Simplifying algebraic fractions

Simplify $\frac{12a^{3}b}{18a{b}^2}\backslash frac\{12a^3\; b\}\{18ab^2\}$.

Solution Conditions: $a\mathrm{\ne }0a\; \backslash ne\; 0$, $b\mathrm{\ne }0b\; \backslash ne\; 0$. The highest common factor of the two terms is $6ab6ab$:

$$\frac{12a^{3}b}{18a{b}^2}=\frac{2a^2\times 6ab}{3b\times 6ab}=\frac{2a^2}{3b},a\mathrm{\ne }0,b\mathrm{\ne }0.\backslash frac\{12a^3\; b\}\{18ab^2\}\; =\; \backslash frac\{2a^2\; \backslash times\; 6ab\}\{3b\; \backslash times\; 6ab\}\; =\; \backslash frac\{2a^2\}\{3b\},\; \backslash qquad\; a\; \backslash ne\; 0,\backslash \; b\; \backslash ne\; 0.$$

Simplify $\frac{x^2-9}{x^2+5x+6}\backslash frac\{x^2\; -\; 9\}\{x^2\; +\; 5x\; +\; 6\}$.

Solution The denominator is $(x+2)(x+3)(x\; +\; 2)(x\; +\; 3)$, so the conditions are $x\mathrm{\ne }-2x\; \backslash ne\; -2$ and $x\mathrm{\ne }-3x\; \backslash ne\; -3$. The numerator is a difference of two squares.

$$\frac{x^2-9}{x^2+5x+6}=\frac{(x-3)\cancel{(x+3)}}{(x+2)\cancel{(x+3)}}=\frac{x-3}{x+2},x\mathrm{\ne }-2,x\mathrm{\ne }-3.\backslash frac\{x^2\; -\; 9\}\{x^2\; +\; 5x\; +\; 6\}\; =\; \backslash frac\{(x\; -\; 3)\backslash cancel\{(x\; +\; 3)\}\}\{(x\; +\; 2)\backslash cancel\{(x\; +\; 3)\}\}\; =\; \backslash frac\{x\; -\; 3\}\{x\; +\; 2\},\; \backslash qquad\; x\; \backslash ne\; -2,\backslash \; x\; \backslash ne\; -3.$$

Check with $x=1x\; =\; 1$: the original gives $\frac{1-9}{1+5+6}=\frac{-8}{12}=-\frac{2}{3}\backslash frac\{1\; -\; 9\}\{1\; +\; 5\; +\; 6\}\; =\; \backslash frac\{-8\}\{12\}\; =\; -\backslash frac\{2\}\{3\}$, and the answer gives $\frac{-2}{3}\backslash frac\{-2\}\{3\}$. ✓ Note that $\frac{x-3}{x+2}\backslash frac\{x\; -\; 3\}\{x\; +\; 2\}$ on its own would allow $x=-3x\; =\; -3$, but the original fraction does not exist there, so the condition $x\mathrm{\ne }-3x\; \backslash ne\; -3$ must be kept.

Simplify $\frac{5-x}{x^2-25}\backslash frac\{5\; -\; x\}\{x^2\; -\; 25\}$.

Solution Conditions: $x\mathrm{\ne }\pm 5x\; \backslash ne\; \backslash pm\; 5$. Write $5-x=-(x-5)5\; -\; x\; =\; -(x\; -\; 5)$, so the numerator and denominator share the factor $x-5x\; -\; 5$:

$$\frac{5-x}{x^2-25}=\frac{-(x-5)}{(x-5)(x+5)}=-\frac{1}{x+5},x\mathrm{\ne }\pm 5.\backslash frac\{5\; -\; x\}\{x^2\; -\; 25\}\; =\; \backslash frac\{-(x\; -\; 5)\}\{(x\; -\; 5)(x\; +\; 5)\}\; =\; -\backslash frac\{1\}\{x\; +\; 5\},\; \backslash qquad\; x\; \backslash ne\; \backslash pm\; 5.$$

Simplify (a) $\frac{2x^2+5x-3}{4x^2-1}\backslash frac\{2x^2\; +\; 5x\; -\; 3\}\{4x^2\; -\; 1\}$ (b) $\frac{ax+ay+2x+2y}{a^2-4}\backslash frac\{ax\; +\; ay\; +\; 2x\; +\; 2y\}\{a^2\; -\; 4\}$.

Solution

(a) $4x^2-1=(2x-1)(2x+1)4x^2\; -\; 1\; =\; (2x\; -\; 1)(2x\; +\; 1)$, so $x\mathrm{\ne }\pm \frac{1}{2}x\; \backslash ne\; \backslash pm\backslash frac\{1\}\{2\}$. For the numerator, $ac=-6ac\; =\; -6$ and $6+(-1)=56\; +\; (-1)\; =\; 5$: $2x^2+6x-x-3=(2x-1)(x+3)2x^2\; +\; 6x\; -\; x\; -\; 3\; =\; (2x\; -\; 1)(x\; +\; 3)$. So

$$\frac{(2x-1)(x+3)}{(2x-1)(2x+1)}=\frac{x+3}{2x+1},x\mathrm{\ne }\pm \frac{1}{2}\mathrm{.}\backslash frac\{(2x\; -\; 1)(x\; +\; 3)\}\{(2x\; -\; 1)(2x\; +\; 1)\}\; =\; \backslash frac\{x\; +\; 3\}\{2x\; +\; 1\},\; \backslash qquad\; x\; \backslash ne\; \backslash pm\backslash tfrac\{1\}\{2\}.$$

(b) Group the numerator: $a(x+y)+2(x+y)=(x+y)(a+2)a(x\; +\; y)\; +\; 2(x\; +\; y)\; =\; (x\; +\; y)(a\; +\; 2)$. The denominator is $(a-2)(a+2)(a\; -\; 2)(a\; +\; 2)$, so $a\mathrm{\ne }\pm 2a\; \backslash ne\; \backslash pm\; 2$.

$$\frac{(x+y)(a+2)}{(a-2)(a+2)}=\frac{x+y}{a-2},a\mathrm{\ne }\pm 2.\backslash frac\{(x\; +\; y)(a\; +\; 2)\}\{(a\; -\; 2)(a\; +\; 2)\}\; =\; \backslash frac\{x\; +\; y\}\{a\; -\; 2\},\; \backslash qquad\; a\; \backslash ne\; \backslash pm\; 2.$$