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\(\textbf{1)}\) \( \displaystyle\frac{5}{x}+\frac{x^2}{7} \)

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The answer is \( \displaystyle\frac{x^3+35}{7x} \)

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\(\,\,\,\,\,\displaystyle\frac{5}{x}+\frac{x^2}{7}\)


\(\,\,\,\,\,\displaystyle\frac{7}{7} \cdot \frac{5}{x}+\frac{x^2}{7} \cdot \frac{x}{x}\)


\(\,\,\,\,\,\displaystyle\frac{35}{7x}+\frac{x^3}{7x}\)


\(\,\,\,\,\,\displaystyle\frac{x^3+35}{7x}\)

\(\textbf{2)}\) \( \displaystyle\frac{x-5}{x+3}+\frac{x+2}{x^2+5x+6} \)

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The answer is \( \displaystyle\frac{x-4}{x+3} \)

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\(\,\,\,\,\,\displaystyle\frac{x-5}{x+3}+\frac{x+2}{x^2+5x+6}\)


\(\,\,\,\,\,\displaystyle\frac{x-5}{x+3}+\frac{x+2}{\left(x+2\right)\left(x+3\right)}\)


\(\,\,\,\,\,\displaystyle \frac{x+2}{x+2} \cdot \frac{x-5}{x+3}+\frac{x+2}{\left(x+2\right)\left(x+3\right)}\)


\(\,\,\,\,\,\displaystyle \frac{\left(x+2\right)\left(x-5\right)}{\left(x+2\right)\left(x+3\right)}+\frac{x+2}{\left(x+2\right)\left(x+3\right)}\)


\(\,\,\,\,\,\displaystyle \frac{x^2+5x+6}{\left(x+2\right)\left(x+3\right)}+\frac{x+2}{\left(x+2\right)\left(x+3\right)}\)

\(\,\,\,\,\,\displaystyle \frac{x^2+6x+8}{\left(x+2\right)\left(x+3\right)}\)


\(\,\,\,\,\,\displaystyle \frac{\left(x+2\right)\left(x-4\right)}{\left(x+2\right)\left(x+3\right)}\)


\(\,\,\,\,\,\displaystyle\frac{x-4}{x+3} \)

\(\textbf{3)}\) \( \displaystyle\frac{5}{x+3}+\frac{x}{x^2-9} \)

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The answer is \( \displaystyle\frac{3(2x-5)}{(x-3)(x+3)} \)

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\(\,\,\,\,\,\displaystyle\frac{5}{x+3}+\frac{x}{x^2-9}\)


\(\,\,\,\,\,\displaystyle\frac{5}{x+3}+\frac{x}{\left(x-3\right)\left(x+3\right)}\)


\(\,\,\,\,\,\displaystyle\frac{x-3}{x-3} \cdot \frac{5}{x+3}+\frac{x}{\left(x-3\right)\left(x+3\right)}\)


\(\,\,\,\,\,\displaystyle\frac{5x-15}{\left(x-3\right)\left(x+3\right)}+\frac{x}{\left(x-3\right)\left(x+3\right)}\)


\(\,\,\,\,\,\displaystyle\frac{6x-15}{\left(x-3\right)\left(x+3\right)}\)


\(\,\,\,\,\,\displaystyle\frac{3(2x-5)}{(x-3)(x+3)} \)

\(\textbf{4)}\) \( \displaystyle\frac{5}{x-2}-\frac{8}{2-x} \)

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The answer is \( \displaystyle\frac{13}{x-2} \)

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\(\,\,\,\,\,\displaystyle\frac{5}{x-2}-\frac{8}{2-x}\)


\(\,\,\,\,\,\displaystyle\frac{5}{x-2}-\frac{8}{-\left(x-2\right)}\)


\(\,\,\,\,\,\displaystyle\frac{5}{x-2}+\frac{8}{\left(x-2\right)}\)


\(\,\,\,\,\,\displaystyle\frac{13}{x-2} \)

\(\textbf{5)}\) \( \displaystyle\frac{1}{2x}+\frac{x}{3x}+\frac{3}{4x} \)

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The answer is \( \displaystyle\frac{4x+15}{12x} \)

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\(\,\,\,\,\,\displaystyle\frac{1}{2x}+\frac{x}{3x}+\frac{3}{4x}\)


