Question

Solve

Answer 1

\[\frac{ 3x }{ x-1 } + 2 = \frac{ 3 }{ x-1 }\]

Answer 2

im not solving nothing

Answer 3

ok, sorry

Answer 4

lol

Answer 5

Can anybody please help?

Answer 6

@RH first make the left side all over the same denominator. Can you do that?

Answer 7

I dont know how:(

Answer 8

Ok observe. I will now get the left side all over the same denominator. To do this, multiply the top and bottom of "2/1" by (x-1):\[\bf 2=\frac{ 2(x-1) }{ 2(x-1) }\]Now we re-write the left side as:\[\bf \frac{ 3x }{ x-1 } +\frac{ 2(x-1)}{ x-1 }=\frac{ 3x+2(x-1) }{ x-1 }\]

Answer 9

After you do that, multiply both sides of the equation by (x - 1) which effectively cancels out the denominators (x-1) of both sides. What you're left with is this:\[\bf 3x+2(x-1)=3\]Can you solve for 'x' now?

Answer 10

does x equal 1?

Answer 11

no

Answer 12

you can't put 1 because you have then 0 in denominator

Answer 13

you see? x-1 cannot be eual to 0

Answer 14

@genius12

Answer 15

Domain:
\[x-1 \neq 0 \rightarrow x \neq 1\]

Answer 16

@Fifciol makes a valid point but that's the only solution the problem.

Answer 17

there is no solution

Answer 18

you can't divide by zero

Answer 19

@RH Are you sure you typed up the question correctly? Because as @Fifciol mentions, our domain is retricted and x can't be 1 and the only solution we get is 1 which isn't part of the domain hence there exists no solution. If this is one of those problems in which 'x' does have a solution then the question wrong otherwise you typed it incorrectly.