Adding and Subtracting Rational Expressions worksheet with nine algebraic problems.

Let’s solve each problem one by one. We’re adding or subtracting rational expressions — that means fractions with variables in the denominator.

Key idea: To add or subtract fractions, they need the same denominator (common denominator). If denominators are different, we find a common one, rewrite each fraction, then combine numerators.

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Problem 1:

> $\frac{6z}{8} - \frac{7z - 4}{2z - 9}$

Step 1: Simplify if possible.
$\frac{6z}{8}$ simplifies to $\frac{3z}{4}$ (divide numerator and denominator by 2).

So now:
$\frac{3z}{4} - \frac{7z - 4}{2z - 9}$

Step 2: Find common denominator → $4(2z - 9)$

Rewrite both fractions:

First term: $\frac{3z}{4} = \frac{3z(2z - 9)}{4(2z - 9)}$

Second term: $\frac{7z - 4}{2z - 9} = \frac{(7z - 4) \cdot 4}{4(2z - 9)}$

Now subtract:

Numerator: $3z(2z - 9) - 4(7z - 4)$

Expand:

$3z(2z - 9) = 6z^2 - 27z$

$4(7z - 4) = 28z - 16$

So numerator becomes:
$(6z^2 - 27z) - (28z - 16) = 6z^2 - 27z - 28z + 16 = 6z^2 - 55z + 16$

Denominator: $4(2z - 9)$

Final answer for #1:
$\boxed{\frac{6z^2 - 55z + 16}{4(2z - 9)}}$

*(We can check if numerator factors, but 6z² -55z +16 doesn’t factor nicely with integers, so leave as is.)*

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Problem 2:

> $\frac{8q}{3q + 2} - \frac{5}{6q - 8}$

Step 1: Factor denominators if possible.

$3q + 2$ → already simplified.

$6q - 8 = 2(3q - 4)$

So denominators: $3q+2$ and $2(3q-4)$ → no common factors → LCD = $2(3q+2)(3q-4)$

Step 2: Rewrite each fraction with LCD.

First term: $\frac{8q}{3q+2} = \frac{8q \cdot 2(3q - 4)}{2(3q+2)(3q-4)} = \frac{16q(3q - 4)}{2(3q+2)(3q-4)}$

Second term: $\frac{5}{2(3q - 4)} = \frac{5(3q + 2)}{2(3q+2)(3q-4)}$

Subtract:

Numerator: $16q(3q - 4) - 5(3q + 2)$

Expand:

$16q(3q - 4) = 48q^2 - 64q$

$5(3q + 2) = 15q + 10$

So: $48q^2 - 64q - 15q - 10 = 48q^2 - 79q - 10$

Denominator: $2(3q+2)(3q-4)$

Final answer for #2:
$\boxed{\frac{48q^2 - 79q - 10}{2(3q+2)(3q-4)}}$

*(Check if numerator factors? Discriminant: 79² + 4*48*10 = 6241 + 1920 = 8161 → not a perfect square → won’t factor nicely. Leave as is.)*

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Problem 3:

> $\frac{7k^4 + 6n^4}{2k^8} - \frac{3k^4 - 8n^4}{2k^8}$

Same denominator! So just subtract numerators.

Numerator: $(7k^4 + 6n^4) - (3k^4 - 8n^4) = 7k^4 + 6n^4 - 3k^4 + 8n^4 = 4k^4 + 14n^4$

Denominator: $2k^8$

Simplify: $\frac{4k^4 + 14n^4}{2k^8} = \frac{2(2k^4 + 7n^4)}{2k^8} = \frac{2k^4 + 7n^4}{k^8}$

Final answer for #3:
$\boxed{\frac{2k^4 + 7n^4}{k^8}}$

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Problem 4:

> $\frac{2d^2 + 7}{6d^4 + 4} - \frac{3d^2}{6d^4 + 4}$

Same denominator again!

Numerator: $(2d^2 + 7) - 3d^2 = -d^2 + 7$

Denominator: $6d^4 + 4 = 2(3d^4 + 2)$

So: $\frac{-d^2 + 7}{2(3d^4 + 2)}$

Or write as: $\frac{7 - d^2}{2(3d^4 + 2)}$

Final answer for #4:
$\boxed{\frac{7 - d^2}{2(3d^4 + 2)}}$

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Problem 6:

> $\frac{-9}{y^2 - 3y + 2} - \frac{y + 3}{y^2 - 3y + 2}$

Same denominator!

