How would you solve (a+3)+(2a-8)+(3a-3)+(a-1) using like terms?

Question

This is based off a question in my algebra 1 book. In the chapter it asks about combing like terms. Please help

 

 

Answer 1

Combining like terms is fairly simple. To begin with, you must know your Order of Operations. This can be remembered by using PEMDAS. P=parentheses, E=exponents, MD=multiplication/division, and AS=addition/subtraction. The Order of Operations tells you which steps you must take first when solving an algebraic problem. So our first step is to get rid of any parentheses. Notice that all of the parentheses have addition signs in front of them, therefore we do not have to distribute a negative sign. Our problem will become:

a+3+2a-8+3a-3+a-1

The next step is to evaluate exponents, but we don't have any of those, so we skip that step. Third, we need to evaluate any multiplication or division, going in order from left to right. In this problem there is no multiplication or division. We move on to addition and subtraction, again, solving from left to right. The problem now is that we cannot add a variable (letter) with a number. I always tell my students to use the "circle and box" system. To do this, you simply circle any numbers that stand alone and put a box around any numbers that have variables attached to them, or any variables that stand alone. Remember to include the sign before the variable or letter in the circle or box. Therefore, we should circle +3, -8, -3, and -1. We would then put a box around the other values: a, +2a, +3a, and +a. 

 

From here we can evaluate the circles and boxes separately. Let's do the circles first. We will have:

+3-8-3-1=-9

 

And then the boxes:

a+2a+3a+a

You hopefully understand that the a will remain constant (will not change) and you will add the values in front of them together. Any variable without a number in front of it has an invisible "1" value. Therefore...

a+2a=3a

3a+3a=6a

6a+a=7a

OR

a+2a+3a+a=7a

 

Our equation would then be 7a-9

You solve algebraically from here, by moving values around until a is by itself.

7a-9=0 is our original equation. To move the nine to the other side, we must do the inverse of subtraction (because the nine is being subtracted). So we add 9 to both sides.

 

7a-9=0

   +9  +9

7a    =9

 

Notice that the two 9s have cancelled out.

Now we must move the 7 by doing the inverse operation of multiplication (the 7 is multiplied to the a). Therefore, we divide both sides by 7:

 

7a=9

7    7

 

a=9/7

 

 

Hopefully this answer has helped you. It's slightly hard to describe this without verbally discussing the situation with you, but the steps above will help you when using like terms. 

Answer 2

Evan,  I just reviewed Bethann's analysis of your problem.  Although she meant well to give you such a detailed answer, I truly believe that the only requirement in this work was to "combine like-terms".  This is such an early stage of the math that you get in school, that it is unlikely that you would get an equation of multiple steps.  I will give you a shorter version below:

(a + 3) + ( 2a - 8) + ( 3a - 3) + ( a - 1) can be "simplified" by combining like-terms.  These terms are expressions that have the same characteristics.  In this problem, you have  A-terms and constants. The "coefficients" are the number values that sit in front of the variable and tell you how many of those A's that you have.  You will have to "collect like-terms" in two different "camps".  Teachers often like what we describe as "descending order", so the A's are done first.  Bethann did indicate how to do this..... add 1a + 2a + 3a + 1a first and place the rsult in the first positon of the answer.  Then, add the "constants " and place that result in the second position......  3 +- 8 + -3 + -1.   Notice how I used a negative sign for each of those types of numbers.  Your answer should be   7a  -  9  .      Please respond should you have any further questions .

 

Thanks, and good luck in the new school year !

 

Charles S.

Arden, NC