Algebra: make a the subject of $x(a+b)=a-1$ This isn't my homework, it's me trying to help my son
So, we have a series of questions with similar forms, the first of which is:
Make a the subject of:
$x(a+b)=a-1$
Maths isn't really my strong point, but I can complete simple operations on this equation:
$xa+xb+1=a$
$a+b=\dfrac{a-1}{x}$
$xb+1=a-xa$
However, I have tried, I can't work out how to isolate "a" so that there is only one. I would appreciate any pointers to techniques to solve this problem.

So, there have been lots of contributions, but I think that I am correct to say that none has identified the “missing” step that I could not pass. I am reminded of something I read in The Guardian a few days prior to asking this question:

There is a belief among certain academics that a subject is less efficiently learned from an adept than from someone who is studying it or has just finished studying it. The adept’s long acquaintance makes it difficult for him or her to see the subject in its simpler terms or to appreciate what it is like to approach the subject as a greenhorn.


https://www.theguardian.com/books/2022/aug/07/could-learning-algebra-in-my-60s-make-me-smarter-alec-wilkinson-a-divine-language-extract

Therefore, I will try to explain how I have come to understand how to solve this problem (and those like it).
Make “a” the subject of:
$x(a+b)=a-1$
As an aside, I think it is more usual to ask the student to make “x” the subject, but I don’t suppose that matters too much…
The relatively easy part here is getting all the “a”s on one side, and everything else on the other.
First step is to put “x” inside the brackets (that‘s “parentheses” to anyone speaking Americanese) and thus eliminate them.
$(xa+xb)=a-1$
Then:
$xa+xb=a-1$
I believe this is called “expanding” the equation.
I will explain why this is a logical step. Rather than taking the adept route, looking at the “$x(a+b)$” portion and saying:
$x(a+b)=xa+xb=x(a+b)$ QED!
I find that using “real” numbers actually illustrates the point in a concrete way:
$x=5, a=4, b=3$
$5(4+3)=35$
Step through it (using the BIDMAS route: https://www.bbc.co.uk/bitesize/topics/znmtsbk/articles/zj29dxs):
$5(7)=35$
$5*7=35$
Therefore, if we take the expanded version of “$5(4+3)=35$” we see the same result:
$5(4+3)=35$
$(5*4)+(5*3)=35$
$(20)+(15)=35$
$20+15=35$
You can substitute in any numbers you like to prove that (note, you can’t substitute in any numbers you like to resolve the whole equation, just to prove the logic of the steps you take):
$x(a+b)=xa+xb=x(a+b)$, so QED!
So, back to our first step:
$xa+xb=a-1$
The second step is to take “$xb$” and put it on the other side of the “=”:
$xa=a-xb-1$
To prove this with real numbers, we just need to have an equation that “balances” in the first place, and forget that we have “a” on both sides:
$5+4=10-1$ ie… $9=9$
$5=10-4-1$ … $5=6-1$ … $5=5$
I hope you are with me so far!
From $xa=a-xb-1$ we have to get “a” to the opposite side of the equation:
$xa=a-xb-1$
$xa-a=-xb-1$
We can prove this again using our “real” numbers:
$5=10-4-1$
$5-10=-4-1$
$-5=-5$
Now for the tricky part, @lulu writes in the comments:

$a−xa=a(1−x)$

This part I really did not get at all, because there is obviously some intermediate step missing, but others are insisting that there is not.
There is a missing step, and this is it:
The part you have to complete is to factorise the lefthand side of the equation, let’s look at it:
$xa-a$
You take the  “a” in “xa” and put it outside some brackets:
$a(x…)$
Now, the standalone “a” was never multiplied by anything so if you put it in there you get something that is not logically correct:
$a(x-a)$ which if you expand you get $xa-aa$, so that cannot be correct. Instead, what you have to do is divide “a” by “a”:
$a(x-\dfrac{a}{a})$
We can demonstrate this again with real numbers:
a=4, x=5
$xa-a$
$(5*4)-4=16$ … $(20)-4=16$
$a(x-\dfrac{a}{a})$:
$4(5-\dfrac{4}{4})=16$ … $4(5-(1))=16$ … $4(4)=16$
So, then we prove that @lulu is correct, because any number divided by itself equals “1”:
$xa-a$ = $a(x-\dfrac{a}{a})$ = $a(x-1)$
We can also prove $a(x-1)$ = $xa-a$ by expanding $a(x-1)$:
$xa-1a$ “1a” is “a” so $xa-a$ is equal to $a(x-1)$
So, our equation is therefore:
$a(x-1)=-xb-1$
I think the next step is straightforward, in that we have to take the lefthand part in barackets, and use it to divide the righthand side, and thus isolate “a”:
$a=\dfrac{-xb-1}{x-1}$
Using real numbers again, we can prove that this is the next logical step:
$4*5=20$ … $4=\dfrac{20}{5}$
So, my answer is: $a=\dfrac{-xb-1}{x-1}$
Now I go back to the answer sheet, and the answer there is:
$a=\dfrac{bx+1}{1-x}$
Um, after all that, did I still get it wrong!?
I’ll step through it again… with a slightly different method…
$x(a+b)=a-1$
$xa+xb=a-1$
$xa+xb+1=a$
$xb+1=a-xa$
$xb+1=a(\dfrac{a}{a}-x)$
$xb+1=a(1-x)$
$a=\dfrac{xb+1}{1-x}$
Are both answers equivalent, or did I make a mistake the first time?
 A: Do everything you can to get the $a$s to one side of an equation and everything else to the other.
$x(a+b)=a-1$
$xa + xb = a-1$
So get everything that has anything to do with $a$ onto one side (say the LHS but it doesn't really matter which) and get everything that doesn't have anything to do with $a$ onto the other side.
$\underbrace{xa}_{\text{something to do with }a} + \underbrace{xb}_{\text{nothing to do with }a}=\underbrace{a}_{\text{something to do with }a} - \underbrace{1}_{\text{nothing to do with }a}$
You have something to do with $a$ on the right hand side.  Get it to the left hand side.
You have something that doesn't have anything to do with $a$ on the left hand side.  Get it to the right hand side.
$\underbrace{xa}_{\text{something to do with }a} -\underbrace{a}_{\text{something to do with }a} = - \underbrace{1}_{\text{nothing to do with }a}-\underbrace{xb}_{\text{nothing to do with }a}$
Can you finish up from there?

 Hint: factor

$a(x-1) = -1 -xb$



 And divide

$a = \frac {-1-xb}{x-1}$

  (It is important to know that we are assuming $x \ne 1$.  If $x = 1$ we can not solve for $a$.... If $x=1$ we have $a+b = a-1$ and we can conclude $b=-1$ but $a$ could be anything.)

A: $$ x = \frac{a-1}{a+b} $$
says that $x$  is the result of a Moebius transformation of $a,$  coefficient matrix
$$
\left(
\begin{array}{rr}
1& -1 \\
1 & b
\end{array}
\right)
$$
So, with the adjoint matrix
$$
\left(
\begin{array}{rr}
b& 1 \\
-1 & 1
\end{array}
\right)
$$
we find
$$  a = \frac{bx+1}{-x+1}  $$
