(Solved) How do you solve a system of equations with variables in the denominator?

mgrebo

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11 years ago

How do you solve a system of equations with variables in the denominator?

like 1/x + 3/y = 3/4
4/x - 2/y = 2.

Please, help me on how would you solve that equation I have no idea how to.

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leezysosteezy

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11 years ago

Clear fractions by multiplying by LCD
4xy(1/x + 3/y = 3/4)... 4y+12x=3xy
xy(4/x - 2/y = 2) ... 4y-2x=2xy...-4y+2x=-2xy

Solve those with elimination to get 14x=xy therefor y=14
4(14)+12x=3x(14)
56+12x=42x
56=30x
x=56/30 reduce

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caitlinritz

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11 years ago

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budd20

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11 years ago

Several ways, but the easiest is to temporarily substitute dummie variables:

Let a = 1/x and let b = 1/y, then we have:

a + 3b = 3/4

4a - 2b = 2.

Multiply the first equation by -4, and then add them:

-4a - 12b = -3
4a - 2b = 2

-14 b = -1
b = 1/14

Remember that this is a subst. variable, so if b = 1/14, y =14.

I'll bet you can take it from here, okay?

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lew

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11 years ago

To eliminate fractions in any equation, multiply each term in the equation (on both sides) by the LCD. Your first equation would become 4y + 12x = 3xy and the 2nd equation would be 4y - 2x = 2xy.

If you line these up and subtract you get the following:

4y + 12x = 3xy
4y - 2x = 2xy
--------------------------
14x = 1xy

subtract and factor:
x(14 - y) = 0

so either x = 0 (impossible, as it would make the original equation invalid) or y = 14. If y = 14, plug in to solve for x:

4(14) - 2x = 2x(14)
56 - 2x = 28x
56 = 30x
x = 56/30

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_vibram_

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11 years ago

-2/y = 2-4/x = (2x-4)/x
Thus y = -2x/(2x-4)
1/x +3(2x-4)/-2x = 3/4
1/x -1.5(2x-4)/x = 3/4
(1-3x +6)/x = 3/4
(7-3x)/x = 3/4
3x = 28-12x
15x = 28
x = 28/15
You can take it from here

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