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Question 175028: I posted this problem and someone was able to proved the algebraic equation for it, but unfortunantly, I still do not understand how to solve it:
"I am thinking of three consecutive numbers. If I multiply the first with the second and then subtract three times the third, the result is 57. What are the numbers?"
The equation given was x(x+1) - 3.(x+2)=57
Can someone please help me to actually solve this? Thanks
Found 3 solutions by scott8148, actuary, jojo14344:
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! distributing gives __ x^2+x-3x-6=57
subtracting 57 __ x^2-2x-63=0
factoring __ (x-9)(x+7)=0
so x equals 9 or -7
which makes the numbers 9, 10, 11 or -7, -6, -5
Answer by actuary(112)  (Show Source):
You can put this solution on YOUR website! Let me give a try.
Let x be the first number, x+2 is the second number and x+3 is the third number.
x*(x+1)-3*(x+2)=57
Simplify the equation.
x^2+x-3*x-6=57
So x^2-2*x-63=0
WE have a quadratic equation so by the quadratic formula
x=-[-(-2)+/-Sqrt(-2^2-4*1*(-63)]/2*1
x=2+/-Sqrt[4+252]/2=2+/-Sqrt[256]/2
x=(2+/-16)/2
There for x =18/2=9
or x = -7
So the the three consectutive numbers are 9,10,11
or -7,-6,-5
I hope that this helps.
Answer by jojo14344(1513) (Show Source):
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