When you add any fractions, you need to find a common denominator. In this case they have no common factors so you just multiply them together. First multiply the first fraction by [tex] \frac{x - 2}{x - 2} [/tex] which gets you [tex] \frac{3x - 6}{ x^{2} - 2x} [/tex]. Then multiply the second fraction by [tex] \frac{x}{x} [/tex] to get [tex] \frac{5x}{ x^{2} - 2x} [/tex]. Now you can add them together to get [tex] \frac{-2x - 6}{ x^{2} - 2x} [/tex].
(b) [tex] \sqrt{18} = \sqrt{9 * 2} = 3 \sqrt{2} [/tex] [tex] \sqrt{72} = \sqrt{36 * 2} = 6 \sqrt{2}[/tex] now you can pull out root 2 and get [tex] \sqrt{2}(3 - 1 + 6) [/tex] which equals [tex]8 \sqrt{2} [/tex]
a) We need a common denominator to be able to subtract the fractions, so by multiplying both the numerator and the denominator of each fraction by the denominator of the other. [tex] \frac{x + 2}{x + 2} ( \frac{3}{x}) - \frac{x}{x} ( \frac{5}{x + 2} )[/tex] Then simplify the fraction: [tex] \frac{6 + 3x}{x^{2} + 2x} - \frac{5x}{x^{2} + 2x} [/tex] Then you can just subtract the nominators from each other, but leave the denominator as it is [tex] \frac{(6 + 3x) - 5x}{x^{2} + 2x} [/tex] to get: [tex] \frac{6 - 2x}{x^{2} + 2x} [/tex] and then you can simplify to get the final answer: [tex] \frac{2(3 - x)}{x(x + 2)}[/tex]
b) you need to have the number inside the root the same to add or subtract: [tex] \sqrt{2(9)} - \sqrt{2} + \sqrt{2(36)} [/tex] Then you can 'take out' the numbers that square root easily: [tex] 3\sqrt{2} - \sqrt{2} + 6\sqrt{2} [/tex] now you can just add and subtract using the whole numbers outside the root of 2 ( since 2 is a surd): [tex]8\sqrt{2} [/tex]
