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).

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

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

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

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

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

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

\(\displaystyle \frac{10}{14}+\frac{11}{14}\)

Then, we add across the numerators and simplify:

\(\displaystyle \frac{10}{14}+\frac{11}{14}=\frac{21}{14}=\frac{3}{2}\)

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:

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

FOIL and simplify.

\(\displaystyle \frac{-x^2 + x + 12}{-x^2 -3x+28} + \frac{x^3 +7x^2}{-x^2 -3x+28}\)

Combine numerators.

\(\displaystyle \frac{-x^2 + x + 12}{-x^2 -3x+28} + \frac{x^3 +7x^2}{-x^2 -3x+28} = \frac{x^3 +6x^2 + x + 12}{-x^2-3x+28}\) 

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

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 \(\displaystyle 3x\) and \(\displaystyle x^2\) is \(\displaystyle 3x^2\). So the first fraction needs to be multiplied by \(\displaystyle x\) and the second by \(\displaystyle 3\):

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

\(\displaystyle \frac{x^3 + 3x^2 - 2x}{3x^2} + \frac{3x-9}{3x^2}\)

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

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

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

Solve the first equation for \(\displaystyle x\).

\(\displaystyle 2x-12=24\)

\(\displaystyle 2x=36\)

\(\displaystyle x=18\)

Solve the second equation for \(\displaystyle y\). 

\(\displaystyle 3y-15=60\)

\(\displaystyle 3y=75\)

\(\displaystyle y=25\)

The final step is to multiply \(\displaystyle x\) and \(\displaystyle y\).

\(\displaystyle xy = (18)(25)=450\)

The table shows that for every two minutes, the temperature of the coffee lowers 2.7ºF. At 1:25, 16 minutes, or eight 2-minute intervals have passed, and the temperature of the coffee has lowered by 8*2.7ºF, reaching a temperature of 165.5ºF.

Amy has bought 7 tickets, x of them at $40/ticket, and the remaining 7-x at $54/ticket. She spends at total of 

\(\displaystyle \\ \\ \$350 = 40x + (54)(7-x) = 40x + 378 – 54x = 378 – 14x\rightarrow \\ 14x = 378-350 = 28 \rightarrow x =2.\)

To find an equivalency we must rationalize the denominator.

To rationalize the denominator multiply the numerator and denominator by the denominator.

\(\displaystyle \frac{3}{\sqrt{6}}\) \(\displaystyle *\frac{\sqrt{6}}{\sqrt{6}}=\)

\(\displaystyle \frac{3*\sqrt{6}}{6}=\)  

To simplify completely, factor out a three from the numerator and denominator resulting in the final solution.

\(\displaystyle \frac{\sqrt{6}}{2}\)

The common denominator of these two fractions simply is the product of the two denominators, namely:

\(\displaystyle (x-3)(x+1)\)

Thus, you will need to multiply each fraction's numerator and denominator by the opposite fraction's denominator:

\(\displaystyle \frac{4}{x-3}-\frac{12}{x+1}=\frac{4}{x-3}*\frac{x+1}{x+1}-\frac{12}{x+1}*\frac{x-3}{x-3}\)

Let's first simplify the numerator:

\(\displaystyle \frac{4(x+1)-12(x-3)}{(x-3)(x+1)}\)

\(\displaystyle \frac{4x+4-12x+36}{(x-3)(x+1)}\)

\(\displaystyle \frac{-8x+40}{(x-3)(x+1)}\), which is the simplest form you will need for this question.

However, the correct answer has the denominator multiplied out. Merely FOIL \(\displaystyle (x-3)(x+1)\): 

\(\displaystyle \frac{40-8x}{x^2-2x-3}\)