Math worksheet for simplifying rational expressions with unlike denominators.

Let’s solve each problem one by one. We’ll simplify rational expressions with unlike denominators by finding a common denominator, combining the numerators, and then simplifying if possible.

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Problem 1:
$\frac{2}{y^2 - 16} + \frac{4}{y + 4}$

Step 1: Factor $y^2 - 16 = (y - 4)(y + 4)$
So expression becomes:
$\frac{2}{(y - 4)(y + 4)} + \frac{4}{y + 4}$

Step 2: Common denominator is $(y - 4)(y + 4)$
Rewrite second fraction: $\frac{4(y - 4)}{(y - 4)(y + 4)}$

Step 3: Add numerators:
$2 + 4(y - 4) = 2 + 4y - 16 = 4y - 14$

Final: $\frac{4y - 14}{(y - 4)(y + 4)}$ → can factor numerator: $2(2y - 7)$
But no common factors with denominator → leave as is.

✔ Final Answer for #1: $\boxed{\frac{4y - 14}{(y - 4)(y + 4)}}$ or $\boxed{\frac{2(2y - 7)}{(y - 4)(y + 4)}}$

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Problem 2:
$\frac{9x}{x - 2} - \frac{8x + 5}{x - 1}$

Step 1: Denominators are different: $(x - 2)$ and $(x - 1)$ → LCD = $(x - 2)(x - 1)$

Step 2: Rewrite both fractions:

First: $\frac{9x(x - 1)}{(x - 2)(x - 1)}$
Second: $\frac{(8x + 5)(x - 2)}{(x - 2)(x - 1)}$

Step 3: Subtract numerators:

$9x(x - 1) - (8x + 5)(x - 2)$

Compute each part:

$9x(x - 1) = 9x^2 - 9x$

$(8x + 5)(x - 2) = 8x(x) + 8x(-2) + 5(x) + 5(-2) = 8x^2 - 16x + 5x - 10 = 8x^2 - 11x - 10$

Now subtract:

$(9x^2 - 9x) - (8x^2 - 11x - 10) = 9x^2 - 9x - 8x^2 + 11x + 10 = x^2 + 2x + 10$

✔ Final Answer for #2: $\boxed{\frac{x^2 + 2x + 10}{(x - 2)(x - 1)}}$

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Problem 3:
$\frac{4x}{x + 1} - \frac{6x - 6}{2x + 2}$

Step 1: Notice $2x + 2 = 2(x + 1)$ → so LCD is $2(x + 1)$

Rewrite first fraction: $\frac{4x \cdot 2}{2(x + 1)} = \frac{8x}{2(x + 1)}$

Second fraction already has denominator $2(x + 1)$

Step 2: Subtract numerators:

$8x - (6x - 6) = 8x - 6x + 6 = 2x + 6$

So: $\frac{2x + 6}{2(x + 1)}$

Factor numerator: $2(x + 3)$

Cancel 2: $\frac{x + 3}{x + 1}$

✔ Final Answer for #3: $\boxed{\frac{x + 3}{x + 1}}$

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Problem 4:
$\frac{2x}{16x^2 - 4x} + \frac{4}{4x - 1}$

Step 1: Factor denominator of first fraction:
$16x^2 - 4x = 4x(4x - 1)$

So expression: $\frac{2x}{4x(4x - 1)} + \frac{4}{4x - 1}$

Simplify first fraction: cancel $x$? Wait — only if $x ≠ 0$, but we can reduce:

$\frac{2x}{4x(4x - 1)} = \frac{2}{4(4x - 1)} = \frac{1}{2(4x - 1)}$ after canceling $x$ and reducing 2/4 to 1/2.

Wait — let's do it carefully:

$\frac{2x}{4x(4x - 1)} = \frac{2}{4(4x - 1)} = \frac{1}{2(4x - 1)}$ ✔

Now add to $\frac{4}{4x - 1}$

Common denominator: $2(4x - 1)$

Rewrite second fraction: $\frac{4 \cdot 2}{2(4x - 1)} = \frac{8}{2(4x - 1)}$

Add: $\frac{1 + 8}{2(4x - 1)} = \frac{9}{2(4x - 1)}$

✔ Final Answer for #4: $\boxed{\frac{9}{2(4x - 1)}}$

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Problem 5:
$\frac{5x + 9}{x^2 - 2x + 1} - \frac{x + 15}{x - 1}$

Step 1: Factor $x^2 - 2x + 1 = (x - 1)^2$

Expression: $\frac{5x + 9}{(x - 1)^2} - \frac{x + 15}{x - 1}$

LCD = $(x - 1)^2$

Rewrite second fraction: $\frac{(x + 15)(x - 1)}{(x - 1)^2}$

Subtract numerators:

$(5x + 9) - (x + 15)(x - 1)$

First compute $(x + 15)(x - 1) = x^2 - x + 15x - 15 = x^2 + 14x - 15$

Now subtract:

$5x + 9 - (x^2 + 14x - 15) = 5x + 9 - x^2 - 14x + 15 = -x^2 - 9x + 24$

So: $\frac{-x^2 - 9x + 24}{(x - 1)^2}$

We can factor out negative: $-\frac{x^2 + 9x - 24}{(x - 1)^2}$ — but quadratic doesn’t factor nicely.

✔ Final Answer for #5: $\boxed{\frac{-x^2 - 9x + 24}{(x - 1)^2}}$

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Problem 6:
$2 - \frac{y}{y - 4}$

Write 2 as $\frac{2(y - 4)}{y - 4}$

Then: $\frac{2(y - 4) - y}{y - 4} = \frac{2y - 8 - y}{y - 4} = \frac{y - 8}{y - 4}$

✔ Final Answer for #6: $\boxed{\frac{y - 8}{y - 4}}$

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Problem 7:
$\frac{x}{2x - 2} - \frac{3x + 3}{2x - 1}$

Wait — denominators: $2x - 2 = 2(x - 1)$, and $2x - 1$ — they don’t match. No common factors.

