SOLUTION: Simplifying rational expressions worksheet algebra ... - Free Printable

Let’s solve each problem step by step. We’ll simplify rational expressions by factoring, canceling common factors, and combining terms.

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Problem 1:
\[
\frac{x^2 + 11x - 12}{x^2 - 1} \cdot \frac{5x + 5}{3x + 1}
\]

Step 1: Factor everything.

- \(x^2 + 11x - 12 = (x - 1)(x + 12)\)
- \(x^2 - 1 = (x + 1)(x - 1)\) → difference of squares
- \(5x + 5 = 5(x + 1)\)

So now we have:
\[
\frac{(x - 1)(x + 12)}{(x + 1)(x - 1)} \cdot \frac{5(x + 1)}{3x + 1}
\]

Step 2: Cancel common factors.

- Cancel \((x - 1)\) top and bottom in first fraction.
- Cancel \((x + 1)\) from denominator of first fraction with numerator of second fraction.

Left with:
\[
\frac{(x + 12)}{1} \cdot \frac{5}{3x + 1} = \frac{5(x + 12)}{3x + 1}
\]

You can also write it as \(\frac{5x + 60}{3x + 1}\), but factored form is usually preferred unless told otherwise.

✔ Final Answer for Problem 1: \(\boxed{\frac{5(x + 12)}{3x + 1}}\)

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Problem 2:
\[
(2x + 16) \cdot \frac{5}{2x^2 + 17x + 8}
\]

Step 1: Factor.

- \(2x + 16 = 2(x + 8)\)
- \(2x^2 + 17x + 8\) → factor by grouping or trial:
Find two numbers that multiply to \(2 \cdot 8 = 16\) and add to 17 → 16 and 1.
So: \(2x^2 + 16x + x + 8 = 2x(x + 8) + 1(x + 8) = (2x + 1)(x + 8)\)

Now expression becomes:
\[
\frac{2(x + 8)}{1} \cdot \frac{5}{(2x + 1)(x + 8)}
\]

Step 2: Cancel \((x + 8)\)

Left with:
\[
\frac{2}{1} \cdot \frac{5}{2x + 1} = \frac{10}{2x + 1}
\]

✔ Final Answer for Problem 2: \(\boxed{\frac{10}{2x + 1}}\)

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Problem 3:
\[
\frac{y - 6}{10} \div \frac{6 - y}{5}
\]

Step 1: Invert and multiply.

\[
\frac{y - 6}{10} \cdot \frac{5}{6 - y}
\]

Note: \(6 - y = -(y - 6)\), so rewrite:

\[
\frac{y - 6}{10} \cdot \frac{5}{-(y - 6)} = \frac{y - 6}{10} \cdot \frac{-5}{y - 6}
\]

Step 2: Cancel \((y - 6)\)

Left with:
\[
\frac{1}{10} \cdot (-5) = -\frac{5}{10} = -\frac{1}{2}
\]

✔ Final Answer for Problem 3: \(\boxed{-\frac{1}{2}}\)

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Problem 4:
\[
\frac{2x^2 - 5x - 3}{4x^2 - 12x - 7} \div \frac{4x + 9}{2x - 7}
\]

Step 1: Invert and multiply.

\[
\frac{2x^2 - 5x - 3}{4x^2 - 12x - 7} \cdot \frac{2x - 7}{4x + 9}
\]

Step 2: Factor numerators and denominators.

- \(2x^2 - 5x - 3\): Find two numbers multiplying to \(2 \cdot -3 = -6\), adding to -5 → -6 and 1.
→ \(2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\)

- \(4x^2 - 12x - 7\): Multiply \(4 \cdot -7 = -28\), find two numbers adding to -12 → -14 and 2.
→ \(4x^2 - 14x + 2x - 7 = 2x(2x - 7) + 1(2x - 7) = (2x + 1)(2x - 7)\)

So now:
\[
\frac{(2x + 1)(x - 3)}{(2x + 1)(2x - 7)} \cdot \frac{2x - 7}{4x + 9}
\]

Step 3: Cancel common factors.

