In order to determine the least common denominator, we will need to multiply both denominators together.  Be careful not to mix up the denominators as similar entities.

\(\displaystyle x^2y(x^2+y) = x^4y+x^2y^2\)

Do not solve the problem.  

The answer is:  \(\displaystyle x^4y+x^2y^2\)

The least common denominator is the term that is both divisible by all denominators and do not have common denominators smaller than this term.

Notice that both denominators can share a \(\displaystyle x^3\).  We can simply change the first denominator to \(\displaystyle x^3\) by multiplying an x-term.

\(\displaystyle 9x^2(x)=9x^3\)

For the second denominator, notice that we can change the three to a nine, similar to the first denominator by multiplying a three.

\(\displaystyle 3(3x^3)= 9x^3\)

Since these terms are common and is the least possible term, this is the least common denominator.

The answer is:  \(\displaystyle 9x^3\)

In order to determine the least common denominator, we will need to break up both denominator with their simplest forms.

\(\displaystyle x^2-x = x(x-1)\)

\(\displaystyle 2x-2 =2(x-1)\)

Notice that both of the values share an \(\displaystyle (x-1)\) term.  The first expression lacks a two coefficient, and the second expression lacks an \(\displaystyle x\) term.

This means that the LCD will be:  \(\displaystyle 2x(x-1)\)

Simplify this expression by distribution.

The answer is:  \(\displaystyle 2x^2-2x\)

This is a more complicated form of \(\displaystyle \frac{1}{2}+\frac{1}{3}= \frac{3}{6}+\frac{2}{6}=\frac{5}{6}\)

Find the least common denominator (LCD) and convert each fraction to the LCD, then add the numerators.  Simplify as needed.

\(\displaystyle \frac{x+2}{x-1}+\frac{x-5}{x+3}=\frac{x+2}{x-1}\cdot \frac{x+3}{x+3}+\frac{x-5}{x+3}\cdot \frac{x-1}{x-1}\)

which is equivalent to \(\displaystyle \frac{(x+2)\cdot (x+3)+(x-5)\cdot (x-1)}{(x+3)\cdot (x-1)}\)

Simplify to get \(\displaystyle \frac{2x^{2}-x+11}{x^{2}+2x-3}\)

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

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 \(\displaystyle \frac{6x}{x+3}\).

a.  Find a common denominator by identifying the Least Common Multiple of both denominators.  The LCM of 3 and 1 is 3.  The LCM of \(\displaystyle x^2\) and \(\displaystyle x^3\) is \(\displaystyle x^3\). Therefore, the common denominator is \(\displaystyle 3x^3\).

b. Write an equivialent fraction to \(\displaystyle \frac{5}{x^2}\) using \(\displaystyle 3x^3\) as the denominator.  Multiply both the numerator and the denominator by \(\displaystyle 3x\) to get \(\displaystyle \frac{15x}{3x^3}\). Notice that the second fraction in the original expression already has \(\displaystyle 3x^3\) as a denominator, so it does not need to be converted.

The expression should now look like: \(\displaystyle \frac{15x}{3x^3}-\frac{1}{3x^3}\). 

c. Subtract the numerators, putting the difference over the common denominator.

\(\displaystyle \frac{15x-1}{3x^3}\)

To combine fractions of different denominators, we must first find a common denominator between the two. We can do this by multiplying the first fraction by \(\displaystyle \small \frac{y}{y}\) and the second fraction by \(\displaystyle \small \frac{25x}{25x}\). We therefore obtain:

\(\displaystyle \small \frac{x^2+y}{25x}\times \frac{y}{y}+\frac{z+3}{y}\times\frac{25x}{25x}\)

\(\displaystyle \small \small \frac{x^2y+y^2}{25xy}+\frac{25xz+75x}{25xy}\)

Since these fractions have the same denominators, we can now combine them, and our final answer is therefore:

\(\displaystyle \small \frac{x^2y+y^2+25xz+75x}{25xy}\)

We start by adjusting both terms to the same denominator which is 2 x 3 = 6

Then we adjust the numerators by multiplying x+1 by 2 and 2x-5 by 3

\(\displaystyle \frac{(x+1)\cdot 2}{3 \cdot 2}+\frac{(2x-5) \cdot 3}{2 \cdot 3}\)

The results are:

\(\displaystyle \frac{2x+2}{6}+\frac{6x-15}{6}\)

So the final answer is,

\(\displaystyle \frac{8x-13}{6}\)

Start by putting both equations at the same denominator. 

2x+4 = (x+2) x 2 so we only need to adjust the first term: 

\(\displaystyle \frac{(3x-4)\cdot 2}{(x+2)\cdot 2}-\frac{5x+2}{2x+4}\)

\(\displaystyle =\frac{6x-8}{2x+4}-\frac{5x+2}{2x+4}\)

Then we subtract the numerators, remembering to distribute the negative sign to all terms of the second fraction's numerator:

\(\displaystyle =\frac{6x -8 -5x-2}{2x+4}=\frac{x-10}{2x+4}\)

(x+5)(x+3) is the common denominator for this problem making the numerators 7(x+3) and 8(x+5).

7(x+3)+8(x+5)= 7x+21+8x+40= 15x+61

A=61