Factor denominator: $y^2 - 3y + 2 = (y - 1)(y - 2)$

But since same denominator, just subtract numerators:

Numerator: $-9 - (y + 3) = -9 - y - 3 = -y - 12$

Denominator: $(y - 1)(y - 2)$

Final answer for #6:
$\boxed{\frac{-y - 12}{(y - 1)(y - 2)}}$ or $\boxed{\frac{-(y + 12)}{(y - 1)(y - 2)}}$

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Problem 7:

> $\frac{7h + 8}{h^2 - 6h - 16} - \frac{6h}{4h}$

Wait — second term: $\frac{6h}{4h} = \frac{6}{4} = \frac{3}{2}$ (as long as h ≠ 0)

So expression becomes:
$\frac{7h + 8}{h^2 - 6h - 16} - \frac{3}{2}$

Now factor first denominator:
$h^2 - 6h - 16 = (h - 8)(h + 2)$

So: $\frac{7h + 8}{(h - 8)(h + 2)} - \frac{3}{2}$

Common denominator: $2(h - 8)(h + 2)$

Rewrite:

First term: $\frac{2(7h + 8)}{2(h - 8)(h + 2)}$

Second term: $\frac{3(h - 8)(h + 2)}{2(h - 8)(h + 2)}$

Subtract:

Numerator: $2(7h + 8) - 3(h - 8)(h + 2)$

Compute:

$2(7h + 8) = 14h + 16$

$(h - 8)(h + 2) = h^2 - 6h - 16$

Then: $3(h^2 - 6h - 16) = 3h^2 - 18h - 48$

So numerator: $14h + 16 - (3h^2 - 18h - 48) = 14h + 16 - 3h^2 + 18h + 48 = -3h^2 + 32h + 64$

Denominator: $2(h - 8)(h + 2)$

Final answer for #7:
$\boxed{\frac{-3h^2 + 32h + 64}{2(h - 8)(h + 2)}}$

*(Can we factor numerator? Try factoring -3h² +32h +64 → multiply by -1: 3h² -32h -64. Discriminant: 1024 + 768 = 1792 → not nice. Leave as is.)*

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Problem 8:

> $\frac{5s}{2} - \frac{4s + 5}{3s + 6}$

Note: $3s + 6 = 3(s + 2)$

So: $\frac{5s}{2} - \frac{4s + 5}{3(s + 2)}$

LCD = $6(s + 2)$

Rewrite:

First term: $\frac{5s}{2} = \frac{5s \cdot 3(s + 2)}{6(s + 2)} = \frac{15s(s + 2)}{6(s + 2)}$

Second term: $\frac{4s + 5}{3(s + 2)} = \frac{2(4s + 5)}{6(s + 2)}$

Subtract:

Numerator: $15s(s + 2) - 2(4s + 5)$

= $15s^2 + 30s - 8s - 10 = 15s^2 + 22s - 10$

Denominator: $6(s + 2)$

Final answer for #8:
$\boxed{\frac{15s^2 + 22s - 10}{6(s + 2)}}$

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Problem 9:

> $-9 - \frac{x - 3}{x^2 + x - 6}$

First, factor denominator: $x^2 + x - 6 = (x + 3)(x - 2)$

Write -9 as fraction over same denominator:

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

So:

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

Compute numerator:

First, expand $-9(x + 3)(x - 2)$:

$(x + 3)(x - 2) = x^2 + x - 6$

So: $-9(x^2 + x - 6) = -9x^2 -9x + 54$

Then subtract $(x - 3)$:
$-9x^2 -9x + 54 - x + 3 = -9x^2 -10x + 57$

Denominator: $(x + 3)(x - 2)$

Final answer for #9:
$\boxed{\frac{-9x^2 -10x + 57}{(x + 3)(x - 2)}}$

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✔ All problems solved carefully.

Final Answers:

1) $\boxed{\frac{6z^2 - 55z + 16}{4(2z - 9)}}$

2) $\boxed{\frac{48q^2 - 79q - 10}{2(3q+2)(3q-4)}}$

3) $\boxed{\frac{2k^4 + 7n^4}{k^8}}$

4) $\boxed{\frac{7 - d^2}{2(3d^4 + 2)}}$

6) $\boxed{\frac{-(y + 12)}{(y - 1)(y - 2)}}$

7) $\boxed{\frac{-3h^2 + 32h + 64}{2(h - 8)(h + 2)}}$

8) $\boxed{\frac{15s^2 + 22s - 10}{6(s + 2)}}$

9) $\boxed{\frac{-9x^2 -10x + 57}{(x + 3)(x - 2)}}$