SOLUTION: I am having a very hard time remembering how to solve this type of system 1/40 + 1/(x+10) = 1/y 1/60 + 1/x = 1/y Thank you, Kristin

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Question 74919: I am having a very hard time remembering how to solve this type of system
1/40 + 1/(x+10) = 1/y
1/60 + 1/x = 1/y
Thank you,
Kristin
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
1%2F40+%2B+1%2F%28x%2B10%29+=+1%2Fy
.
One method is to look at the denominators of the three terms in this equation. The denominators
are 40, (x+10), and y. Suppose we made a product of all three of these ... that product
being 40*(x+10)*y. Next suppose we multiplied this product times all three of the terms in
the equation. This will get rid of all the denominators as shown below:
.

.
Now cancel the common terms in the numerators and denominators:
.

.
What you are then left with is:
.
%28%28x%2B10%29%2Ay%29+%2B+40%2Ay+=+40%2A%28x%2B10%29
.
Multiply out the left side to get:
.
xy+%2B+10%2Ay+%2B40%2Ay+=+40%2A%28x%2B10%29.
.
Factor the common y out on the left side:
.
y%2A%28x+%2B+10+%2B+40%29=+40%2A%28x%2B10%29
.
combine the 10 and the 40 on the left side:
.
y%2A%28x%2B50%29+=+40%2A%28x%2B10%29
.
divide both sides by %28x%2B50%29 and you get:
.
y+=+%2840%2A%28x%2B10%29%29%2F%28x%2B50%29
.
That may be as far as you want to go. Maybe you can multiply out the numerator, but that's
about it.
.
Your second problem is almost the same, and you can use the above technique.
.
1%2F60+%2B+1%2Fx+=+1%2Fy
.
Multiply all the terms in this equation by 60*x*y, the product of the three denominators.
.
When you do that and cancel like terms in the numerators and denominators you get:
.
%28xy%29+%2B+60y+=+60%2Ax
.
On the left side, factor the common y to get:
.
y%2A%28x%2B60%29+=+60%2Ax
.
divide both sides by %28x%2B60%29 and you end up with:
.
y+=+%2860%2Ax%2F%28x%2B60%29%29
.
Hope this helps you. Multiplying every term on both sides by a common denominator
enables you to eliminate the denominators entirely, and then you can work what's left just
as you would an ordinary equation. I hope the form of the answers above reflects what you were
asked to do ... namely solve for y in terms of x.