# Thread: I have problems with algebraic fractions.

1. ## I have problems with algebraic fractions.

Good day everyone.

I have three questions relating to simplifying algebraic questions, and what intimidates me the most are the denominators. Here are the questions:

1. $\dfrac{4}{(x -2)(x -4)}$ $- \dfrac{2}{(x -2)(x -3)}$

2. $\dfrac{3x}{x -3}$ $- \dfrac{x}{x^2 -9}$

3. $\dfrac{x -4}{(x +1)(x -5)}$ $-\dfrac{x +5}{(x +1)(x +3)}
$

Here are my workings respectively:

1. $\dfrac{4}{(x -2)(x -4)}$ $- \dfrac{2}{(x -2)(x -3)}$

$=\dfrac{4(x -3) -2(x-4)}{(x -2)(x -3)(x -4)}$

$=\dfrac{4x -12 -2x +8}{(x-2)(x-3)(x-4)}$

$=\dfrac{4x-2x-12+8}{(x-2)(x-3)(x-4)}$

$=\dfrac{2x -4}{(x-2)(x-3)(x-4)}$ ?

-------------------------------------------------------

2. $\dfrac{3x}{x -3}$ $- \dfrac{x}{x^2 -9}$

$=\dfrac{3x}{x-3}$ $-\dfrac{x}{(x-3)(x+3)}$

$=\dfrac{3x(x-3)(x+3) -x(x-3)}{(x-3)^2(x+3)}$

$=\dfrac{3x^3+9x^2-9x^2-27x-x^2+3x}{(x-3)^2(x+3)}$

$=\dfrac{3x^3-24x-x^2}{(x-3)^2(x+3)}$ ?

I am so sorry that I couldn't do it further. I felt mortified when my answers are wrong. Please find my faults and explain how I should have done. Thank you.

PythagorasNeophyte

2. Originally Posted by PythagorasNeophyte
2. $\dfrac{3x}{x -3}$ $- \dfrac{x}{x^2 -9}$

$=\dfrac{3x}{x-3}$ $-\dfrac{x}{(x-3)(x+3)}$

$=\dfrac{3x(x-3)(x+3) -x(x-3)}{(x-3)^2(x+3)}$
$(x - 3)(x + 3)$ is the LCM, not $(x - 3)^2(x + 3)$, so it's better to write this:

$=\dfrac{(3x)(x + 3)}{(x-3)(x + 3)} - \dfrac{x}{(x-3)(x+3)}$

$=\dfrac{3x^2 + 9x -x}{(x-3)(x + 3)}$

$=\dfrac{3x^2 + 8x}{(x-3)(x + 3)}$

3. Hello, PythagorasNeophyte!

$(1)\;\;\dfrac{4}{(x -2)(x -4)}$ $- \dfrac{2}{(x -2)(x -3)}$

$\dfrac{4}{(x -2)(x -4)}- \dfrac{2}{(x -2)(x -3)} \;= \;\dfrac{4(x -3) -2(x-4)}{(x -2)(x -3)(x -4)}$

. . $= \;\dfrac{4x -12 -2x +8}{(x-2)(x-3)(x-4)} \;=\;\dfrac{2x -4}{(x-2)(x-3)(x-4)}$

This can be reduced:

. .Factor: . $\dfrac{2(x-2)}{(x-2)(x-3)(x-4)}$

. . Cancel: . $\dfrac{2}{(x-3)(x-4)}$

$(3)\;\;\dfrac{x -4}{(x +1)(x -5)}$ $-\dfrac{x +5}{(x +1)(x +3)}$

$\dfrac{(x-4)(x+3) - (x+5)(x-5)}{(x+1)(x-5)(x+3)}$

. . $=\;\dfrac{(x^2-x-12) - (x^2-25)}{(x+1)(x-5)(x+3)}$

. . $=\; \dfrac{13 - x}{(x+1)(x-5)(x+3)}
$

4. ## Instructions

The first and third problems you can simplify with partial fractions. Have you learned partial fractions?

The second problem, just multiply the numerator and denominator by x + 3 of the first fraction.

Now you can complete combining the fractions with my instructions.

5. I'm not sure why you suggested partial fractions. In partial fractions you are usually given a single fraction and you want to decompose it into a sum/difference of fractions. What the OP wants to do is the other way around -- combine multiple fractions into a single fraction.

6. Originally Posted by eumyang
I'm not sure why you suggested partial fractions. In partial fractions you are usually given a single fraction and you want to decompose it into a sum/difference of fractions. What the OP wants to do is the other way around -- combine multiple fractions into a single fraction.
The poster did say that he/she is intimidated by the denominator so by breaking it down further, it would be easier to follow before combining.

7. Thank you so much everyone!

To eumyang:

In your post, you state that $(x-3)(x+3)$ is the LCM, and not $(x-3)^2(x+3)$.

At first I thought that the LCM is $(x-3)^2(x+3)^2$. Why isn't $(x-3)(x-3) = (x-3)^2$ and $(x+3)(x+3) = (x+3)^2$ in this case?

Why is $(x-3)(x+3)$ the denominator instead of squaring both of the terms? Where did the another $(x-3)(x+3)$ gone to?

8. The other expressions you suggest are common denominators; they are just not the LEAST common denominator.

Look at what we do with finding the LCM of two numbers Consider 2 and 10.
The prime factors of 2 is just 2.
The prime factors of 10 is 2 x 5.

2 goes into 10. If one number goes into another then the LCM is just the larger number, or in our case, 10.

If we were to apply what you suggested to this case, you would be claiming that the LCM is $2^2 \cdot 5$ or 20. You want to "grab" the 2 factor from both numbers. But we don't need to do that.

It's the same with expressions.
The factors of x - 3 is just (x - 3).
The factors of $x^2 - 9$ is (x - 3)(x + 3).

x - 3 "goes into" [maht]x^2 - 9[/tex], so the LCM is the "larger" expression, or [maht]x^2 - 9[/tex].