Before you can add the two numerators from the fractions, the fractions must have a common denominator. The common denominator will be the Lowest Common Denominator (LCD). When you are dealing with variables, you get this by multiplying the two denominators together:

\(\displaystyle \frac{2x}{x+3} + \frac{4}{x} = \frac{?}{x(x+3)}\)

To follow through with this LCD, you must multiply each fraction by the other fraction's denominator so that they end up with common denominators. The first fraction needs to be multiplied by \(\displaystyle \frac{x}{x}\) and the second fraction needs to be multiplied by \(\displaystyle \frac{x+3}{x+3}\). Note that both of these are equivalent to ONE, so the value of the fraction does not change:

\(\displaystyle \frac{(2x)(x)}{(x+3)(x)} + \frac{(4)(x+3)}{(x)(x+3)} = \frac{2x^2}{x(x+3)} + \frac{4x+12}{x(x+3)}\)

Now that the two fractions have common denominators, you can add the two numerators:

\(\displaystyle \frac{2x^2+4x+12}{x(x+3)}\)

Before you can subtract the two numerators from the fractions, the fractions must have a common denominator. The common denominator will be the Lowest Common Denominator (LCD). When you are dealing with variables, you get this by multiplying the two denominators together:

\(\displaystyle \frac{4x}{7} - \frac{2x}{2x+4} = \frac{?}{7(2x+4)}\)

To follow through with this LCD, you must multiply each fraction by the other fraction's denominator so that they end up with common denominators. The first fraction needs to be multiplied by \(\displaystyle \frac{2x+4}{2x+4}\) and the second fraction needs to be multiplied by \(\displaystyle \frac{7}{7}\). Note that both of these are equivalent to ONE, so the value of the fraction does not change:

\(\displaystyle \frac{4x(2x+4)}{7(2x+4)} - \frac{2x(7)}{2x+4(7)} = \frac{8x^2+16x}{7(2x+4)} - \frac{14x}{7(2x+4)}\)

Now that the two fractions have common denominators, you can subtract the two numerators. You can also distribute the denominator so you can simplify later:

\(\displaystyle \frac{8x^2+2x}{14x+28}\)

Finally you can divide out each term in the numerator and denominator by 2 to fully simplify the answer:

\(\displaystyle \frac{4x^2+x}{7x+14}\)

\(\displaystyle \frac{4x}{x+3}+\frac{2x}{x+3}\)

Because the two rational expressions have the same denominator, we can simply add straight across the top.  The denominator stays the same.

Therefore the answer is \(\displaystyle \frac{6x}{x+3}\).

Multiply both sides by \(\displaystyle (x-1) (x-3)\):

\(\displaystyle \frac{3}{x-1} + \frac{4}{x- 3 } = 5\)

\(\displaystyle \frac{3}{x-1} \cdot (x-1) (x-3)+ \frac{4}{x- 3 } \cdot (x-1) (x-3) = 5\cdot (x-1) (x-3)\)

\(\displaystyle 3 \cdot (x-3)+4\cdot (x-1) = 5\cdot (x-1) (x-3)\)

\(\displaystyle 3x-9 + 4x-4 = 5 (x ^{2 } -4x+3)\)

\(\displaystyle 7x-13= 5 x ^{2 } -20x+15\)

\(\displaystyle 7x -13 -7x + 13= 5 x ^{2 } -20x -7x +15+ 13\)

\(\displaystyle 5 x ^{2 } -27x + 28 = 0\)

Factor this using the \(\displaystyle ac\)-method. We split the middle term using two integers whose sum is \(\displaystyle -27\) and whose product is \(\displaystyle 5 \cdot28 = 140\). These integers are \(\displaystyle -20,-7\):

\(\displaystyle 5 x ^{2 } -20x -7x + 28 = 0\)

\(\displaystyle \left (5 x ^{2 } -20x \right ) - \left (7x - 2 8\right ) = 0\)

\(\displaystyle 5x \left (x-4 \right ) - 7 \left (x-4 \right ) = 0\)

