\(\textbf{1)}\) \( \displaystyle\frac{3x}{4y}\cdot \displaystyle\frac{2y^2}{5x} \) The answer is \( \displaystyle\frac{3y}{10} \)
\(\textbf{2)}\) \( \displaystyle\frac{3x^4 y^3 z}{4x^2 z^2}\cdot\displaystyle\frac{x^2 y^2 z^4}{9x^2} \) The answer is \( \displaystyle\frac{x^2 y^5 z^3}{12} \)
\(\textbf{3)}\) \( \displaystyle\frac{x^2+3x-4}{(x+4)(x+5)}\cdot \displaystyle\frac{x+5}{x-1} \) The answer is \( 1 \)
\(\textbf{4)}\) \( \displaystyle\frac{3x^4 y^3}{4z^3}\div \displaystyle\frac{x^2 y^2}{9z^2} \) The answer is \( \displaystyle\frac{27x^2 y}{4z} \)
\(\textbf{5)}\) \( \displaystyle\frac{x^2-1}{x+4}\div \displaystyle\frac{x-1}{x+2} \) The answer is \( \displaystyle\frac{(x+1)(x+2)}{(x+4)} \)
\(\textbf{6)}\) \( \displaystyle \displaystyle\frac{\displaystyle\frac{x^2-2x-8}{x+2}}{\displaystyle\frac{x^2-4}{x-4}} \) The answer is \( \displaystyle\frac{(x-4)^2}{(x+2)^2 (x-2)} \)
\(\textbf{7)}\) \( \displaystyle\frac{\displaystyle\frac{x^2-2x-15}{x^4-16}}{\displaystyle\frac{5-x}{x^2+4}} \) The answer is \( -\displaystyle\frac{x+3}{x^2-4} \)
\(\,\,\,\,\,\,\displaystyle\frac{\frac{x^2-2x-15}{x^4-16}}{\frac{5-x}{x^2+4}}\)
\(\,\,\,\,\,\,\frac{x^2-2x-15}{x^4-16}\div \frac{5-x}{x^2+4}\)
\(\,\,\,\,\,\,\frac{(x-5)(x+3)}{(x-2)(x+2)(x^2+4)}\div \frac{5-x}{x^2+4}\)
\(\,\,\,\,\,\,\frac{(x-5)(x+3)}{(x-2)(x+2)(x^2+4)}\cdot\frac{x^2+4}{5-x}\)
\(\,\,\,\,\,\,\frac{(x-5)(x+3)}{(x-2)(x+2)}\cdot\frac{1}{5-x}\)
\(\,\,\,\,\,\,\frac{(x-5)(x+3)}{(x-2)(x+2)}\cdot\frac{1}{-(x-5)}\)
\(\,\,\,\,\,\,-\frac{x+3}{(x-2)(x+2)}\)
\(\,\,\,\,\,\,\textbf{Restrictions: } x\neq\pm2,\; x\neq5\)
The answer is \(-\frac{x+3}{(x-2)(x+2)}\)
\(\,\,\,\,\,\,\displaystyle\frac{\frac{x^2-2x-15}{x^4-16}}{\frac{5-x}{x^2+4}}\)
\(\,\,\,\,\,\,\frac{x^2-2x-15}{x^4-16}\div \frac{5-x}{x^2+4}\)
\(\,\,\,\,\,\,\frac{(x-5)(x+3)}{(x-2)(x+2)(x^2+4)}\div \frac{5-x}{x^2+4}\)
\(\,\,\,\,\,\,\frac{(x-5)(x+3)}{(x-2)(x+2)(x^2+4)}\cdot\frac{x^2+4}{5-x}\)
\(\,\,\,\,\,\,\frac{(x-5)(x+3)}{(x-2)(x+2)}\cdot\frac{1}{5-x}\)
\(\,\,\,\,\,\,\frac{(x-5)(x+3)}{(x-2)(x+2)}\cdot\frac{1}{-(x-5)}\)
\(\,\,\,\,\,\,-\frac{x+3}{(x-2)(x+2)}\)
\(\,\,\,\,\,\,\textbf{Restrictions: } x\neq\pm2,\; x\neq5\)
The answer is \(-\frac{x+3}{(x-2)(x+2)}\)
See Related Pages\(\)
\(\bullet\text{ Rational Expression Calculator }\)
\(\,\,\,\,\,\,\,\,\text{(Symbolab.com)}\)
\(\bullet\text{ Ratios and Proportions}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{4}{3}=\frac{d-4}{12}…\)
\(\bullet\text{ Rational Expressions- Multiplying and Dividing}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{x^2+3x-4}{(x+4)(x+5)}\cdot \displaystyle\frac{x+5}{x-1}…\)
\(\bullet\text{ Rational Expressions- Adding and Subtracting}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{x-5}{x+3}+\frac{x+2}{x^2+5x+6}…\)
\(\bullet\text{ Direct, Inverse, and Joint Variation}\)
\(\,\,\,\,\,\,\,\,\)
\(…\)
\(\bullet\text{ Complex Fractions}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{\frac{x}{5}+\frac{1}{3}}{\frac{1}{5}-\frac{1}{6}}…\)
\(\bullet\text{ Partial Fraction Decomposition}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{8x+10}{x^2+2x}=\displaystyle\frac{5}{x} + \frac{3}{x+2}…\)