1 2/a is NOT "(1 2)/a" which would be 1/a.1. ax = a + 2. How to find "a"?

2. ax = a  2. How to find "x"?
Here's how I calculated
Divide both sides by "a"
x = a/a  2/a
x = 1  2/a
x = 1/a
Is there anything wrong?
One obvious thing that is wrong is that you solved for x and not a!!!1. ax = a + 2. How to find "a"?

2. ax = a  2. How to find "x"?
Here's how I calculated
Divide both sides by "a"
x = a/a  2/a
x = 1  2/a
x = 1/a
Is there anything wrong?
As I read it, the second question asks to solve for x, not a.One obvious thing that is wrong is that you solved for x and not a!!!
Thank you!As I read it, the second question asks to solve for x, not a.
Of course, it could be a typo, since it's odd to have these two problems in succession; but I assume the OP is doing what it says to do.
There will, of course, be no solution if a = 0.
ax = a + 2For the first, collect terms with a on one side, then factor out the a. Please show work.
For the second, your error is in forgetting the order of operations. It is not true that 1  2/a = 1/a, because it means 1  (2/a), not (1  2)/a. Just don't make that last step. You could also leave the answer as x = (a  2)/a.
x  2 = a/aThe solution can't have \(\displaystyle a\) on both sides.
First collect the terms with \(\displaystyle a\) on one side, and then \(\displaystyle a\).
Also, is the sign in the equation positive or negative? Check for typos in your work.
So first you said to solve for x, then solve for a, and now back to solve for x. What EXACTLY does the problem say?1. ax = a + 2. How to find "a"?
ax  a = 2
a(x  a) = 2
a = 2/(x  a)
2. ax = a  2. How to find "x"?
x = a/a  2/a
x = a  2 / a
Thank you very much!Given that we now know that your first problem is to solve for x and your second problem is to solve for a, we can definitively say that your answers in post 12 are WRONG. For the first problem, you solved for a correctly, but that is not what the problem requires. For the second problem, you tried to solved for x, which is not what the problem requires, but you did it incorrectly.
In our guidelines, we ask that you give one problem per thread and that you state each problem completely and correctly. Think how much time you would have saved if you had followed those two guidelines.
With respect to solving for x in the second problem
\(\displaystyle ax = a + 2 \implies \dfrac{ax}{a} = \dfrac{a + 2}{a} = \dfrac{a}{a} + \dfrac{2}{a} \implies x = 1 + \dfrac{2}{a}.\)
If you saw 5/5, you would never say 5/5 = 5. You would say 5/5 = 1. But a is just a number that you don't know yet. It follows all the rules that apply to numbers generally such as that every number (except zero) divided by itself equals 1.
a/a is 1 or 05/5 is not 5, 17/17 is not 17,... so a/a is not a! So what is a/a??