Question

How do you solve `\frac { 1} { 2x - 4} + \frac { 3} { 5} = \frac { 3} { 10}`?

Answer 1

See a solution process below:

Explanation:

First, subtract `color(red)(3/5)` from each side of the equation to isolate the fraction with the `x` term while keeping the equation balanced:

`1/(2x - 4) + 3/5 - color(red)(3/5) = 3/10 - color(red)(3/5)`

`1/(2x - 4) + 0 = 3/10 - (2/2 xx color(red)(3/5))`

`1/(2x - 4) = 3/10 - color(red)(6/10)`

`1/(2x - 4) = -3/10`

Next, multiply each side of the equation by `(color(red)(2x) - color(red)(4))` to get the `x` term in the numerator while keeping the equation balanced:

`(color(red)(2x) - color(red)(4)) xx 1/(2x - 4) = -3/10(color(red)(2x) - color(red)(4))`

`cancel((color(red)(2x) - color(red)(4))) xx 1/color(red)(cancel(color(black)(2x - 4))) = (-3/10 xx color(red)(2x)) - (-3/10 xx color(red)(4))`

`1 = (-6x)/10 - (-12/10)`

`1 = (-6x)/10 + 12/10`

`1 = (-6x + 12)/10`

Then, multiply each side of the equation by `color(red)(10)` to eliminate the fractions while keeping the equation balanced:

`color(red)(10) xx 1 = color(red)(10) xx (-6x + 12)/10`

`10 = cancel(color(red)(10)) xx (-6x + 12)/color(red)(cancel(color(black)(10)))`

`10 = -6x + 12`

Next, subtract `color(red)(12)` from each side of the equation to isolate the `x` term while keeping the equation balanced:

`10 - color(red)(12) = -6x + 12 - color(red)(12)`

`-2 = -6x + 0`

`-2 = -6x`

Now, divide each side of the equation by `color(red)(-6)` to solve for `x` while keeping the equation balanced:

`(-2)/color(red)(-6) = (-6x)/color(red)(-6)`

`1/3 = (color(red)(cancel(color(black)(-6)))x)/cancel(color(red)(-6))`

`1/3 = x`

`x = 1/3`