Question

What is the solution to the rational equation 5 over x¨Csquared minus 3x plus 2 - 1 over x minus 2 = 1 over 3x minus 3 ?

Answer 1

let me see if i can write this, and you tell me if it is correct

Answer 2

\[\frac{5}{x^2-3x+2}-\frac{1}{x}=\frac{1}{3x-3}\]

Answer 3

maybe this
\[\frac{5}{x^2-3x+2}-\frac{1}{x-2}=\frac{1}{3x-3}\]

Answer 4

maybe?

Answer 5

yes thats the equation

Answer 6

ok then we can do it
first you have to add on the left hand side

Answer 7

\[\frac{5}{x^2-3x+2}-\frac{1}{x-2}=\frac{1}{3x-3}\]factor the first denominator
\[\frac{5}{(x-1)(x-2)}-\frac{1}{x-2}=\frac{1}{3x-3}\]
then least common denominator on the left is \((x-1)(x-2)\) so subtract via
\[\frac{5}{(x-1)(x-2)}-\frac{1}{x-2}=\frac{5-(x-1)}{(x-1)(x-2)}\]
\[=\frac{6-x}{(x-1)(x-2)}\]

Answer 8

giving us
\[\frac{6-x}{(x-1)(x-2)}=\frac{1}{3x-3}\] then we can "cross multiply" to get
\[(3x-3)(6-x)=(x-1)(x-2)\]

Answer 9

we can even simplify a little before solving because the left hand side now factors as
\[3(x-1)(6-x)=(x-1)(x-2)\] cancel the \((x-1)\) which we are allowed to do because that factor cannot be zero, and we get \[3(6-x)=x-2\]
\[18-x=x-2\]
\[20=2x\]
\[10=x\]

Answer 10

hold on i made a mistake somewhere!

Answer 11

wow a really bad bush league mistake
start with
\[3(6-x)=x-2\]
\[18-3x=x-2\]
\[20=4x\]
\[x=5\] now i am embarrassed

Answer 12

thanks !