ADA 2007 online test QR# 32

[rchristense]

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Can someone please explain how to find the answer to QR question# 32? Thanks

 

[duodenum]

y=(x+2)/(x-3)

set it equal to x
(x-3)👍=(x+2)
(xy-3y-2)=x

now substitute the value u got for x into the original
y=[(xy-3y-2)+2]/[(xy-3y-2)+3]
now simplify
y=(xy-3y)/(xy-3y-5)

get rid of the denominator of the above by multiplying both sides by the denominator (notice it cancels out on the right side)
👍(xy-3y-5)=(xy-3y)

simplify again
(xy^2)-(3y^2)-(5y)=(xy-3y)

separate the like items
(xy^2)-(xy)=(3y^2)-(3y)+(5y)

simplify again
(xy^2)-(xy)=(3y^2)+(2y)

take out an x from the left equation and you are left with
(x)(y^2-y)=(3y^2)+(2y)

divide both sides by (y^2-y) and simplify

you get x=(3y+2)/(y-1)

 

[rchristense]

you da man.

 

[UCB05]

y=(x+2)/(x-3)

set it equal to x
(x-3)👍=(x+2)
(xy-3y-2)=x

now substitute the value u got for x into the original
y=[(xy-3y-2)+2]/[(xy-3y-2)+3]
now simplify
y=(xy-3y)/(xy-3y-5)

get rid of the denominator of the above by multiplying both sides by the denominator (notice it cancels out on the right side)
👍(xy-3y-5)=(xy-3y)

simplify again
(xy^2)-(3y^2)-(5y)=(xy-3y)

separate the like items
(xy^2)-(xy)=(3y^2)-(3y)+(5y)

simplify again
(xy^2)-(xy)=(3y^2)+(2y)

take out an x from the left equation and you are left with
(x)(y^2-y)=(3y^2)+(2y)

divide both sides by (y^2-y) and simplify

you get x=(3y+2)/(y-1)

You can cut out so much of that unnecessary substitution and factoring after the first step:

(xy-3y-2)=x **just move the x terms to one side, move the y & constant to the other**
xy - x = 3y+2 **factor out x
x(y-1) = 3y+2 **divide
x= (3y+2)/(y-1)

 

[duodenum]

oh snap, good call!