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.

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(39617)   (Show Source): You can put this solution on YOUR website!
x=6
y=9
Answer by ikleyn(52779)   (Show Source): You can put this solution on YOUR website!
.
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 !

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