\(\,\,\,\,\,\displaystyle\frac{1}{2x} \cdot\frac{6}{6} +\frac{x}{3x} \cdot\frac{4}{4} +\frac{3}{4x} \cdot\frac{3}{3} \)


\(\,\,\,\,\,\displaystyle\frac{6}{12x} +\frac{4x}{12x} +\frac{9}{12x} \)


\(\,\,\,\,\,\displaystyle\frac{6+4x+9}{12x}\)


\(\,\,\,\,\,\displaystyle\frac{4x+15}{12x} \)

\(\textbf{6)}\) \( \displaystyle\frac{5}{x+3}+4 \)

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The answer is \( \displaystyle\frac{4x+17}{x+3} \)

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\(\,\,\,\,\,\displaystyle\frac{5}{x+3}+4\)


\(\,\,\,\,\,\displaystyle\frac{5}{x+3}+\frac{4}{1} \cdot \frac{x+3}{x+3}\)


\(\,\,\,\,\,\displaystyle\frac{5}{x+3}+\frac{4x+12}{x+3}\)


\(\,\,\,\,\,\displaystyle\frac{4x+17}{x+3} \)

\(\textbf{7)}\) \( \displaystyle\frac{x}{x+5}+\frac{3x+1}{x+2} \)

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The answer is \( \displaystyle\frac{4x^2+18x+5}{(x+2)(x+5)} \)

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\(\,\,\,\,\,\displaystyle\frac{x}{x+5}+\frac{3x+1}{x+2}\)


\(\,\,\,\,\,\displaystyle\frac{x+2}{x+2} \cdot \frac{x}{x+5}+\frac{3x+1}{x+2} \cdot \frac{x+5}{x+5}\)


\(\,\,\,\,\,\displaystyle\frac{x^2+2x}{\left(x+2\right)\left(x+5\right)}+\frac{\left(3x+1\right)\left(x+5\right)}{\left(x+2\right)\left(x+5\right)}\)


\(\,\,\,\,\,\displaystyle\frac{x^2+2x}{\left(x+2\right)\left(x+5\right)}+\frac{3x^2+16x+5}{\left(x+2\right)\left(x+5\right)}\)


\(\,\,\,\,\,\displaystyle\frac{4x^2+18x+5}{\left(x+2\right)\left(x+5\right)}\)

\(\textbf{8)}\) \( \displaystyle\frac{2x+5}{x^2+3x+2}+\frac{x-1}{x^2+7x+10} \)

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The answer is \( \displaystyle\frac{3x^2+15x+24}{\left(x+5\right)\left(x+1\right)\left(x+2\right)} \)

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\(\,\,\,\,\,\displaystyle\frac{2x+5}{x^2+3x+2}+\frac{x-1}{x^2+7x+10}\)


\(\,\,\,\,\,\displaystyle\frac{2x+5}{\left(x+1\right)\left(x+2\right)}+\frac{x-1}{\left(x+5\right)\left(x+2\right)}\)


\(\,\,\,\,\,\displaystyle\frac{x+5}{x+5} \cdot \frac{2x+5}{\left(x+1\right)\left(x+2\right)}+\frac{x-1}{\left(x+5\right)\left(x+2\right)} \cdot \frac{x+1}{x+1}\)


\(\,\,\,\,\,\displaystyle\frac{\left(x+5\right)\left(2x+5\right)}{\left(x+5\right)\left(x+1\right)\left(x+2\right)}+\frac{\left(x-1\right)\left(x+1\right)}{\left(x+5\right)\left(x+2\right)\left(x+1\right)}\)


\(\,\,\,\,\,\displaystyle\frac{2x^2+15x+25}{\left(x+5\right)\left(x+1\right)\left(x+2\right)}+\frac{x^2-1}{\left(x+5\right)\left(x+2\right)\left(x+1\right)}\)


\(\,\,\,\,\,\displaystyle\frac{3x^2+15x+24}{\left(x+5\right)\left(x+1\right)\left(x+2\right)}\)


\(\,\,\,\,\,\displaystyle\frac{3x^2+15x+24}{\left(x+5\right)\left(x+1\right)\left(x+2\right)}\)

\(\textbf{9)}\) \( \displaystyle\frac{x+7}{x^2+4x+4}+\frac{x+1}{x^2-4} \)

