y=1/(3X-1) Find the inverse
Here's what I did:
X= 1/(3y-1)
X(3y-1)=1
3y-1= 1/X
y= 1/(3X) +1
However the answer is y= (1+X)/3X and I can't figure out why. Can anyone explain? Thanks
2006-10-22 00:58:33 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
When you divided by the 3 to solve for y, you didn't divide the 1 also. The result is
y=1/(3x)+ 1/3.
If you put this over a common denominator, you get the answer you are looking for.
2006-10-22 02:11:14 · answer #1 · answered by mathematician 7 · 1⤊ 0⤋
You are fine up to the step
3y - 1 = 1/x.
Then you moved the 1 from the left-hand side to the right-hand side, and divided only the first term 1/x by 3. You have to do all operations to all terms of the equation.
You should have added 1 to both sides:
3y = 1/x + 1
and then divided both sides by 3
y = 1/3x + 1/3 = (1+x)/3x.
2006-10-22 08:33:45 · answer #2 · answered by ninasgramma 7 · 0⤊ 1⤋
You went wrong in the last line. You need to add 1 and then divide by 3. You divided and then added.
2006-10-22 08:04:12 · answer #3 · answered by coolman9999uk 2 · 0⤊ 0⤋
y=1/(3X-1) Find the inverse
1=y/(3-1)x
x=y/2-1
x= y/1
so
y= (1+X)/3X
2006-10-22 08:08:47 · answer #4 · answered by ehsan 2 · 0⤊ 0⤋
y=1/(3x-1)
x=1/(3y-1)
3y-1=1/x
3y=(x+1)/x
y=3(x+1)/x
2006-10-22 08:05:52 · answer #5 · answered by raj 7 · 0⤊ 0⤋