Simplifying algebraic fractions - Variation Theory - Free Printable

Let’s solve this step by step.

We are given a list of rational expressions (fractions with polynomials). The goal is to simplify each one as much as possible — that means factoring numerator and denominator, then canceling any common factors.

We’ll go row by row, left column first, then right column.

---

Left Column:

1. First expression:
\[
\frac{2(x + 3)}{3(x + 3)}
\]
→ Both numerator and denominator have (x + 3). Cancel them.
→ Left with: \( \frac{2}{3} \)

2. Second expression:
\[
\frac{2(x + 3)}{(x + 3)}
\]
→ Cancel (x + 3) top and bottom.
→ Left with: \( 2 \)

3. Third expression:
\[
\frac{(x + 3)}{2(x + 3)}
\]
→ Cancel (x + 3).
→ Left with: \( \frac{1}{2} \)

4. Fourth expression:
\[
\frac{x + 3}{2x + 6}
\]
→ Factor denominator: 2x + 6 = 2(x + 3)
→ So: \( \frac{x + 3}{2(x + 3)} \)
→ Cancel (x + 3)
→ Left with: \( \frac{1}{2} \)

5. Fifth expression:
\[
\frac{(x + 3)(x + 4)}{2x + 6}
\]
→ Denominator: 2x + 6 = 2(x + 3)
→ So: \( \frac{(x + 3)(x + 4)}{2(x + 3)} \)
→ Cancel (x + 3)
→ Left with: \( \frac{x + 4}{2} \)

6. Sixth expression:
\[
\frac{(x + 3)(x + 4)}{2x + 8}
\]
→ Denominator: 2x + 8 = 2(x + 4)
→ So: \( \frac{(x + 3)(x + 4)}{2(x + 4)} \)
→ Cancel (x + 4)
→ Left with: \( \frac{x + 3}{2} \)

7. Seventh expression:
\[
\frac{x^2 + 7x + 12}{2x + 8}
\]
→ Factor numerator: x² + 7x + 12 = (x + 3)(x + 4)
→ Denominator: 2x + 8 = 2(x + 4)
→ So: \( \frac{(x + 3)(x + 4)}{2(x + 4)} \)
→ Cancel (x + 4)
→ Left with: \( \frac{x + 3}{2} \)

---

Right Column:

8. Eighth expression:
\[
\frac{x^2 + 7x + 12}{(x + 4)(x - 9)}
\]
→ Numerator: x² + 7x + 12 = (x + 3)(x + 4)
→ So: \( \frac{(x + 3)(x + 4)}{(x + 4)(x - 9)} \)
→ Cancel (x + 4)
→ Left with: \( \frac{x + 3}{x - 9} \)

9. Ninth expression:
\[
\frac{x^2 + 7x + 12}{x^2 - 5x - 36}
\]
→ Numerator: (x + 3)(x + 4)
→ Denominator: x² - 5x - 36 → factor: look for two numbers that multiply to -36 and add to -5 → -9 and +4 → so (x - 9)(x + 4)
→ So: \( \frac{(x + 3)(x + 4)}{(x - 9)(x + 4)} \)
→ Cancel (x + 4)
→ Left with: \( \frac{x + 3}{x - 9} \)

10. Tenth expression:
\[
\frac{2x^2 + 14x + 24}{x^2 - 5x - 36}
\]
→ Numerator: factor out 2 first → 2(x² + 7x + 12) = 2(x + 3)(x + 4)
→ Denominator: same as above → (x - 9)(x + 4)
→ So: \( \frac{2(x + 3)(x + 4)}{(x - 9)(x + 4)} \)
→ Cancel (x + 4)
→ Left with: \( \frac{2(x + 3)}{x - 9} \)

11. Eleventh expression:
\[
\frac{2x^2 + 14x + 24}{3x^2 - 15x - 108}
\]
→ Numerator: 2(x² + 7x + 12) = 2(x + 3)(x + 4)
→ Denominator: factor out 3 → 3(x² - 5x - 36) = 3(x - 9)(x + 4)
→ So: \( \frac{2(x + 3)(x + 4)}{3(x - 9)(x + 4)} \)
→ Cancel (x + 4)
→ Left with: \( \frac{2(x + 3)}{3(x - 9)} \)