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The answer is \( \displaystyle\frac{2(x-8)}{(x-2)(x+2)^2} \)

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\(\,\,\,\,\,\displaystyle\frac{x+7}{x^2+4x+4}+\frac{x+1}{x^2-4}\)


\(\,\,\,\,\,\displaystyle\frac{x+7}{\left(x+2\right)\left(x+2\right)}+\frac{x+1}{\left(x+2\right)\left(x-2\right)}\)


\(\,\,\,\,\,\displaystyle\frac{x-2}{x-2} \cdot \frac{x+7}{\left(x+2\right)\left(x+2\right)}+\frac{x+1}{\left(x+2\right)\left(x-2\right)} \cdot \frac{x+2}{x+2}\)


\(\,\,\,\,\,\displaystyle\frac{x^2+5x-14}{\left(x-2\right)\left(x+2\right)^2}+\frac{x^2+3x+2}{\left(x-2\right)\left(x+2\right)^2}\)


\(\,\,\,\,\,\displaystyle\frac{2x^2+8x-12}{\left(x-2\right)\left(x+2\right)^2}+\frac{x^2+3x+2}{\left(x-2\right)\left(x+2\right)^2}\)

\(\textbf{10)}\) \( \displaystyle\frac{2x}{x+2}+\frac{4}{x+2} \)

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The answer is \( 2 \)

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\(\,\,\,\,\,\displaystyle\frac{2x}{x+2}+\frac{4}{x+2}\)


\(\,\,\,\,\,\displaystyle\frac{2x+4}{x+2}\)


\(\,\,\,\,\,\displaystyle\frac{2\left(x+2\right)}{x+2}\)


\(\,\,\,\,\,2 \)

\(\textbf{11)}\) \( \displaystyle\frac{3x+1}{\left(x+2\right)^2}+\frac{3}{x+2} \)

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The answer is \( \displaystyle\frac{6x+7}{(x+2)^2} \)

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\(\,\,\,\,\,\displaystyle\frac{3x+1}{(x+2)^2}+\frac{3}{x+2}\)


\(\,\,\,\,\,\displaystyle\frac{3x+1}{(x+2)^2}+\frac{3}{x+2} \cdot \frac{x+2}{x+2}\)


\(\,\,\,\,\,\displaystyle\frac{3x+1}{\left(x+2\right)^2}+\frac{3x+6}{\left(x+2\right)^2}\)


\(\,\,\,\,\,\displaystyle\frac{6x+7}{(x+2)^2} \)

\(\textbf{12)}\) \( \displaystyle\frac{4}{x}+\frac{3}{x+1} \)

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The answer is \( \displaystyle\frac{7x+4}{x(x+1)} \)

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\(\,\,\,\,\,\displaystyle\frac{4}{x}+\frac{3}{x+1}\)


\(\,\,\,\,\,\displaystyle\frac{x+1}{x+1} \cdot \frac{4}{x}+\frac{3}{x+1} \cdot \frac{x}{x}\)


\(\,\,\,\,\,\displaystyle\frac{4x+4}{x\left(x+1\right)}+\frac{3x}{x\left(x+1\right)}\)


\(\,\,\,\,\,\displaystyle\frac{7x+4}{x(x+1)} \)

See Related Pages\(\)

\(\bullet\text{ Rational Expression Calculator }\)
\(\,\,\,\,\,\,\,\,\text{(Symbolab.com)}\)

\(\bullet\text{ Ratios and Proportions}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{4}{3}=\frac{d-4}{12}…\)

\(\bullet\text{ Rational Expressions- Multiplying and Dividing}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{x^2+3x-4}{(x+4)(x+5)}\cdot \displaystyle\frac{x+5}{x-1}…\)

\(\bullet\text{ Rational Expressions- Adding and Subtracting}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{x-5}{x+3}+\frac{x+2}{x^2+5x+6}…\)

\(\bullet\text{ Direct, Inverse, and Joint Variation}\)
\(\,\,\,\,\,\,\,\,\) \(…\)

\(\bullet\text{ Complex Fractions}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{\frac{x}{5}+\frac{1}{3}}{\frac{1}{5}-\frac{1}{6}}…\)

\(\bullet\text{ Partial Fraction Decomposition}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{8x+10}{x^2+2x}=\displaystyle\frac{5}{x} + \frac{3}{x+2}…\)

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