When adding variables together, you must first make sure you are combining the same variable.  So, in this case

\(\displaystyle a+4a\)

we can see that both terms contain the variable a.  Therefore, we can combine them. 

Now, when we combine them, we can think of the variables as objects.  So, we can say were are combining an apple and 4 apples together.  So,

\(\displaystyle \text{apple} + 4\text{ apples } = 5\text{ apples}\)

We can simplify our problem the same way.

\(\displaystyle a +4a = 5a\)

Combine the following variables:

\(\displaystyle 5x^6+6x^5+4x^5+3x^2+4x^6+12x^2\)

First, group the like terms. This means we will put terms with the same exponent next to each other. This will make it easier to be sure we are adding correctly.

\(\displaystyle 5x^6+4x^6+6x^5+4x^5+3x^2+12x^2\)

Next, add the coefficients, but keep the exponents the same. Only combine terms with the same exponents. Notice that we have three different exponents (6,5 and 2) This means we will have three terms in our final answer.

So we get the following:

\(\displaystyle 9x^6+10x^5+15x^2\)

To add variables, we will think of the variables as objects.  So, in the problem

\(\displaystyle 3x+7x\)

let's think of the variable x as books.  So, we can look at the problem as we currently have 3 books.  We go to the library and borrow 7 more books.  How many books do we have now?  The answer is 10.  We now have 10 books.  

We can add the variables in the same way.  So,

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

To add variables, we will look at the variables as objects.  So, 

\(\displaystyle 6b + 4b\)

we can look at it as the following.  We have 6 dollars.  We do all of our chores, and we get 4 more dollars.  How many dollars do we have now?  We now have 10 dollars.  We add variables in the same way. So

\(\displaystyle 6b+4b = 10b\)

When we add variables, we can think of the variables as objects.  So in the problem,

\(\displaystyle 5a+6a\)

we can think of the variable a as apples.  We can say that we currently have 5 apples.  We go to the store and buy 6 more apples.  How many apples do we have now?  We have 11 apples.  

We can add variables the same way.  So,

\(\displaystyle 5a+6a=11a\)

Combine the following terms.

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

We need to realize that we can only add variables together if they have the same exponent. Then, all we need to do is add the coefficients (numbers in front)

So, let's reorganize our statement to have variables of the same power together

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

Now, we can add the first two and last two to get our answer

\(\displaystyle 9x^6+7x^5\)

To add variables, we will look at the variables like objects.  So in the problem

\(\displaystyle 3a+6a\)

we will think of a as being apples.  So we can read this problem as

\(\displaystyle 3 \text{ apples} + 6 \text{ apples}\)

So, if we have 3 apples, and we go to the market and we buy 6 more apples, how many apples do we now have?

\(\displaystyle 3 \text{ apples} + 6 \text{ apples} = 9 \text{ apples}\)

We add variables the same way.  So,

\(\displaystyle 3a+6a\)

We can look at this as we have 3 of something, we want to add 6 more somethings, so how many do we have now?

\(\displaystyle 3a+6a = 9a\)

In the problem

\(\displaystyle 5a+a\)

we will first insert a coefficient on the second term.  When a variable stands alone, it has a coefficient of 1.  So,

\(\displaystyle 5a+1a\)

When adding variables, we can look at the variables as objects, then combine those objects.

Now, we can think of the variable a as cars. So, we can look at it as

\(\displaystyle 5 \text{ cars} + 1 \text{ car}\)

We can think of it like this:  If we see 5 cars pass by us on the road, and we see 1 more car pass us, how many cars have passed us?  The answer is 6.  There were 6 cars that passed us.  So,

\(\displaystyle 5 \text{ cars} + 1 \text{ car} = 6\text{ cars}\)

We add variables the same way.

\(\displaystyle 5a+1a=6a\)

When adding variables, we can look at the variables as objects.  So, in the problem

\(\displaystyle 4x + 2x\)

we can think of x as books.  So, we can write it like this

\(\displaystyle 4 \text{ books} + 2 \text{ books}\)

We can think of it like this.  We have 4 books and we borrow 2 more books.  How many books have we borrowed?  

\(\displaystyle 4 \text{ books} + 2 \text{ books} = 6 \text{ books}\)

We have borrowed 6 books.  

We can add variables the same way.

\(\displaystyle 4x + 2x\)

\(\displaystyle 4x + 2x = 6x\)

When adding variables, we can look at the variables as objects.

So, in the problem

\(\displaystyle 2a+3a\)

we can think of the variable a as apples.  So, we can write it like

\(\displaystyle 2\text{ apples} + 3\text{ apples}\)

We can think of the problem like this:  We have 2 apples, and we go to the store and buy 3 more apples.  How many apples do we have now?  We have 5 apples.  So, we can solve it like

\(\displaystyle 2\text{ apples} + 3\text{ apples} = 5\text{ apples}\)

We add variables the same way.  So,

\(\displaystyle 2a+3a = 5a\)