LCD = $2(x - 1)(2x - 1)$

Rewrite both:

First: $\frac{x(2x - 1)}{2(x - 1)(2x - 1)}$

Second: $\frac{(3x + 3) \cdot 2(x - 1)}{2(x - 1)(2x - 1)}$ → wait, no! To get LCD, multiply numerator and denominator by what’s missing.

Actually:

First fraction needs to multiply top and bottom by $(2x - 1)$ → already did.

Second fraction: denominator is $2x - 1$, need to multiply by $2(x - 1)$ to get LCD.

So:

Numerator of second fraction becomes: $(3x + 3) \cdot 2(x - 1)$

But note: $3x + 3 = 3(x + 1)$, so:

Total numerator: $x(2x - 1) - 2(x - 1)(3x + 3)$

Compute:

First part: $x(2x - 1) = 2x^2 - x$

Second part: $2(x - 1)(3x + 3) = 2[3x(x) + 3x(-1) + (-1)(3x) + (-1)(3)]$ → better:

$(x - 1)(3x + 3) = 3x(x) + 3x(-1) + (-1)(3x) + (-1)(3) = 3x^2 - 3x - 3x - 3 = 3x^2 - 6x - 3$

Multiply by 2: $6x^2 - 12x - 6$

Now subtract:

$(2x^2 - x) - (6x^2 - 12x - 6) = 2x^2 - x - 6x^2 + 12x + 6 = -4x^2 + 11x + 6$

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

Check if numerator factors: $-4x^2 + 11x + 6$

Try factoring: multiply -4 * 6 = -24, find two numbers that multiply to -24 and add to 11 → 12 and -2

Split middle term: $-4x^2 + 12x - 2x + 6 = -4x(x - 3) -2(x - 3) = (-4x - 2)(x - 3)$ → or factor out -2: $-2(2x + 1)(x - 3)$

Denominator: $2(x - 1)(2x - 1)$ — no common factors.

✔ Final Answer for #7: $\boxed{\frac{-4x^2 + 11x + 6}{2(x - 1)(2x - 1)}}$ or $\boxed{\frac{-2(2x + 1)(x - 3)}{2(x - 1)(2x - 1)}} = \boxed{\frac{-(2x + 1)(x - 3)}{(x - 1)(2x - 1)}}$

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Problem 8:
$\frac{4x}{3x + 3} - \frac{x - 3}{2x + 2}$

Factor denominators:

$3x + 3 = 3(x + 1)$
$2x + 2 = 2(x + 1)$

LCD = $6(x + 1)$

Rewrite:

First: $\frac{4x \cdot 2}{6(x + 1)} = \frac{8x}{6(x + 1)}$

Second: $\frac{(x - 3) \cdot 3}{6(x + 1)} = \frac{3(x - 3)}{6(x + 1)}$

Subtract: $\frac{8x - 3(x - 3)}{6(x + 1)} = \frac{8x - 3x + 9}{6(x + 1)} = \frac{5x + 9}{6(x + 1)}$

✔ Final Answer for #8: $\boxed{\frac{5x + 9}{6(x + 1)}}$

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Problem 9:
$\frac{2x}{x^2 - 4x} + \frac{8}{4x - 16}$

Factor:

$x^2 - 4x = x(x - 4)$
$4x - 16 = 4(x - 4)$

LCD = $4x(x - 4)$

Rewrite:

First: $\frac{2x \cdot 4}{4x(x - 4)} = \frac{8x}{4x(x - 4)}$

Second: $\frac{8 \cdot x}{4x(x - 4)} = \frac{8x}{4x(x - 4)}$

Add: $\frac{8x + 8x}{4x(x - 4)} = \frac{16x}{4x(x - 4)}$

Cancel $4x$: $\frac{4}{x - 4}$

✔ Final Answer for #9: $\boxed{\frac{4}{x - 4}}$

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Problem 10:
$\frac{9}{10x + 5} - \frac{10x + 8}{2x + 1}$

Factor:

$10x + 5 = 5(2x + 1)$
So LCD = $5(2x + 1)$

Rewrite first fraction: already good.

Second fraction: multiply numerator and denominator by 5:

$\frac{5(10x + 8)}{5(2x + 1)}$

Now subtract:

$\frac{9 - 5(10x + 8)}{5(2x + 1)} = \frac{9 - 50x - 40}{5(2x + 1)} = \frac{-50x - 31}{5(2x + 1)}$

Can write as $-\frac{50x + 31}{5(2x + 1)}$

✔ Final Answer for #10: $\boxed{\frac{-50x - 31}{5(2x + 1)}}$

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## Final Answers:

1) $\boxed{\frac{4y - 14}{(y - 4)(y + 4)}}$
2) $\boxed{\frac{x^2 + 2x + 10}{(x - 2)(x - 1)}}$
3) $\boxed{\frac{x + 3}{x + 1}}$
4) $\boxed{\frac{9}{2(4x - 1)}}$
5) $\boxed{\frac{-x^2 - 9x + 24}{(x - 1)^2}}$
6) $\boxed{\frac{y - 8}{y - 4}}$
7) $\boxed{\frac{-(2x + 1)(x - 3)}{(x - 1)(2x - 1)}}$
8) $\boxed{\frac{5x + 9}{6(x + 1)}}$
9) $\boxed{\frac{4}{x - 4}}$
10) $\boxed{\frac{-50x - 31}{5(2x + 1)}}$