12. Twelfth expression:
\[
\frac{2x^2 + 14x + 24}{3x^2 + 4x - 15}
\]
→ Numerator: 2(x + 3)(x + 4)
→ Denominator: 3x² + 4x - 15 → factor: need two numbers that multiply to 3*(-15)= -45 and add to 4 → 9 and -5
→ Split middle term: 3x² + 9x - 5x - 15 = 3x(x + 3) -5(x + 3) = (3x - 5)(x + 3)
→ So: \( \frac{2(x + 3)(x + 4)}{(3x - 5)(x + 3)} \)
→ Cancel (x + 3)
→ Left with: \( \frac{2(x + 4)}{3x - 5} \)

13. Thirteenth expression:
\[
\frac{14x - 24 - 2x^2}{3x^2 + 4x - 15}
\]
→ First, rearrange numerator in standard form: -2x² + 14x - 24
→ Factor out -2: -2(x² - 7x + 12)
→ Factor inside: x² - 7x + 12 = (x - 3)(x - 4)
→ So numerator: -2(x - 3)(x - 4)
→ Denominator: from before → (3x - 5)(x + 3)
→ No common factors → cannot simplify further.
→ But let’s check if we can write it differently? Actually, maybe factor numerator differently?

Wait — let me double-check:

Original numerator: 14x - 24 - 2x² → rewrite as -2x² + 14x - 24

Factor out -2: -2(x² - 7x + 12) → yes, which is -2(x - 3)(x - 4)

Denominator: 3x² + 4x - 15 = (3x - 5)(x + 3) — no common factors with numerator.

So answer is: \( \frac{-2(x - 3)(x - 4)}{(3x - 5)(x + 3)} \)

But perhaps we can leave it as is or factor negative sign? Let’s see next one.

Actually, wait — maybe I made a mistake? Let me check the last one too.

14. Fourteenth expression:
\[
\frac{14x - 24 - 2x^2}{3x^2 - 4x - 15}
\]
→ Numerator: same as above → -2x² + 14x - 24 = -2(x² - 7x + 12) = -2(x - 3)(x - 4)
→ Denominator: 3x² - 4x - 15 → factor: multiply 3*(-15)= -45, find two numbers that add to -4 → -9 and +5
→ 3x² - 9x + 5x - 15 = 3x(x - 3) + 5(x - 3) = (3x + 5)(x - 3)
→ So: \( \frac{-2(x - 3)(x - 4)}{(3x + 5)(x - 3)} \)
→ Cancel (x - 3)
→ Left with: \( \frac{-2(x - 4)}{3x + 5} \) or \( \frac{2(4 - x)}{3x + 5} \)

Now back to #13 — since denominator was (3x - 5)(x + 3), and numerator has (x-3)(x-4), no cancellation → so it stays as is.

But let me confirm all answers again carefully.

---

Let me list all simplified forms clearly:

Left Column:

1. \( \frac{2}{3} \)

2. \( 2 \)

3. \( \frac{1}{2} \)

4. \( \frac{1}{2} \)

5. \( \frac{x + 4}{2} \)

6. \( \frac{x + 3}{2} \)

7. \( \frac{x + 3}{2} \)

Right Column:

8. \( \frac{x + 3}{x - 9} \)

9. \( \frac{x + 3}{x - 9} \)

10. \( \frac{2(x + 3)}{x - 9} \)

11. \( \frac{2(x + 3)}{3(x - 9)} \)

12. \( \frac{2(x + 4)}{3x - 5} \)

13. \( \frac{-2(x - 3)(x - 4)}{(3x - 5)(x + 3)} \) — but maybe we can write numerator as -2(x² -7x +12) but better to leave factored.

Wait — actually, in problem 13, denominator is 3x² + 4x -15 = (3x -5)(x+3), numerator is -2x² +14x -24 = -2(x² -7x +12) = -2(x-3)(x-4). No common factors → correct.

