How to add rational expressions with different denominators

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ACT Math › How to add rational expressions with different denominators

Questions 1 - 5

Simplify the following expression:

$\frac{5}{7}+\frac{11}{14}$

$\frac{3}{2}$

$\frac{8}{7}$

$\frac{16}{7}$

$\frac{16}{21}$

Explanation

In order to add fractions, we must first make sure they have the same denominator.

So, we multiply $\frac{5}{7}$ by $\frac{2}{2}$ and get the following:

$\frac{5}{7}\left(\frac{2}{2}\right)+\frac{11}{14}$

$\frac{10}{14}+\frac{11}{14}$

Then, we add across the numerators and simplify:

$\frac{10}{14}+\frac{11}{14}=\frac{21}{14}=\frac{3}{2}$

Simplify the following $\frac{3x}{x-3} + \frac{7}{x-4}$

$\frac{3x^{2} - 5x -21}{x^{2}-7x+12}$

$\frac{3x^{2} -12x +21}{x^{2}-7x+12}$

$\frac{3x+7}{x^{2}-7x+12}$

$\frac{3x^{2}-5x-21}{x^{2}-12}$

$\frac{3x^{2}+7}{x^{2}-12}$

Explanation

Find the least common denominator between x-3 and x-4, which is (x-3)(x-4). Therefore, you have $\frac{3x(x-4)}{(x-3)(x-4)} + \frac{7(x-3)}{(x-4)(x-3)}$ . Multiplying the terms out equals $\frac{3x^{2}-12x+ 7x-21}{x^{2}-7x+12}$ . Combining like terms results in $\frac{3x^{2}-5x-21}{x^{2}-7x+12}$ .

Combine the following two expressions if possible.

$\frac{x+3}{x+7} + \frac{x^2}{4-x}$

$\frac{x^3 +6x^2 + x + 12}{-x^2-3x+28}$

$\frac{(x-3)^3}{-x^2-3x+28}$

$\frac{x^3 +3x^2 -7x - 22}{x^2-3x-28}$

$\frac{-x^3 +4x^2 + 8x -4}{x^2-3x-28}$

$\frac{2x^3 -5x^2 + 2x -30}{-x^2-3x+28}$

Explanation

For binomial expressions, it is often faster to simply FOIL them together to find a common trinomial than it is to look for individual least common denominators. Let's do that here:

$\frac{x+3}{x+7} + \frac{x^2}{4-x} = \frac{x+3}{x+7} * \frac{(4-x)}{(4-x)} + \frac{x^2}{4-x} * \frac{(x+7)}{(x+7)}$

FOIL and simplify.

$\frac{-x^2 + x + 12}{-x^2 -3x+28} + \frac{x^3 +7x^2}{-x^2 -3x+28}$

Combine numerators.

$\frac{-x^2 + x + 12}{-x^2 -3x+28} + \frac{x^3 +7x^2}{-x^2 -3x+28} = \frac{x^3 +6x^2 + x + 12}{-x^2-3x+28}$

Thus, our answer is $\frac{x^3 +6x^2 + x + 12}{-x^2-3x+28}$

Select the expression that is equivalent to

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

$\frac{x^3 + 3x^2 +x -9}{3x^2}$

$\frac{x^3 - x^2 -9}{3x^2}$

$\frac{x^3 + 3x^2 +x -9}{3x + x^2}$

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

$\frac{x^2 + 3x +x -9}{3x^2}$

Explanation

To add the two fractions, a common denominator must be found. With one-term denominators, it is easier to simply find the least common denominator between them and multiply each side to obtain it.

In this case, the least common denominator between and is . So the first fraction needs to be multiplied by and the second by :

$\frac{x^2 + 3x - 2}{3x} + \frac{x-3}{x^2} = \frac{x^2 + 3x - 2}{3x}* \left(\frac{x}{x}\right) + \frac{x-3}{x^2} * \left(\frac{3}{3}\right)$

$\frac{x^3 + 3x^2 - 2x}{3x^2} + \frac{3x-9}{3x^2}$

Now, we can add straight across, remembering to combine terms where we can.

$\frac{x^3 + 3x^2 - 2x}{3x^2} + \frac{3x-9}{3x^2} = \frac{x^3 +3x^2 -2x +3x - 9}{3x^2} = \frac{x^3 + 3x^2 +x -9}{3x^2}$

So, our simplified answer is $\frac{x^3 + 3x^2 +x -9}{3x^2}$

Simplify the following:

$\frac{x+3}{x-2}+\frac{x-2}{x+3}$

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

$\frac{2(x+3)(x-2)}{(x+3)(x-2)}$

Explanation

To simplify the following, a common denominator must be achieved. In this case, the first term must be multiplied by (x+2) in both the numerator and denominator and likewise with the second term with (x-3).

$\frac{(x+3)(x+3)}{(x-2)(x+3)}+\frac{(x-2)(x-2)}{(x+3)(x-2)}$

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

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