Question

How do you add `\frac { 4x + 2} { 3x - 12} + \frac { 2x } { x - 4}`?

Answer 1

See a solution process below:

Explanation:

To add or subtract fractions the must be over a common denominator. First, multiply the fraction on the right by the appropriate form of `1` to put it over a common denominator while not changing the value of the fraction:

`(4x + 2)/(3x - 12) + (3/3 xx (2x)/(x - 4)) =>`

`(4x + 2)/(3x - 12) + (3 xx 2x)/(3(x - 4)) =>`

`(4x + 2)/(3x - 12) + (6x)/((3 xx x) - (3 xx 4)) =>`

`(4x + 2)/(3x - 12) + (6x)/(3x - 12)`

We can now add the numerators over the common denominator:

`(4x + 2 + 6x)/(3x - 12) =>`

`(4x + 6x + 2)/(3x - 12) =>`

`((4 + 6)x + 2)/(3x - 12) =>`

`(10x + 2)/(3x - 12)`

Answer 2

`color(blue)((10x+2)/(3x-12)`

Explanation:

`(4x+2)/(3x-12)+(2x)/(x-4)`

`:.=(4x+2)/(3(x-4))+(2x)/(x-4)`

`:.=(1(4x+2)+3(2x))/(3(x-4))`

`:.=(4x+2+6x)/(3(x-4))`

`:.color(blue)(=(10x+2)/(3x-12)`

Answer 3

Clarifying some points

`(10x+2)/(3x-12)`

Explanation:

`color(blue)("Preamble")`

Using an example consider `6+3=9`

We have added these numbers directly. So why can we do that.

We can not always add fractions directly.

This is why: 6, 3 and 9 are all what is called rational numbers. This means that they can (and may if you so choose) write them in the form of `a/b` where both `a and b` are integer values. What we really have is:

`6/1+3/1+9/1 larr` These are what is called 'improper fraction'

In the same way that `23/2` is an improper fraction.

This means that for years you have been using the rule: To directly add or subtract fraction numerators the denominators must be the same.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Answering the question")`

We need to determine, as best as we can, some common features that already exist within the denominators. Notice that `3x-12` is the same as `3(x-4)` and the `x-4` is a common feature so this is our starting point.

`color(green)([(4x+2)/(3(x-4))]+[(2x)/(x-4)color(red)(xx1)]`

`color(green)([(4x+2)/(3(x-4))]+[(2x)/(x-4)color(red)(xx3/3)]`

`color(green)([(4x+2)/(3(x-4))]+[(6x)/(3(x-4))]`

`color(green)((4x+2+6x)/(3(x-4))`

`color(green)((10x+2)/(3(x-4)) ->(10x+2)/(3x-12)`