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!

Eliminate the parentheses and combine like-terms.

$\displaystyle -x^2+2x^2=x^2$

$\displaystyle -3x+(-3x)=-6x$

$\displaystyle 6+1=7$

Combine all the terms.

The answer is:  $\displaystyle x^2-6x+7$

When adding trinomials we combine together coefficients of like terms. 

$\displaystyle (5x^2-8x+3)+(6x^2+4x-5)=(5+6)x^2+(-8+4)x+3+(-5)$

$\displaystyle =11x^2-4x-2$

Combine like terms:

$\displaystyle (4x^{2} + 6x +14) + (5x^{2} - 3x -13)$

$\displaystyle (4x^{2} + 5x^{2}) + (6x - 3x) + (14 - 13)$

$\displaystyle 9x^{2} + 3x + 1$

Multiply the top equation by 3 on both sides, then add the second equation to eliminate the $\displaystyle y$ terms:

$\displaystyle 2x + y = 12$

$\displaystyle 3\cdot \left (2x + y \right )= 3\cdot 12$

$\displaystyle 6x+3y = 36$

$\displaystyle \underline{x - 3y = 13}$

$\displaystyle 7x\;\;\;\; \; =49$

$\displaystyle 7x \div 7 =49 \div 7$

$\displaystyle x = 7$

To subtract these two trinomials, you first need to flip the sign on every term in the second trinomial, since it is being subtrated:

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

$\displaystyle 2x^2+\frac{3}{4}x-5 - x^2-x+10$

Next you can combine like terms. You have two terms with $\displaystyle x^2$, two terms with $\displaystyle x$, and two terms with no variable:

$\displaystyle x^2-\frac{1}{4}x+5$

When subtracting trinomials, first distribute the negative sign to the expression being subtracted, and then remove the parentheses: $\displaystyle (x^2+15x-3)-(x^2-5x-3)=x^2+15x-3-x^2+5x+3$

Next, identify and group the like terms in order to combine them: $\displaystyle (x^2-x^2)+(15x+5x)+(3-3)=20x$.