But let me check if the original problem might have typo? Or maybe I should present it simplified as much as possible.

Alternatively, perhaps factor numerator differently? No, it's correct.

For consistency, let’s write all answers in simplest form.

Problem 13: \( \frac{-2(x - 3)(x - 4)}{(3x - 5)(x + 3)} \) — this is fully simplified.

Problem 14: \( \frac{-2(x - 4)}{3x + 5} \) or \( \frac{2(4 - x)}{3x + 5} \) — both are fine. Usually we prefer positive leading coefficient, so \( \frac{2(4 - x)}{3x + 5} \) is nicer.

But mathematically equivalent.

Now, let me verify problem 12 again:

Numerator: 2x² +14x +24 = 2(x² +7x +12) = 2(x+3)(x+4)

Denominator: 3x² +4x -15 = (3x -5)(x+3) — yes, because 3x*x=3x², 3x*3=9x, -5*x=-5x, -5*3=-15 → 9x-5x=4x → correct.

Cancel (x+3): get \( \frac{2(x+4)}{3x -5} \) — correct.

All others seem correct.

So final simplified answers:

Left column (top to bottom):

1. 2/3
2. 2
3. 1/2
4. 1/2
5. (x+4)/2
6. (x+3)/2
7. (x+3)/2

Right column (top to bottom):

8. (x+3)/(x-9)
9. (x+3)/(x-9)
10. 2(x+3)/(x-9)
11. 2(x+3)/(3(x-9))
12. 2(x+4)/(3x-5)
13. [-2(x-3)(x-4)] / [(3x-5)(x+3)]
14. [2(4-x)] / (3x+5) OR [-2(x-4)]/(3x+5) — I'll use the second for consistency.

Actually, for problem 14, after canceling (x-3), we had:

\( \frac{-2(x - 4)}{3x + 5} \)

That’s acceptable.

To make it look cleaner, sometimes people write it as \( \frac{2(4 - x)}{3x + 5} \), but either is fine.

I think for the answer, we’ll present them as simplified fractions.

Now, compiling all:

Final Answer:

Left Column:
1. \(\frac{2}{3}\)
2. \(2\)
3. \(\frac{1}{2}\)
4. \(\frac{1}{2}\)
5. \(\frac{x + 4}{2}\)
6. \(\frac{x + 3}{2}\)
7. \(\frac{x + 3}{2}\)

Right Column:
8. \(\frac{x + 3}{x - 9}\)
9. \(\frac{x + 3}{x - 9}\)
10. \(\frac{2(x + 3)}{x - 9}\)
11. \(\frac{2(x + 3)}{3(x - 9)}\)
12. \(\frac{2(x + 4)}{3x - 5}\)
13. \(\frac{-2(x - 3)(x - 4)}{(3x - 5)(x + 3)}\)
14. \(\frac{-2(x - 4)}{3x + 5}\)

Note: For problem 13, since no simplification beyond factoring, this is the simplest form.

For problem 14, we could also write it as \(\frac{2(4 - x)}{3x + 5}\), but I’ll stick with the direct result from cancellation.

Double-check problem 14:

After canceling (x-3), we have:

Numerator: -2(x - 4)

Denominator: 3x + 5

Yes.

All done.

Final Answer:
\[
\boxed{
\begin{array}{c|c}
\text{Left Column} & \text{Right Column} \\
\hline
\dfrac{2}{3} & \dfrac{x + 3}{x - 9} \\
2 & \dfrac{x + 3}{x - 9} \\
\dfrac{1}{2} & \dfrac{2(x + 3)}{x - 9} \\
\dfrac{1}{2} & \dfrac{2(x + 3)}{3(x - 9)} \\
\dfrac{x + 4}{2} & \dfrac{2(x + 4)}{3x - 5} \\
\dfrac{x + 3}{2} & \dfrac{-2(x - 3)(x - 4)}{(3x - 5)(x + 3)} \\
\dfrac{x + 3}{2} & \dfrac{-2(x - 4)}{3x + 5} \\
\end{array}
}
\]