Question

How do you solve `x+ \frac { 3} { x - 1} = \frac { 3x } { x - 1}`?

Answer 1

`x = 3`

`x = 1` is not a solution because a denominator of `0`, (1 - 1 = 0) is irrational.

Explanation:

First, you need to get all terms over a common denominator, in this case `x - 1`.

`(x-1)/(x-1) * x + 3/(x-1) = (3x)/(x - 1)`

`(x^2 - x)/(x - 1) + 3/(x - 1) = (3x)/(x - 1)`

We can now get this in terms of a quadratic equation over a fraction equaling zero:L

`(x^2 - x + 3)/(x - 1) - (3x)/(x - 1) = (3x)/(x - 1) - (3x)/(x - 1)`

`(x^2 - x - 3x + 3)/(x-1) = 0`

`(x^2 - 4x + 3)/(x-1) = 0`

Now, we can multiple by (x - 1) on each side of the equation to eliminate the fraction and keep the equation balanced:

`(x - 1)(x^2 - 4x + 3)/(x -1 ) = 0*(x - 1)`

`(x^2 - 4x + 3) = 0`

Factoring the quadratic equation gives:

`(x - 3)(x - 1) = 0`

Solving for `(x - 3)` gives:

`(x - 3)(x - 1)/(x - 1) = 0/(x - 1)`

`x - 3 = 0`

`x - 3 + 3 = 0 + 3`

`x = 3`

Solving for `(x - 1)` gives:

`(x - 3)/(x -3) (x - 1) = 0/(x - 3)`

`x - 1 = 0`

`x - 1 + 1 = 0 + 1`

`x = 1`

Answer 2

`x = 3`

Explanation:

When you have an equation with fractions, you can get rid of the denominators by multiplying each term by the LCM of the denominators, in this case `(x-1)`

`x xxcolor(blue)((x-1))+ (3xxcolor(blue)(cancel(x-1)))/(cancel(x-1))= (3x xx(color(blue)cancel(x-1)))/(cancel(x-1)) " " x!=1`

`x^2-x +3 = 3x" "larr`a quadratic, set equal to 0

`x^2-4x +3 = 0" "larr` find factors

`(x-3)(x-1)= 0`

Let each factor be equal to 0.

`x-3 = 0 rarr x = 3`

`x-1 = 0 rarr x = 1`

Reject x = 1, denominator may not be 0.

`x = 3`