Add the numbers and keep the variable:

\(\displaystyle 7n+3n=10n\)

Answer: \(\displaystyle 10n\)

Add the numbers and keep the variable:\(\displaystyle 14s+3s=17s\)

Answer: \(\displaystyle 17s\)

In order to add variables the terms must be like. In order for terms to be like, the variables must be exactly alike also being raised to the same power by the exponent.

In this case the like terms are \(\displaystyle x^{1}y^{3}\) and \(\displaystyle xy^{3}\). Just because there is a 1 in the exponent for the first term doesnt mean it is different from the second term. With exponents if a variable does not show an exponent, that means it is still to the first power. 

We add the coefficients of the like terms. The coefficient is the number in front of the first variable, in this case it is 1 for both terms because of the identity property of multiplication stating any variable, term, or number multiplied by 1 is itself.

\(\displaystyle x^{1}y^{3}= 1(x^1y^3)\)    \(\displaystyle xy^3= 1(xy^3)\)

Our last term is not like because the \(\displaystyle x\) variable is raised to a different power than the other two. In this case we do not combine it to the like terms, we just add it to the end of the term. 

Remember, for exponent problems, you group together different exponents and different combinations of variables as though each were a different type of variable.  Therefore, you can group your problem as follows:

\(\displaystyle (2x + 15x) + (2x^{2} + 3x^{2}) + (15y + 2y) + (14y^{2} + 3y^{2})\)

Now, just combine like terms:

\(\displaystyle 17x + 5x^{2} + 17y + 17y^{2}\)

You should begin by distributing \(\displaystyle 2\) through the whole group that it precedes:

\(\displaystyle 23x + 22y + 8x + 6y\)

Now, move your like variables next to each other:

\(\displaystyle 23x + 8x + 22y + 6y\)

Finally, combine the like terms:

\(\displaystyle 31x + 28y\)

First, group together your like variables:

\(\displaystyle 2x + 4x - 3x + 4y + 15xy - 5xz\)

The only like variables needing to be combined are the x-variables.  You can do this in steps or all at once:

\(\displaystyle 2x + x + 4y + 15xy - 5xz\)

\(\displaystyle 3x + 4y + 15xy - 5xz\)

First, move the like terms to be next to each other:

\(\displaystyle 20x - 14x + 3y + 12y - 4z - 2xy\)

Now, combine the x-variables and the y-variables:

\(\displaystyle 6x + 15y - 4z - 2xy\)

Let's begin by moving the like terms toward each other.  Notice the following: zy is the same as yz.  (Recall the commutative property of multiplication.)

\(\displaystyle 14xy + x + 15x + 12yz + 3yz\)

Now, all you have to do is combine the x-variables and the yz-terms:

\(\displaystyle 14xy + 16x + 15yz\)

Notice that you do not end up with any exponent changes.  That would only happen if you multiplied those variables.

Remember, when you have exponents like this, you will treat each exponented variable as though it were its own "type."  Likewise, pairs of variables are to be grouped together.  Therefore, group the problem as follows:

\(\displaystyle (x^{2} + 15x^{2}) + 3y^{2} + 5y^{4} + 12x^{2}y^{4}\)

Notice that the only thing to be combined are the \(\displaystyle x^{2}\) terms.

Therefore, your answer will be:

\(\displaystyle 16x^{2} + 5y^{4} + 3y^{2} + 12x^{2}y^{4}\)

Remember, for exponent problems, you group together different exponents and different combinations of variables as though each were a different type of variable.  Therefore, you can group your problem as follows:

\(\displaystyle 3x + (5x^{2} + 12x^{2})+ 15xy + 4y^{2}\)

Then, all you need to do is to combine the \(\displaystyle x^{2}\) terms:

\(\displaystyle 3x + (17x^{2})+ 15xy + 4y^{2}\)