PreCal 11 – Adding and Subtracting Rational expressions

This week we learnt and practiced adding and subtracting rational expressions

Adding and subtracting rational expressions works a lot like normal integer fractions except you must state restrictions and be aware of variables.

With both, you get a common denominator with the lowest common multiple of the denominator.

Ex.

$\frac{6x}{4xy^2} + \frac{2}{x^3}$

$4xy^2 -> 2 \cdot 2 \cdot x \cdot y \cdot y$

$x^3 -> x \cdot x \cdot x$

LCD -> $4x^3y^2$

With adding, you must get the common denominator then add the numerators.

Ex.

$\frac{3x}{4x^2} + \frac{x^2}{3}$

$\frac{3x(3)}{4x^2(3)} + \frac{x^2(4x^2)}{3(4x^2)}$

$\frac{9x}{12x^2} + \frac{4x^4}{12x^2}$

$\frac{9x + 4x^4}{12x^2}$

$\frac{4x^3 + 9}{12x}$

$x \ne 0$

With subtracting, it’s the same as adding but to make it easier, distribute the negative sign into the next fraction.

Ex.

$\frac{a^2}{(3 + a)} - \frac{3a}{2a}$

$\frac{a^2(2a)}{(3 + a)(2a)} - \frac{3a(3 + a)}{2a(3 + a)}$

$\frac{2a^3}{(6a + 2a^2} - \frac{(9a + 3a^2)}{(6a + 2a^2)}$

$\frac{2a^3}{(6a + 2a^2} + \frac{(-9a - 3a^2)}{(6a + 2a^2)}$

$\frac{2a^3 - 3a^2 - 9a}{6a + 2a^2}$

$\frac{2a^2 - 3a - 9}{6 + 2a}$

$x \ne 0, -3, -\frac{a^2}{3}$