\(\displaystyle \left (5x - 7 \right )\left (x-4 \right ) = 0\)

Set each factor equal to 0 and solve separately:

\(\displaystyle 5x - 7 = 0\)

\(\displaystyle 5x - 7 + 7 = 0 + 7\)

\(\displaystyle 5x = 7\)

\(\displaystyle 5x \div 5 = 7 \div 5\)

\(\displaystyle x = \frac{7}{5} = 1 \frac{2}{5}\)

 or

\(\displaystyle x - 4 = 0\)

\(\displaystyle x - 4 + 4 = 0+ 4\)

\(\displaystyle x = 4\)

Since the two fractions already have a common denominator, you can add the fractions by adding up the two numerators and keeping the common denominator:

\(\displaystyle \frac{3x+10}{x+3} = 5\)

Next you will algebraically solve for \(\displaystyle x\) by isolating it on one side of the equation. The first step is to multiply each side by \(\displaystyle x+3\):

\(\displaystyle (x+3)\frac{3x+10}{x+3} = 5(x+3)\)

Cancel out the \(\displaystyle x+3\) on the left and distribute out on the right. Then solve for \(\displaystyle x\) by subtracting \(\displaystyle 5x\) to the left and subtracting 10 to the right. Finally divide each side by negative 2:

\(\displaystyle 3x+10 = 5x+15\)

\(\displaystyle -2x+10=15\)

\(\displaystyle -2x=5\)

\(\displaystyle x=-\frac{5}{2}\)

With rational equations we must first note the domain, which is all real numbers except \(\displaystyle \small \small y=0\). (If \(\displaystyle \small y=0\), then the term \(\displaystyle \small \frac{1}{y}\) will be undefined.) Next, the least common denominator is \(\displaystyle \small 2y\), so we multiply every term by the LCD in order to cancel out the denominators. The resulting equation is \(\displaystyle \small 2 +y=16y\). Subtract \(\displaystyle \small y\) on both sides of the equation to collect all variables on one side: \(\displaystyle \small 2=15y\). Lastly, divide by the constant to isolate the variable, and the answer is \(\displaystyle \small y=\frac{2}{15}\). Be sure to double check that the solution is in the domain of our equation, which it is. 

With rational equations we must first note the domain, which is all real numbers except \(\displaystyle \small \tiny x=2\). (Recall, the denominator cannot equal zero. Thus, to find the domain set each denominator equal to zero and solve for what the variable cannot be.) 

The least common denominator or \(\displaystyle \small 2\) and \(\displaystyle \small 2x-4\) is \(\displaystyle \small 2(x-2)\). Multiply every term by the LCD to cancel out the denominators. The equation reduces to \(\displaystyle \small \small (x-2)(x+2)-3=6(x-2)\). We can FOIL to expand the equation to \(\displaystyle \small \small x^2 -4-3=6x-12\). Combine like terms and solve: \(\displaystyle \small x^2-6x+5=0\). Factor the quadratic and set each factor equal to zero to obtain the solution, which is \(\displaystyle \small x=5\) or \(\displaystyle \small x=1\). These answers are valid because they are in the domain. 

There are two ways to solve this problem using LCD (Least Common Denominator). The first method requires that you convert all denominators to the LCD by multiplying appropriately, and then follow the operations the equation requests. The second method allows you to cancel out terms using the LCD by mutiplying each term by the LCD. This is the method used for this problem and sometimes the simpler method since it tends to eliminate some of the fractions.

\(\displaystyle \frac{8}{x(x+5)}-\frac{4}{x+5}=\frac{12}{x}\)

Find the LCD:

The LCD will be based of the denominator of the first fraction (with 8 in the numerator). The middle term, based on this, is missing an x while the third term is missing \(\displaystyle (x+5)\). Our LCD will be \(\displaystyle x(x+5)\) since it has all the parts of each denominator. Multiply each term by this LCD.

