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get x if the given is y=x+2 divided by x-2

2007-08-28 20:12:53 · 5 answers · asked by Maria Katheryn 1 in Science & Mathematics Mathematics

5 answers

U can't obtain "x" if u don't give "y" a value...

For example if y = 2

then,
2 = x + 2 / x - 2
2( x - 2) = x + 2
2x - 4 = x + 2
2x - x = 2 + 4
x = 6

or do u want the equation of x??

which will be

y = x + 2 / x - 2
y (x - 2) = x + 2
xy - 2y = x + 2
xy - x = 2 + 2y
x(y - 1) = 2 (1 + y)
x = 2 (1 + y) / y - 1

hope this help..

2007-08-28 20:31:12 · answer #1 · answered by Slitwrist 1 · 0 0

y = (x+2) / (x -2)
first thing you do is multiply both sides by x-2 to clear the fraction:

y(x -2) = (x +2) / (x-2) * (x-2)
now the x-2 and the x-2 will concel each other out by division. giving you:

y(x -2) = (x +2)
now clear the paranthesis on the left side term: by multiplying:

xy - 2y = x +2
use additive inverse to get x to the left side of the = sign

xy - 2y - x = x -x +2
and
xy -2y -x = 2
now used additive inverse again this time to get 2y on right side of = sign.

xy - 2y +2y -x = 2 + 2y
and
xy - x = 2 + 2y
now factor both sides
which gives you:

x(y -1) = 2(1 + y) since both 1 & y are positive rearrange to standard form which is (y +1)
now tranpose (y-1) by dividing both sides by (y-1_

x(y -1) / (y-1) = 2( y +1) / (y-1)

x = 2( y +1) / (y - 1)
hope that helped. show you how to work the problem and similiar problems.

2007-08-28 20:45:43 · answer #2 · answered by JUAN FRAN$$$ 7 · 0 0

one cannot determine the value of x untill the value of y is given. for each unknown veriable we need an equation. if you want to get x in terms of y then:
y = (x+2) / (x-2)
y(x-2) = x + 2
yx - 2y = x - 2
yx - 2y + 2= x
x - xy = -2(y+2)
x(1-y) = -2(y+2)
x = -2(y+2) / (1-y)

2007-08-28 20:38:36 · answer #3 · answered by cOoL @$ FiRe~~ 1 · 0 0

x = (2 + 2y) / ( y - 1)

= 2(1 + y) / ( y - 1 )

2007-08-28 20:26:05 · answer #4 · answered by Anonymous · 0 0

y=(x+2)/(x-2)
x+2=y(x-2)
x=xy-2y-2
x-xy=-2(y+2)
x(1-y)=-2(y+2)
x=-2(y+2)/(1-y)

2007-08-28 20:31:32 · answer #5 · answered by Anonymous · 0 0

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