\(\displaystyle \left (\frac{1}{3}x^{2}+\frac{1}{6}xy-\frac{5}{4}y^{2} \right )-\left (\frac{1}{9}x^{2}-\frac{4}{3}xy+y^{2} \right )\)

Since we are only adding and subtracting (there is no multiplication or division), we can remove the parentheses.

\(\displaystyle \frac{1}{3}x^{2}+\frac{1}{6}xy-\frac{5}{4}y^{2}-\left \frac{1}{9}x^{2}+\frac{4}{3}xy-y^{2} \right\)

Regroup the expression so that like variables are together. Remember to carry positive and negative signs.

\(\displaystyle (\frac{1}{3}x^{2}-\left \frac{1}{9}x^{2})+(\frac{1}{6}xy+\frac{4}{3}xy)-(\frac{5}{4}y^{2}-y^{2} \right)\)

For all fractional terms, find the least common multiple in order to add and subtract the fractions.

\(\displaystyle (\frac{3}{9}x^{2}-\left \frac{1}{9}x^{2})+(\frac{1}{6}xy+\frac{8}{6}xy)-(\frac{5}{4}y^{2}-\frac{4}{4}y^{2} \right)\)

Combine like terms and simplify.

\(\displaystyle \frac{2}{9}x^{2}+\frac{9}{6}xy-\frac{9}{4}y^{2}\)

\(\displaystyle \frac{2}{9}x^{2}+\frac{3}{2}xy-\frac{9}{4}y^{2}\)

\(\displaystyle \frac{1}{2}(4x^2-5x+10)+\frac{1}{4}(7x^2-5x+12)\)

First distribute the \(\displaystyle \frac{1}{2}\):

\(\displaystyle 2x^2-\frac{5}{2}x+5+\frac{1}{4}(7x^2-5x+12)\)

Then distribute the \(\displaystyle \frac{1}{4}\):

\(\displaystyle 2x^2-\frac{5}{2}x+5+\frac{7}{4}x^2-\frac{5}{4}x+\frac{12}{4}\)

Finally combine like terms:

\(\displaystyle 2x^2-\frac{10}{4}x+5+1\frac{3}{4}x^2-\frac{5}{4}x+3\)

\(\displaystyle 3\frac{3}{4}x^2-\frac{15}{4}x+8\)

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

With this problem, you need to take the trinomials out of parentheses and combine like terms. Since the two trinomials are being added together, you can remove the parentheses without needing to change any signs:

\(\displaystyle (2x^2+3x-4) + (3x^2-6x+7)\)

\(\displaystyle 2x^2+3x-4 + 3x^2-6x+7\)

The next step is to combine like terms, based on the variables. You have two terms with \(\displaystyle x^2\), two terms with \(\displaystyle x\), and two terms with no variable. Make sure to pay attention to plus and minus signs with each term when combining like terms:

\(\displaystyle 5x^2-3x+3\)

With this problem, you need to distribute the two fractions across each of the trinomials. To do this, you multiply each term inside the parentheses by the fraction outside of it:

\(\displaystyle \frac{1}{2}(6x^2+2x-3) + \frac{1}{4}(12x^2-24x+6)\)

\(\displaystyle 3x^2+x-\frac{3}{2} + 3x^2-6x+\frac{3}{2}\)

The next step is to combine like terms, based on the variables. You have two terms with \(\displaystyle x^2\), two terms with \(\displaystyle x\), and two terms with no variable. Make sure to pay attention to plus and minus signs with each term when combining like terms. Since you have a positive and negative \(\displaystyle \frac{3}{2}\), those two terms will cancel out:

\(\displaystyle 6x^2-5x\)

To add these two trinomials, you will first begin by combining like terms. You have two terms with \(\displaystyle x^2\), two terms with \(\displaystyle x\), and two terms with no variable. For the two fractions with \(\displaystyle x^2\), you can immediately add because they have common denominators:

\(\displaystyle (\frac{1}{2}x^2 +2x - 4) + (\frac{3}{2}x^2 -5x -4)\)

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

\(\displaystyle 2x^2 -3x - 8\)

To add trinomials, identify and group together the like-terms: \(\displaystyle (3x^2+x^2) +(14x-2x)+(9+4)\). Next, factor out what is common between the like-terms:\(\displaystyle (3+1)x^2+(14-2)x+(9+4)\). Finally, add what is left inside the parentheses to obtain the final answer of \(\displaystyle 4x^2+12x+13\).

First, we want to eliminate the variable \(\displaystyle y\) from the set of equations. To do this, we need to make the coefficients of the two \(\displaystyle y\)'s equal but opposite. This way, when we add the equations, we will be able to eliminate them. To make the \(\displaystyle -2y\) equal but opposite to the \(\displaystyle 4y\), we need to multiply the top equation by 2 on both sides. This gives us the equation:

\(\displaystyle 6x - 4y = 28\)

Then, we combine the two equations by adding them. We add the like terms together and get this equation:

\(\displaystyle 7x = 49\)

Using our knowledge of algebra, we know we can divide both sides by 7 to isolate the \(\displaystyle x\). Doing so leaves us with our answer, \(\displaystyle 7\).

To solve this problem, simply add the terms with like exponents and variables:

\(\displaystyle \\(2x^2+7x+4) + (x^2-3x+2) \\= (2x^2 +x^2) + (7x-3x) + (4+2) \\= 3x^2 - 4x +6\)

Thus, \(\displaystyle 3x^2 - 4x +6\) is our answer.

The first step is to combine all terms with like exponents and variables. Watch for negative signs!

\(\displaystyle 2x+5x^2 -2 - 3x^2 = 18 - 4x\rightarrow\)  \(\displaystyle 6x+2x^2 -20\)

Next, rearrange into standard quadratic form \(\displaystyle ax^2 + bx + c\):

\(\displaystyle 6x+2x^2 -20 = 2x^2 + 6x - 20\)

Thus, our answer is \(\displaystyle 2x^2 + 6x - 20\).

Let's solve this problem the long way, to see how it's done. Then we can look at a shortcut.

First, FOIL the binomial combinations:

FOIL stands for the multiplication between the first terms, outer terms, inner terms, and then the last terms.

\(\displaystyle (x+3)(x-2) = (x^2+3x-2x-6) = (x^2 + x - 6)\)

\(\displaystyle (x+3)(x+2) = (x^2 + 3x + 2x + 6) = (x^2 + 5x + 6)\)

Lastly, add the compatible terms in our trinomials:

\(\displaystyle (x^2 + x - 6) +\) \(\displaystyle (x^2 + 5x + 6) =\) \(\displaystyle 2x^2 + 6x\)

So, our answer is \(\displaystyle 2x^2 + 6x\).

Now, let's look at a potentially faster way.

Look at our initial problem.

\(\displaystyle (x+3)(x-2) + (x+3)(x+2)\)

Notice how \(\displaystyle (x+3)\) can be found in both terms? Let's factor that out:

\(\displaystyle (x+3)(x-2) + (x+3)(x+2) = (x+3)(x-2+x+2)\)

Simpify the second term:

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

Now, perform a much easier multiplication:

\(\displaystyle (x+3)(2x) = 2x^2 + 6x\)

So, our answer is \(\displaystyle 2x^2 + 6x\), and we had a much easier time getting there!