    (where x is the larger number)
y^2 - x^2 = 45.     (2)


From eq(2), you have

(y-x)*(x+y) = 45.


Replace (y-x) in the left side by -3, based on eq(1). You will get

-3*(x+y) = 45,   or

x + y = -45/3 = -15.


Thus you have two equatiuons

x - y =   3,      (1)
x + y = -15       (3)    <<<---=== you were fortune to reduce the given system with a quadratic equation to system of two linear equations !


Add them to get  2x = 3+(-15) = -12  ====>  x = -12/2 = -6.


Then from eq(1),  y = x - 3 = -6-3 = -9.


Answer.  The numbers are  -6 and -9.


Check.   (-9)^2 - (-6)^2 = 81 - 36 = 45.     ! Correct !

Solved.

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Be aware:   the answer by  @josgarithmetic  "6 and 9"  is   INCORRECT !