x+x=?

[Norloonda]

The following example solutions/explanations (*) are from the "basic operation" of algebra from my math book. *A. 4x+x=5x. (x means 1x); *B. 2ab+ab=3ab. (Not 3a2b?). In a later section on the distribution principle, *C. x²+5x/x(+x)=x+5+x=2x+5. In the example "B", I concluded that in the absence of a more detailed explanation that should have accompanied *B example, that in combining "like" terms, matching variables (both sides of the plus sign: "b" in example"*B") with coefficients of ONE(1) on both sides of the plus sign are to remain "as is" ie, 5ab2c+abc=6ab3c, 2bc+6bc=8bc, until I reached the "distribution principle" section. Why has the x+x in "*C" now 2x? In (*B) above, why the "b" (b+b) is not 2b? Is "b" isolated in some way from combining? What are the rules?

Both *A and *B are rules I have to go by before I came to the "Distribution Principle" section.

The comment (Not 3a2b?) is a statement of my confusion.

P.S. In my book, each section has problems at the end and not one of them have just adding two(2) variables alone. All of them have at least one(1) coefficient on one side.

 

[Norloonda]

1 apple + 1 apple = 2 apples
1 x + 1 x = 2 x

Are you a student attending math classes?

No I'm not a student. I'm trying to keep an active mind by studying my poorest subject (math) when I was in school. Classroom settings has always went too fast for me to grasp something I didn't quite understand and if I asked question, the answer might go in one ear and out the other.

The 1x+1x=2x is so easy to understand; matter of fact, my textbook starts out with kangaroos and apples in defining "like" and "unlike" terms. In my textbook - 2ab+ab= 3ab (remember: without further explanation, this is the answer). My question is - why "b" (b+b) doesn't become 2b?

In another textbook, it explain what the x,y & z's; a,b & c's and all the other letters are used for. I once thought the answer had something to do with that; (just guessing).

 

[Norloonda]

Center on 2ab

1 apple + 1 apple = 2 apples
1 x + 1 x = 2 x

Are you a student attending math classes?

No I'm not a student. I just want to learn more about the subject I had trouble with when I was in school. To be more direct, the examples of *A & *B is all I have in this math book to go on for rules. My question is if x+x=2x, then why (as illustrated in my math book) is 2ab+ab=3ab and not 3a2b as this would seem logically sound to me? This math book(textbook) is named Arithmetic and Algebra Again by Brita Immergut and Jean Burr Smith.

 

[ksdhart2]

It seems this post got lost in the shuffle for several days. I apologize for the delay. That said, I believe I may be able to ease your woes in this matter. The transition to algebra is an area of mathematics I've seen many, many students struggle with over the years. My best advice to suss out why specific rules for manipulating algebraic expressions work (or perhaps why your intuition doesn't work) is to substitute "real" concrete values for the variables. The expression you're having difficulty with is 2ab + ab = 3ab, yes? In this case, as you hopefully know, the term "ab" contains an implicit multiplication, such that the expression means: 2*a*b + 1*a*b = 3*a*b. Here we have two main options for seeing why this holds true. I'll detail the method of substituting a fixed value for the variables first. Let a = 2 and b = 3:

\(\displaystyle 2 \cdot a \cdot b + 1 \cdot a \cdot b = 3 \cdot a \cdot b \implies 2 \cdot 2 \cdot 3 + 1 \cdot 2 \cdot 3 = 3 \cdot 2 \cdot 3 \implies 2 \cdot 6 + 1 \cdot 6 = 3 \cdot 6 \implies 12 + 6 = 18\)

You can try with a = 3 and b = 5, or any other values (they can even be the same if you'd like) and it will still work. Now, the other method is a bit more abstract, but if you can master it, it will serve you well in the future. Working with the original equation and looking only at the left hand side, we see that there are two terms, and each of them contains a multiplication by b. By the distribution principle, we can factor out that b term, like so: 2*a*b + 1*a*b = b*(2*a + 1*a). If so desired, you can substitute some values for b to verify that this does indeed work. With that done, we can apply the same logic to also factor out the common multiplication of a to get a*b*(2+1), which is of course the same as 3*a*b.

As for why your intuition doesn't work, if we write out the implicit (some might say "invisible") multiplication signs, your solution becomes: 3*a*2*b or 6*a*b. And hopefully now you can see why that can't be correct