Algebra 1 Worksheets | Rational Expressions Worksheets - Free Printable

Let’s solve each problem one by one. Remember: when dividing rational expressions, you flip the second fraction (take its reciprocal) and then multiply. Then simplify by canceling common factors.

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Problem 1:

\[
\frac{7q}{q + 11} \div \frac{7q}{8q + 88}
\]

Step 1: Flip the second fraction → multiply by reciprocal:
\[
\frac{7q}{q + 11} \times \frac{8q + 88}{7q}
\]

Step 2: Factor where possible. Notice \(8q + 88 = 8(q + 11)\):
\[
\frac{7q}{q + 11} \times \frac{8(q + 11)}{7q}
\]

Step 3: Cancel common factors:
- \(7q\) cancels with \(7q\)
- \(q + 11\) cancels with \(q + 11\)

Left with: \(8\)

✔ Final Answer for #1: 8

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Problem 2:

\[
\frac{7s^2}{6} \div \frac{8s}{2}
\]

Step 1: Flip second fraction:
\[
\frac{7s^2}{6} \times \frac{2}{8s}
\]

Step 2: Multiply numerators and denominators:
Numerator: \(7s^2 \cdot 2 = 14s^2\)
Denominator: \(6 \cdot 8s = 48s\)

So: \(\frac{14s^2}{48s}\)

Step 3: Simplify:
- Divide numerator and denominator by 2: \(\frac{7s^2}{24s}\)
- Cancel one \(s\): \(\frac{7s}{24}\)

✔ Final Answer for #2: \(\frac{7s}{24}\)

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Problem 3:

\[
\frac{b^2 + 18b + 72}{b^2 + 13b + 42} \div \frac{1}{b + 7}
\]

Step 1: Flip second fraction → multiply by \(b + 7\):
\[
\frac{b^2 + 18b + 72}{b^2 + 13b + 42} \times (b + 7)
\]

Step 2: Factor both quadratics.

First numerator: \(b^2 + 18b + 72\) → find two numbers that multiply to 72 and add to 18 → 6 and 12 → \((b + 6)(b + 12)\)

First denominator: \(b^2 + 13b + 42\) → 6 and 7 → \((b + 6)(b + 7)\)

So now:
\[
\frac{(b + 6)(b + 12)}{(b + 6)(b + 7)} \times (b + 7)
\]

Step 3: Cancel common factors:
- \(b + 6\) cancels
- \(b + 7\) cancels

Left with: \(b + 12\)

✔ Final Answer for #3: \(b + 12\)

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Problem 4:

\[
\frac{c + 9}{c + 13c + 36} \div \frac{10c}{c + 3}
\]

Wait — look at denominator: \(c + 13c + 36 = 14c + 36\)? That seems odd. Probably a typo — likely meant \(c^2 + 13c + 36\). Let’s assume that (since otherwise it doesn’t factor nicely and isn’t typical for this level).

Assume: \(\frac{c + 9}{c^2 + 13c + 36} \div \frac{10c}{c + 3}\)

Step 1: Flip second fraction:
\[
\frac{c + 9}{c^2 + 13c + 36} \times \frac{c + 3}{10c}
\]

Step 2: Factor denominator: \(c^2 + 13c + 36 = (c + 4)(c + 9)\)

So:
\[
\frac{c + 9}{(c + 4)(c + 9)} \times \frac{c + 3}{10c}
\]

Step 3: Cancel \(c + 9\):
\[
\frac{1}{c + 4} \times \frac{c + 3}{10c} = \frac{c + 3}{10c(c + 4)}
\]

✔ Final Answer for #4: \(\frac{c + 3}{10c(c + 4)}\)

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Problem 5:

\[
\frac{12r^2}{11} \div \frac{3r}{4}
\]

Step 1: Flip second fraction:
\[
\frac{12r^2}{11} \times \frac{4}{3r}
\]

Step 2: Multiply:
Numerator: \(12r^2 \cdot 4 = 48r^2\)
Denominator: \(11 \cdot 3r = 33r\)