\(\displaystyle \frac{8}{x(x+5)}*[\frac{x(x+5)}{1}]-\frac{4}{x+5}*[\frac{x(x+5)}{1}]=\frac{12}{x}*[\frac{x(x+5)}{1}]\)

Cancel terms that are present in our denominators and the LCD (same terms) we are multiplying by (red numbers will be canceled out):

\(\displaystyle \frac{8}{\color{Red} {x(x+5)}}*[\frac{\color{Red} {x(x+5)}}{1}]-\frac{4}{\color{Red} {x+5}}*[\frac{x{\color{Red} (x+5)}}{1}]=\frac{12}{\color{Red} {x}}*[\frac{{\color{Red} x}(x+5)}{1}]\)

Rewrite equation:

\(\displaystyle 8-4x=12(x+5)\)

Distribute the right side of the equation:

\(\displaystyle 8-4x=12x+60\)

Move x's to one side:

\(\displaystyle 8=16x+60\)

completely isolate x:

\(\displaystyle -52=16x\)

\(\displaystyle x=-\frac{13}{4}\)       \(\displaystyle or\)    \(\displaystyle x=-3.25\)

In order to simplify this problem, we are going to need to factor the ploynimials and then get the two terms to have the same denominator.

Starting out with factoring:

\(\displaystyle \frac{4}{x^2-4}+\frac{7x}{2x^3+8x^2+8x}\)

\(\displaystyle \frac{4}{(x+2)(x-2)}+\frac{7x}{2x(x^2+4x+4)}\)

the x's on the right term cancel:

\(\displaystyle \frac{4}{(x+2)(x-2)}+\frac{7}{2(x^2+4x+4)}\)

factoring one more time:

\(\displaystyle \frac{4}{(x+2)(x-2)}+\frac{7}{2(x+2)^2}\)

Now we need to make the denominators on both terms the same. Let's start by multiplying the left term by (x+2)/(x+2): 

\(\displaystyle \frac{(x+2)}{(x+2)}\cdot\frac{4}{(x+2)(x-2)}+\frac{7}{2(x+2)^2}\)

\(\displaystyle \frac{4(x+2)}{(x+2)^2(x-2)}+\frac{7}{2(x+2)^2}\)

**Note: We can do this because (x+2)/(x+2)=1 and multiplying something by 1 does not change its value

Now let's multiply the right term by (x-2)/(x-2):

\(\displaystyle \frac{4(x+2)}{(x+2)^2(x-2)}+\frac{7}{2(x+2)^2}\cdot \frac{(x-2)}{(x-2)}\)

\(\displaystyle \frac{4(x+2)}{(x+2)^2(x-2)}+\frac{7(x-2)}{2(x+2)^2(x-2)}\)

Now the only difference in the denominators is the 2 on the right term. Let's multipl the left term by 2/2 in order to solve this:

\(\displaystyle \frac{2}{2}\cdot\frac{4(x+2)}{(x+2)^2(x-2)}+\frac{7(x-2)}{2(x+2)^2(x-2)}\)

\(\displaystyle \frac{8(x+2)}{2(x+2)^2(x-2)}+\frac{7(x-2)}{2(x+2)^2(x-2)}\)

Our denominators are finally the same! Let's combine the fractions:

\(\displaystyle \frac{8(x+2)+7(x-2)}{2(x+2)^2(x-2)}\)

Now let's distribute and combine like terms (we are almost done):

\(\displaystyle \frac{8x+16+7x-14}{2(x+2)^2(x-2)}\)

\(\displaystyle \frac{15x+2}{2(x+2)^2(x-2)}\)

This expression can't be simplified any further. We have our answer!

\(\displaystyle \frac{x}{3}-\frac{4}{15}=3\)

Find a common denominator:

\(\displaystyle \frac{5}{5}*\frac{x}{3}-\frac{4}{15}=3\rightarrow \frac{5x}{15}-\frac{4}{15}=3\)

Subtract the fractions and multiply to cancel the denominator:

\(\displaystyle \frac{5x-4}{15}=3\rightarrow 5x-4=45\)

Move over the constant and isolate x:

\(\displaystyle 5x=49\rightarrow x=\frac{49}{5}\)