Addition and subtraction of rational expressions

Example: Simplify, $\dfrac{4x-5}{8x+7}+\dfrac{11x}{8x+7}$.

Solution:

The denominators of the two rational expressions are the same. Therefore,

$\dfrac{4x-5}{8x+7}+\dfrac{11x}{8x+7}=\dfrac{4x-5+11x}{8x+7}$.

Combining the $x$ terms of the numerator,

$\hphantom{00000}=\dfrac{15x-5}{8x+7}$.

Factoring the numerator,

$\hphantom{00000}=\boxed{\dfrac{5(3x-1)}{8x+7}}$.

Example 1: Find the least common multiple of the numbers $12,18$ and $30$.

Solution: The LCM is the smallest integer that is divisible by all the given integers, $12,18$ and $30$. Find an integer that divides all three numbers evenly, without a remainder. Start with a small number, then work upward, dividing until no integer divides all the numbers further.

$\begin{align*} 2&\vert \underline{12,18,30}\\ 3&\vert \underline{\hphantom{0}6,\hphantom{0}9,15}\\ &\hphantom{00}2,\hphantom{0}3,\hphantom{0}5\\ \end{align*}$

Now, multiply all the numbers on the left of the vertical line and all the numbers below the last horizontal line. The product is the least common multiple.

$2\cdot3\cdot2\cdot3\cdot5=\boxed{180}$

Example 2: Find the least common multiple of $7u^4$ and $5u^6$

Solution:

First find the least common multiple of the constants, $7$ and $5$. There is no common factor between these two numbers, so just multiply them: $7\cdot 5=35$. This is the LCM of the constants.

Next, find the LCM of the variables. We have $u^4$ and $u^6$; take the one with the larger exponent. That is, the LCM is $u^6$.

Multiply the two LCMs together: $\boxed{35u^6}$. This is the LCM of the two expressions.

Example 3: Find the least common denominator of $\dfrac{7x}{2x-6}$ and $\dfrac{5x+1}{8x-24}$

Solution:

The denominators are $2x-6$ and $8x-24$. Factoring these denominators, we get

$2(x-3)$ and $8(x-3)$.

For the constants $2$ and $8$, the LCM is $8$. For the variable factors, $x-3$ appears in both denominators, so their LCM is $x-3$.

Multiplying the two LCMs gives the LCM of the two denominators.

So the answer is LCD $=8(x-3)$.

Example 4: Subtract and simplify: $\dfrac{8x+2}{11x-3}-\dfrac{2}{3-11x}$

Solution:

We can make the two denominators the same by factoring a $-1$ out of either denominator. Factoring a $-1$ out of the second one,

$\dfrac{8x+2}{11x-3}-\dfrac{2}{3-11x}=\dfrac{8x+2}{11x-3}-\dfrac{2}{-(11x-3)}$

$\hphantom{00000}=\dfrac{8x+2}{11x-3}+\dfrac{2}{11x-3}$

Both now have the same denominator, so put it as the common denominator and add the numerators:

$\hphantom{00000}=\dfrac{8x+2+2}{11x-3}$

$\hphantom{00000}=\boxed{\dfrac{8x+4}{11x-3}}$

Example 5: Add and simplify as much as possible, $\dfrac{2}{2x^2-x-21}+\dfrac{3}{3x+9}$

Solution:

Factoring the denominators $2x^2-x-21$ and $3x+9$, we get

$2x^2-x-21=(x+3)(2x-7)$

and

$3x+9=3(x+3)$

Now,

$\dfrac{2}{2x^2-x-21}+\dfrac{3}{3x+9}=\dfrac{2}{(x+3)(2x-7)}+\dfrac{3}{3(x+3)}$

$\hphantom{00000}=\dfrac{2}{(x+3)(2x-7)}+\dfrac{1}{x+3}$

The LCD of the two rational expressions is $(x+3)(2x-7)$. With this LCD, we can write

$\hphantom{00000}=\dfrac{2}{(x+3)(2x-7)}+\dfrac{2x-7}{(x+3)(2x-7)}$

$\hphantom{00000}=\dfrac{2+2x-7}{(x+3)(2x-7)}$

$\hphantom{00000}=\boxed{\dfrac{2x-5}{(x+3)(2x-7)}}$