→ \(\frac{48r^2}{33r}\)

Step 3: Simplify:
Divide numerator and denominator by 3: \(\frac{16r^2}{11r}\)
Cancel one \(r\): \(\frac{16r}{11}\)

✔ Final Answer for #5: \(\frac{16r}{11}\)

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Problem 6:

\[
\frac{7v}{4} \div \frac{6}{10}
\]

Step 1: Flip second fraction:
\[
\frac{7v}{4} \times \frac{10}{6}
\]

Step 2: Multiply:
Numerator: \(7v \cdot 10 = 70v\)
Denominator: \(4 \cdot 6 = 24\)

→ \(\frac{70v}{24}\)

Step 3: Simplify by dividing numerator and denominator by 2:
\(\frac{35v}{12}\)

✔ Final Answer for #6: \(\frac{35v}{12}\)

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Problem 7:

\[
\frac{9n^2}{10} \div \frac{6n}{11}
\]

Step 1: Flip second fraction:
\[
\frac{9n^2}{10} \times \frac{11}{6n}
\]

Step 2: Multiply:
Numerator: \(9n^2 \cdot 11 = 99n^2\)
Denominator: \(10 \cdot 6n = 60n\)

→ \(\frac{99n^2}{60n}\)

Step 3: Simplify:
Divide numerator and denominator by 3: \(\frac{33n^2}{20n}\)
Cancel one \(n\): \(\frac{33n}{20}\)

✔ Final Answer for #7: \(\frac{33n}{20}\)

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Problem 8:

\[
\frac{5p}{p + 7} \div \frac{5p}{3p + 21}
\]

Step 1: Flip second fraction:
\[
\frac{5p}{p + 7} \times \frac{3p + 21}{5p}
\]

Step 2: Factor \(3p + 21 = 3(p + 7)\)

So:
\[
\frac{5p}{p + 7} \times \frac{3(p + 7)}{5p}
\]

Step 3: Cancel:
- \(5p\) cancels
- \(p + 7\) cancels

Left with: 3

✔ Final Answer for #8: 3

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Problem 9:

\[
\frac{10g}{11} \div \frac{6}{8}
\]

Step 1: Flip second fraction:
\[
\frac{10g}{11} \times \frac{8}{6}
\]

Step 2: Multiply:
Numerator: \(10g \cdot 8 = 80g\)
Denominator: \(11 \cdot 6 = 66\)

→ \(\frac{80g}{66}\)

Step 3: Simplify by dividing by 2:
\(\frac{40g}{33}\)

✔ Final Answer for #9: \(\frac{40g}{33}\)

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Problem 10:

\[
\frac{77d^2 - 43d - 72}{14d^2 - 46d + 36} \div \frac{d^2}{12d^2 d - 48}
\]

Wait — last term in divisor: “12d² d - 48” → probably typo. Likely meant \(12d^2 - 48\) or \(12d^3 - 48\)? But written as “12d² d” which is \(12d^3\). Let’s assume it’s \(12d^3 - 48\), but that seems messy. Alternatively, maybe it’s \(12d^2 - 48\)? Let me check context.

Actually, looking again: “÷ \(\frac{d^2}{12d^2 d - 48}\)” — likely typo. Probably meant \(12d^2 - 48\). I’ll assume that because otherwise it’s too complex.

But let’s try factoring first part.

Numerator: \(77d^2 - 43d - 72\)

Try factoring: Need two numbers that multiply to \(77 \cdot (-72) = -5544\) and add to -43.

This is hard. Maybe use quadratic formula? Or perhaps it factors nicely.

Alternatively, maybe the problem has a typo. Let me see if we can factor numerator and denominator.

Try grouping or AC method.

For \(77d^2 - 43d - 72\):

AC = 77 * -72 = -5544

Find factors of 5544 that differ by 43? This is tedious.

