Question

How do you simplify `(3x+1)/(x-2) - (4x+1)/(x-3)`?

Answer 1

You have to make the denominators the same, so that you can add the two fractions.

One thing you can do to help determine what the final answer should look like is to graph it. If we set `y = (3x+1)/(x-3) - (4x+1)/(x-2)`, and graph it, we get:

graph{y = (3x+1)/(x-3) - (4x+1)/(x-2) [-3, 10, -100, 80]}

As you can see, there are (at least) two zeros (or "roots") to this equation. At approximately `x=0` and `x=6`. There are also the asymptotes around `x=3` and `x=2`. These asymptotes appear because these values will make the denominator equal to zero, exploding the graph to infinite values at these points. This tells you that whatever you simplify to should have roots around `x=0,6`. How shall we simplify this then?

First, multiply both fractional parts by a lowest common denominator, namely `(x-3)*(x-2)`, which is the same as multiplying each fraction by 1:

`((x-2))/((x-2))(3x+1)/(x-3) - (4x+1)/(x-2)((x-3))/((x-3))`

With the same denominator, you can now safely add the numerators in a single fraction, like so:

`((x-2)(3x+1)-(4x+1)(x-3))/((x-2)(x-3))`

Now, let's use the principle of F.O.I.L. to multiply out the numerator:

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

Next, we must combine like terms, and simplify:

`(-x^2+6x+1)/((x-2)(x-3))`

Many times, at this point, you would try to determine if the numerator factors. Unfortunately, it does not factor into anything neat. In fact, if you apply the quadratic formula to the numerator, you find that `x~=-0.1623` and `6.1623`, which you can see on the close up of the graph at those points:

First, near `x~=-0.1623`:

graph{y = (3x+1)/(x-3) - (4x+1)/(x-2) [-.4, 0, -1, 1]}

And also near `x~=6.1623`:

graph{y = (3x+1)/(x-3) - (4x+1)/(x-2) [5.5, 6.5, -1, 1]}.

Therefore, `(-x^2+6x+1)/((x-2)(x-3))`, is the simplified form!