# Two questions

#### 38175425

##### New member
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?

#### Dr.Peterson

##### Elite Member
For 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.

#### HallsofIvy

##### Elite Member
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?
1- 2/a is NOT "(1- 2)/a" which would be -1/a.
Instead, from x= 1- 2/a subtract 1 from both sides to get x- 1= 2/a.
From there, you can do either of two things. you could multiply both sides by a to get
a(x- 1)= 2 (note the parentheses! ax-1= 2 would be wrong), then divide both sides by x- 1 to get
a= 2/(x- 1). Or you could "invert" both sides, 1/(x-1)= a/2, and the multiply both sides by 2 to get a= 2/(x- 1) again. (And, again, the parentheses are important. This is NOT a= 2/x- 1!)

But I would have started differently. Instead of dividing by a, I would first subtract a from both sides:
ax- a= a(x- 1)= 2. Then (as long as x is not 1) divide by x- 1 to get a= 2/(x- 1) yet again.

#### Jomo

##### Elite Member
a/a is NOT 1. No! a/a is what is called a piecewise function. It has different results for different values of a. For example, if a = 0, then a/a is NOT 1, it is said to be indetermined, while if a id not 0, then a/a = 1.

In your problem if a=0 was the solution you never would have gotten that answer since when you computed a/a you got the wrong result. I would never divide by an unknown. Instead combine the a's by adding or subtracting NOT dividing.

#### Jomo

##### Elite Member
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!!!

#### Dr.Peterson

##### Elite Member
One obvious thing that is wrong is that you solved for x and not a!!!
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.

#### Jomo

##### Elite Member
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.
Thank you!

#### 38175425

##### New member
I AM SO SORRY MY SECOND QUESTION IS WRONG!
The question is to find "a" not "x"!

#### 38175425

##### New member
For 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.
ax = a + 2
a = a/x - 2/x

#### Dr.Peterson

##### Elite Member
The solution can't have $$\displaystyle a$$ on both sides.

First collect the terms with $$\displaystyle a$$ on one side, and then isolate $$\displaystyle a$$.

Also, is the sign in the equation positive or negative? Check for typos in your work.

#### 38175425

##### New member
The 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.
x - 2 = a/a

#### 38175425

##### New member
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

#### JeffM

##### Elite Member
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
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?

#### 38175425

##### New member
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?

#### JeffM

##### Elite Member
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.

#### 38175425

##### New member
2. ax = a - 2. How to find "x"?
x = a/a - 2/a
x = a - 2 / a
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.
Thank you very much!

#### Jomo

##### Elite Member
5/5 is not 5, 17/17 is not 17,... so a/a is not a! So what is a/a??

#### 38175425

##### New member
5/5 is not 5, 17/17 is not 17,... so a/a is not a! So what is a/a??
a/a is 1 or 0

Staff member

#### HallsofIvy

##### Elite Member
No, a/a= 1 for all non-zero a. If a= 0, a/a does not exist.