Multiplying and dividing algebraic fractions

Simplify $\frac{6x^3}{5y}\times \frac{10y^2}{9x}\backslash frac\{6x^3\}\{5y\}\; \backslash times\; \backslash frac\{10y^2\}\{9x\}$.

Solution Conditions: $x\mathrm{\ne }0x\; \backslash ne\; 0$, $y\mathrm{\ne }0y\; \backslash ne\; 0$. Cancel number factors and powers across the two fractions:

$$\frac{6x^3}{5y}\times \frac{10y^2}{9x}=\frac{6\times 10}{5\times 9}\times \frac{x^3}{x}\times \frac{y^2}{y}=\frac{4}{3}\times x^2\times y=\frac{4x^{2}y}{3}\mathrm{.}\backslash frac\{6x^3\}\{5y\}\; \backslash times\; \backslash frac\{10y^2\}\{9x\}\; =\; \backslash frac\{6\; \backslash times\; 10\}\{5\; \backslash times\; 9\}\; \backslash times\; \backslash frac\{x^3\}\{x\}\; \backslash times\; \backslash frac\{y^2\}\{y\}\; =\; \backslash frac\{4\}\{3\}\; \backslash times\; x^2\; \backslash times\; y\; =\; \backslash frac\{4x^2y\}\{3\}.$$

Simplify $\frac{x^2-4}{x^2+x}\times \frac{x+1}{x-2}\backslash frac\{x^2\; -\; 4\}\{x^2\; +\; x\}\; \backslash times\; \backslash frac\{x\; +\; 1\}\{x\; -\; 2\}$.

Solution Denominators: $x^2+x=x(x+1)x^2\; +\; x\; =\; x(x\; +\; 1)$ and $x-2x\; -\; 2$, so $x\mathrm{\ne }0,-1,2x\; \backslash ne\; 0,\backslash \; -1,\backslash \; 2$.

$$\frac{\cancel{(x-2)}(x+2)}{x\cancel{(x+1)}}\times \frac{\cancel{x+1}}{\cancel{x-2}}=\frac{x+2}{x},x\mathrm{\ne }0,-1,2.\backslash frac\{\backslash cancel\{(x\; -\; 2)\}(x\; +\; 2)\}\{x\backslash cancel\{(x\; +\; 1)\}\}\; \backslash times\; \backslash frac\{\backslash cancel\{x\; +\; 1\}\}\{\backslash cancel\{x\; -\; 2\}\}\; =\; \backslash frac\{x\; +\; 2\}\{x\},\; \backslash qquad\; x\; \backslash ne\; 0,\backslash \; -1,\backslash \; 2.$$

Simplify $\frac{x^2-9}{2x+4}÷\frac{x-3}{x^2+2x}\backslash frac\{x^2\; -\; 9\}\{2x\; +\; 4\}\; \backslash div\; \backslash frac\{x\; -\; 3\}\{x^2\; +\; 2x\}$.

Solution Factorise: $x^2-9=(x-3)(x+3)x^2\; -\; 9\; =\; (x\; -\; 3)(x\; +\; 3)$, $2x+4=2(x+2)2x\; +\; 4\; =\; 2(x\; +\; 2)$, $x^2+2x=x(x+2)x^2\; +\; 2x\; =\; x(x\; +\; 2)$.

Conditions: $x\mathrm{\ne }-2x\; \backslash ne\; -2$ and $x\mathrm{\ne }0x\; \backslash ne\; 0$ (denominators), and $x\mathrm{\ne }3x\; \backslash ne\; 3$ (the divisor's numerator cannot be 0).

$$\frac{(x-3)(x+3)}{2(x+2)}\times \frac{x(x+2)}{x-3}=\frac{x(x+3)}{2},x\mathrm{\ne }-2,0,3.\backslash frac\{(x\; -\; 3)(x\; +\; 3)\}\{2(x\; +\; 2)\}\; \backslash times\; \backslash frac\{x(x\; +\; 2)\}\{x\; -\; 3\}\; =\; \backslash frac\{x(x\; +\; 3)\}\{2\},\; \backslash qquad\; x\; \backslash ne\; -2,\backslash \; 0,\backslash \; 3.$$

A rectangle has length $\frac{x^2-1}{x}\backslash frac\{x^2\; -\; 1\}\{x\}$ metres and width $\frac{2x}{x+1}\backslash frac\{2x\}\{x\; +\; 1\}$ metres, where $x>1x\; \backslash gt\; 1$. Find its area in simplest form, and its value when $x=5x\; =\; 5$.

Solution

$$\text{Area}=\frac{(x-1)(x+1)}{x}\times \frac{2x}{x+1}=2(x-1){\mathrm{m}}^2\mathrm{.}\backslash text\{Area\}\; =\; \backslash frac\{(x\; -\; 1)(x\; +\; 1)\}\{x\}\; \backslash times\; \backslash frac\{2x\}\{x\; +\; 1\}\; =\; 2(x\; -\; 1)\backslash \; \backslash mathrm\{m^2\}.$$

When $x=5x\; =\; 5$: the length is $\frac{24}{5}=4.8\backslash frac\{24\}\{5\}\; =\; 4.8$ m, the width is $\frac{10}{6}=\frac{5}{3}\backslash frac\{10\}\{6\}\; =\; \backslash frac\{5\}\{3\}$ m, and the area is $2\times 4=8{\mathrm{m}}^{2}2\; \backslash times\; 4\; =\; 8\backslash \; \backslash mathrm\{m^2\}$. Check: $4.8\times \frac{5}{3}=84.8\; \backslash times\; \backslash frac\{5\}\{3\}\; =\; 8$. ✓

Simplify ${\displaystyle \frac{1+\frac{1}{x}}{1-\frac{1}{x^2}}}\backslash dfrac\{1\; +\; \backslash frac\{1\}\{x\}\}\{1\; -\; \backslash frac\{1\}\{x^2\}\}$.

Solution Write the top and the bottom as single fractions, then divide:

$$\frac{\frac{x+1}{x}}{\frac{x^2-1}{x^2}}=\frac{x+1}{x}\times \frac{x^2}{(x-1)(x+1)}=\frac{x}{x-1}\mathrm{.}\backslash frac\{\backslash frac\{x\; +\; 1\}\{x\}\}\{\backslash frac\{x^2\; -\; 1\}\{x^2\}\}\; =\; \backslash frac\{x\; +\; 1\}\{x\}\; \backslash times\; \backslash frac\{x^2\}\{(x\; -\; 1)(x\; +\; 1)\}\; =\; \backslash frac\{x\}\{x\; -\; 1\}.$$

Conditions: $x\mathrm{\ne }0x\; \backslash ne\; 0$ (from $\frac{1}{x}\backslash frac\{1\}\{x\}$), and the bottom $1-\frac{1}{x^2}\mathrm{\ne }01\; -\; \backslash frac\{1\}\{x^2\}\; \backslash ne\; 0$, so $x\mathrm{\ne }\pm 1x\; \backslash ne\; \backslash pm\; 1$. Answer: $\frac{x}{x-1}\backslash frac\{x\}\{x\; -\; 1\}$, $x\mathrm{\ne }0,\pm 1x\; \backslash ne\; 0,\backslash \; \backslash pm\; 1$.