Perhaps the problem intended simpler numbers. Wait — maybe “12d² d” is “12d² - 48”? Let me proceed assuming divisor is \(\frac{d^2}{12d^2 - 48}\)

And factor denominator of first fraction: \(14d^2 - 46d + 36\)

Factor out 2: \(2(7d^2 - 23d + 18)\)

Now factor \(7d^2 - 23d + 18\): need two numbers that multiply to 7*18=126, add to -23 → -14 and -9

So: \(7d^2 -14d -9d +18 = 7d(d - 2) -9(d - 2) = (7d - 9)(d - 2)\)

So denominator: \(2(7d - 9)(d - 2)\)

Now numerator: \(77d^2 -43d -72\)

Try same: 77*(-72)= -5544, find factors adding to -43.

After some trial: 56 and -99? 56*-99=-5544, 56-99=-43 → yes!

So split middle term:

\(77d^2 + 56d - 99d - 72\)

Group: \(7d(11d + 8) -9(11d + 8) = (7d - 9)(11d + 8)\)

Yes! So numerator: \((7d - 9)(11d + 8)\)

Denominator: \(2(7d - 9)(d - 2)\)

So first fraction simplifies to:
\[
\frac{(7d - 9)(11d + 8)}{2(7d - 9)(d - 2)} = \frac{11d + 8}{2(d - 2)} \quad \text{(after canceling } 7d - 9\text{)}
\]

Now divisor: \(\frac{d^2}{12d^2 - 48} = \frac{d^2}{12(d^2 - 4)} = \frac{d^2}{12(d - 2)(d + 2)}\)

So overall expression:
\[
\frac{11d + 8}{2(d - 2)} \div \frac{d^2}{12(d - 2)(d + 2)} = \frac{11d + 8}{2(d - 2)} \times \frac{12(d - 2)(d + 2)}{d^2}
\]

Cancel \(d - 2\):
\[
\frac{11d + 8}{2} \times \frac{12(d + 2)}{d^2}
\]

Multiply:
Numerator: \((11d + 8) \cdot 12(d + 2)\)
Denominator: \(2 \cdot d^2\)

Simplify 12/2 = 6:
\[
\frac{6(11d + 8)(d + 2)}{d^2}
\]

We can leave it like this, or expand numerator:

\((11d + 8)(d + 2) = 11d(d) + 11d(2) + 8(d) + 8(2) = 11d^2 + 22d + 8d + 16 = 11d^2 + 30d + 16\)

Then times 6: \(66d^2 + 180d + 96\)

So final answer: \(\frac{66d^2 + 180d + 96}{d^2}\)

But better to leave factored unless asked to expand.

However, since the original had typos, and this is very complex, perhaps the intended divisor was different. But based on our assumption, this is correct.

But let me double-check the original: “÷ \(\frac{d^2}{12d^2 d - 48}\)” — if “12d² d” means \(12d^3\), then divisor is \(\frac{d^2}{12d^3 - 48} = \frac{d^2}{12(d^3 - 4)}\), which doesn't factor nicely. So likely typo, and my assumption is best.

To match typical problems, perhaps it's \(12d^2 - 48\), so I'll go with that.

Final simplified form: \(\frac{6(11d + 8)(d + 2)}{d^2}\)

Or expanded: \(\frac{66d^2 + 180d + 96}{d^2}\)

But we can also write as \(66 + \frac{180}{d} + \frac{96}{d^2}\), but that’s not standard.

I think leaving as \(\frac{6(11d + 8)(d + 2)}{d^2}\) is fine.

But let me see if we can simplify more — no common factors with denominator.

✔ Final Answer for #10: \(\frac{6(11d + 8)(d + 2)}{d^2}\)

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## Final Answers:

1) 8
2) \(\frac{7s}{24}\)
3) \(b + 12\)
4) \(\frac{c + 3}{10c(c + 4)}\)
5) \(\frac{16r}{11}\)
6) \(\frac{35v}{12}\)
7) \(\frac{33n}{20}\)
8) 3
9) \(\frac{40g}{33}\)
10) \(\frac{6(11d + 8)(d + 2)}{d^2}\)