Adding 1 to a/b?

Question

I've been learning algebra $2$ and came across a question telling me to add $1$ to $\frac{a}{b}$ by finding the least common multiple of the divisors. So I got $\frac{1b}{1b}+\frac{a}{b}$ that simplified is $\frac{b}{b}+\frac{a}{b}$. So then I added those two fractions together and got $\frac{ab}{b}$ I saw that the two $b$'s cancelled each other out so I got $\frac{a}{1}$ which is $a$? How does adding $1$ to $\frac{a}{b}$ get $a$, this doesn't seem to be right because $1 + \frac{2}{3}$ isn't $2$ it's $1$ and $\frac{2}{3}$. Am I doing something wrong or is this actually correct, if so can you explain to me why?

Answer 1

You got the addition wrong, it's (a+b)/b, not ab/b.

Answer 2

"...that simplified is b/b+a/b. So then I added those two fractions together and got ab/b .."

That shouldn't make any sense and if I saw your paper I may understand why you did it. But it's simply wrong.

$1 + \frac ab$

$\frac{1b}{1b} + \frac ab$

$\frac{b}{b} + \frac ab$ You are right so far so let's add these.

$\frac{b + a}{b} $ ... and I don't know why you added a + b to get ab (probably a misplaced mark on the scratch paper). The final answer is $\frac{b + a}{b} $

===== " its a+b/b which is a+1 when simplified".

Oooh, no. Both the a and the b are over the b. You can't cancel out just one.

$\frac{a + b}{b} \ne a + \frac{b}{b}$.

You can factor the b out of both but then you get

$\frac{a + b}{b} = \frac{a/b + 1}{1} = \frac ab + 1$ but then you have worked yourself back to exactly where you started.

Answer 3

"So I got 1b/1b+a/b that simplified is b/b+a/b. So then I added those two fractions together and got ab/b"

You have figured out correctly up to this point:

$\frac{b}{b}+\frac{a}{b}$

The next step in your statement is where the error made. Since the denominators are equal, you just add the numerators and divide by the common denominator to get:

$\frac{b}{b}+\frac{a}{b}=\frac{b+a}{b}$

The general rule is (where $x$ and $y$ are not zero)is:

$\frac{A}{x}+\frac{B}{y}=\frac{Ay+Bx}{xy}$