Adding and Subtracting Rational Expressions - Free Printable

Let’s solve each problem one by one. We’ll simplify each expression by finding a common denominator, combining the fractions, and simplifying if possible.

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Problem 26:
\[
\frac{4}{4x - 1} + \frac{8x - 15}{4x^2 + 11x - 3}
\]

First, factor the quadratic in the second denominator:
\[
4x^2 + 11x - 3
\]
We look for two numbers that multiply to \(4 \cdot (-3) = -12\) and add to 11 → those are 12 and -1.

So:
\[
4x^2 + 12x - x - 3 = 4x(x + 3) -1(x + 3) = (4x - 1)(x + 3)
\]

Now rewrite the expression:
\[
\frac{4}{4x - 1} + \frac{8x - 15}{(4x - 1)(x + 3)}
\]

Common denominator is \((4x - 1)(x + 3)\). Multiply first fraction by \(\frac{x+3}{x+3}\):

\[
\frac{4(x + 3)}{(4x - 1)(x + 3)} + \frac{8x - 15}{(4x - 1)(x + 3)} = \frac{4x + 12 + 8x - 15}{(4x - 1)(x + 3)} = \frac{12x - 3}{(4x - 1)(x + 3)}
\]

Factor numerator: \(12x - 3 = 3(4x - 1)\)

So:
\[
\frac{3(4x - 1)}{(4x - 1)(x + 3)} = \frac{3}{x + 3}, \quad \text{as long as } 4x - 1 \ne 0
\]

✔ Final Answer for 26: \(\boxed{\frac{3}{x + 3}}\)

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Problem 27:
\[
\frac{9}{2y + 3} - \frac{5}{4y}
\]

Denominators: \(2y + 3\) and \(4y\). They have no common factors, so LCD is \(4y(2y + 3)\)

Rewrite both fractions:

First term: \(\frac{9}{2y + 3} \cdot \frac{4y}{4y} = \frac{36y}{4y(2y + 3)}\)

Second term: \(\frac{5}{4y} \cdot \frac{2y + 3}{2y + 3} = \frac{5(2y + 3)}{4y(2y + 3)} = \frac{10y + 15}{4y(2y + 3)}\)

Now subtract:

\[
\frac{36y - (10y + 15)}{4y(2y + 3)} = \frac{36y - 10y - 15}{4y(2y + 3)} = \frac{26y - 15}{4y(2y + 3)}
\]

Can’t simplify further.

✔ Final Answer for 27: \(\boxed{\frac{26y - 15}{4y(2y + 3)}}\)

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Problem 28:
\[
\frac{4}{5x + 7} + \frac{7}{6x}
\]

Denominators: \(5x + 7\) and \(6x\) — no common factors. LCD = \(6x(5x + 7)\)

First term: \(\frac{4}{5x + 7} \cdot \frac{6x}{6x} = \frac{24x}{6x(5x + 7)}\)

Second term: \(\frac{7}{6x} \cdot \frac{5x + 7}{5x + 7} = \frac{7(5x + 7)}{6x(5x + 7)} = \frac{35x + 49}{6x(5x + 7)}\)

Add them:

\[
\frac{24x + 35x + 49}{6x(5x + 7)} = \frac{59x + 49}{6x(5x + 7)}
\]

No simplification possible.

✔ Final Answer for 28: \(\boxed{\frac{59x + 49}{6x(5x + 7)}}\)

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Problem 29:
\[
\frac{1}{3x^2 - 8x} + \frac{6}{8 - 3x}
\]

First, factor denominators.

\(3x^2 - 8x = x(3x - 8)\)

Note: \(8 - 3x = -(3x - 8)\), so we can write:

\[
\frac{1}{x(3x - 8)} + \frac{6}{-(3x - 8)} = \frac{1}{x(3x - 8)} - \frac{6}{3x - 8}
\]

LCD is \(x(3x - 8)\)

Second term becomes: \(\frac{6}{3x - 8} \cdot \frac{x}{x} = \frac{6x}{x(3x - 8)}\)

So:

\[
\frac{1 - 6x}{x(3x - 8)} = \frac{-6x + 1}{x(3x - 8)}
\]

Or written as:

\[
\frac{1 - 6x}{x(3x - 8)}
\]

We can also factor out negative from numerator and denominator if desired, but it's fine as is.

Alternatively, note that \(3x - 8 = -(8 - 3x)\), but since we already simplified, let’s leave it.

✔ Final Answer for 29: \(\boxed{\frac{1 - 6x}{x(3x - 8)}}\)

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Problem 30:
\[
\frac{x + 4}{3x^2 - 4x} - \frac{6}{4 - 3x}
\]

Factor first denominator: \(3x^2 - 4x = x(3x - 4)\)

Note: \(4 - 3x = -(3x - 4)\), so:

\[
\frac{x + 4}{x(3x - 4)} - \frac{6}{-(3x - 4)} = \frac{x + 4}{x(3x - 4)} + \frac{6}{3x - 4}
\]

LCD is \(x(3x - 4)\)

Second term: \(\frac{6}{3x - 4} \cdot \frac{x}{x} = \frac{6x}{x(3x - 4)}\)

Now add:

\[
\frac{x + 4 + 6x}{x(3x - 4)} = \frac{7x + 4}{x(3x - 4)}
\]

No simplification.

✔ Final Answer for 30: \(\boxed{\frac{7x + 4}{x(3x - 4)}}\)

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Final Answers:

26. $\boxed{\frac{3}{x + 3}}$
27. $\boxed{\frac{26y - 15}{4y(2y + 3)}}$
28. $\boxed{\frac{59x + 49}{6x(5x + 7)}}$
29. $\boxed{\frac{1 - 6x}{x(3x - 8)}}$
30. $\boxed{\frac{7x + 4}{x(3x - 4)}}$