Question 1091861: Directions: Find all values of x satisfying the given condition.
y1= 1/x, y2= 1/9x, 1/x-5 for y3., and the sum of 6 times y1 and 8 times y2 is the product of 8 and y3.
I am using / to represent a fraction. To the left of the / is the numerator and to the left of the / is the denominator.
I started by putting together the second part.
6(y1) + 8(y2) = 8(y3)
now plugging the equations in for the y's:
6(1/x) + 8 (1/9x) = 8(1/x-5)
multiplying:
(6/x) + (8/9x) = (8/x-5)
Since this is an addition problem, I need to find a common denominator which is (9x)(x-5):
9x(x-5) (6/x) + (8/9x) = (8/x-5) (9x)(x-5)
the first fraction, the single x is cancelled, the second fraction 9x is cancelled and in the answer x-5 is cancelled and then you multiply:
54x-30 + 8x-40 = 72x
now you combine like terms:
62x - 70 = 72x
then you subtract 62x on each side
-70 = 10x
Once you divide you get -7
I input this into the program where my online math assignment is on and it says it is wrong and the answer is -31.
I don't understand how they got -31.
Found 2 solutions by josmiceli, Boreal:
Answer by josmiceli(19441)   (Show Source):
Answer by Boreal(15235)  (Show Source):
You can put this solution on YOUR website! Go to your step where you are multiplying.
The first term is 9x(x-5)(6/x). The x s do cancel, but you have 9(x-5)*6, which is 54x-270
Subtract an additional 240 from the result, and you get 10x=-310, or x=-31.
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