- Cancel \((2x + 1)\)
- Cancel \((2x - 7)\)

Left with:
\[
\frac{(x - 3)}{1} \cdot \frac{1}{4x + 9} = \frac{x - 3}{4x + 9}
\]

✔ Final Answer for Problem 4: \(\boxed{\frac{x - 3}{4x + 9}}\)

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Problem 5:
\[
\frac{4s^2 + 8s - 12}{5 - 6s + s^2} \div \frac{2s^2 + 7s + 3}{4s^2 - 8s - 5}
\]

First, rearrange denominator of first fraction: \(s^2 - 6s + 5\)

Invert and multiply:
\[
\frac{4s^2 + 8s - 12}{s^2 - 6s + 5} \cdot \frac{4s^2 - 8s - 5}{2s^2 + 7s + 3}
\]

Factor all:

- \(4s^2 + 8s - 12 = 4(s^2 + 2s - 3) = 4(s + 3)(s - 1)\)
- \(s^2 - 6s + 5 = (s - 1)(s - 5)\)
- \(4s^2 - 8s - 5\): Multiply \(4 \cdot -5 = -20\), find two numbers adding to -8 → -10 and 2.
→ \(4s^2 - 10s + 2s - 5 = 2s(2s - 5) + 1(2s - 5) = (2s + 1)(2s - 5)\)
- \(2s^2 + 7s + 3\): Multiply \(2 \cdot 3 = 6\), add to 7 → 6 and 1.
→ \(2s^2 + 6s + s + 3 = 2s(s + 3) + 1(s + 3) = (2s + 1)(s + 3)\)

Now plug in:
\[
\frac{4(s + 3)(s - 1)}{(s - 1)(s - 5)} \cdot \frac{(2s + 1)(2s - 5)}{(2s + 1)(s + 3)}
\]

Cancel:

- \((s - 1)\)
- \((s + 3)\)
- \((2s + 1)\)

Left with:
\[
\frac{4}{s - 5} \cdot (2s - 5) = \frac{4(2s - 5)}{s - 5}
\]

✔ Final Answer for Problem 5: \(\boxed{\frac{4(2s - 5)}{s - 5}}\)

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Problem 6:
This is written as a complex fraction:
\[
\frac{\frac{x^2 + 3x + 2}{x^2 - 1}}{\frac{4x^2 + 8x}{5(x - 1)}}
\]

Which means:
\[
\frac{x^2 + 3x + 2}{x^2 - 1} \div \frac{4x^2 + 8x}{5(x - 1)}
\]

Invert and multiply:
\[
\frac{x^2 + 3x + 2}{x^2 - 1} \cdot \frac{5(x - 1)}{4x^2 + 8x}
\]

Factor:

- \(x^2 + 3x + 2 = (x + 1)(x + 2)\)
- \(x^2 - 1 = (x + 1)(x - 1)\)
- \(4x^2 + 8x = 4x(x + 2)\)

Now:
\[
\frac{(x + 1)(x + 2)}{(x + 1)(x - 1)} \cdot \frac{5(x - 1)}{4x(x + 2)}
\]

Cancel:

- \((x + 1)\)
- \((x - 1)\)
- \((x + 2)\)

Left with:
\[
\frac{1}{1} \cdot \frac{5}{4x} = \frac{5}{4x}
\]

✔ Final Answer for Problem 6: \(\boxed{\frac{5}{4x}}\)

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Final Answers Summary:

1. \(\boxed{\frac{5(x + 12)}{3x + 1}}\)
2. \(\boxed{\frac{10}{2x + 1}}\)
3. \(\boxed{-\frac{1}{2}}\)
4. \(\boxed{\frac{x - 3}{4x + 9}}\)
5. \(\boxed{\frac{4(2s - 5)}{s - 5}}\)
6. \(\boxed{\frac{5}{4x}}\)