Asked by Yaya
I need help with this problem:
6/v-1=6/(v-3)
I tried working it out like this:
>>>6/v-1=6/(v-3)
>>>(6-v)/v= 6/(v-3)
I cross multiplied to get:
>>>3v+18-v^2= 6v
>>> -3v+18-v^2=0
^Here's where I became stuck. I would appreciate it if you would point out where I have gone wrong in this problem.
6/v-1=6/(v-3)
I tried working it out like this:
>>>6/v-1=6/(v-3)
>>>(6-v)/v= 6/(v-3)
I cross multiplied to get:
>>>3v+18-v^2= 6v
>>> -3v+18-v^2=0
^Here's where I became stuck. I would appreciate it if you would point out where I have gone wrong in this problem.
Answers
Answered by
Reiny
6/v-1=6/(v-3)
cross-multiply from here
6(v-3) = 6(v-1)
6v - 18 =6v - 6
-18 = -6
contradiction!
so your equation has no solution.
did you type it correctly?
cross-multiply from here
6(v-3) = 6(v-1)
6v - 18 =6v - 6
-18 = -6
contradiction!
so your equation has no solution.
did you type it correctly?
Answered by
bobpursley
The first step is wrong. How did you get (6-v)/v?
Answered by
Yaya
@Reiny: It is 6/v -1= 6/(v+3) I made a mistake there.
@bobpursley: From the 6/v-1 I made it one fraction
6/v -1/1 --> (6 -(1*v))/v
@bobpursley: From the 6/v-1 I made it one fraction
6/v -1/1 --> (6 -(1*v))/v
Answered by
Reiny
ok, multiply each term by v(v-3)
6(v-3) -v(v-3) = 6v
6v - 18 - v^2 + 3v = 6v
v^2 - 3v + 18 = 0
use the quadratic equation, but just looking at it, I can see that you will not get a real answer.
Your two roots will be complex numbers
6(v-3) -v(v-3) = 6v
6v - 18 - v^2 + 3v = 6v
v^2 - 3v + 18 = 0
use the quadratic equation, but just looking at it, I can see that you will not get a real answer.
Your two roots will be complex numbers
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