What are some fast ways to solve a system of 3 linear equations with this special form?

Perhaps an algebraic matrix ?

Try using Gauss-Jordan Elimination.

Your equations are three mutually perpendicular planes intersecting at a point, which is the solution. I’d solve for Y in the first equation (eq1) and substitute into eq2. Now solve for Z and substitute into eq3, which will then have only X as a variable, which is solved directly. I didn’t actually work it out, however.

I will look at the solutions later, but I suspect that if you told me the solution for X, I could tell you the solutions for Y and Z by symmetry. For example, I am guessing that the coefficients of a and b in the solution of X are the same respectively as the coefficients of c and d in the solution of Y.

In the extreme case where a=c=e and b=d=f, the solution is easy to find, because we must have X=Y=Z.

I overcame my laziness and used Cramer’s rule to solve for X.
X = d – ae + abf

Based only on symmetry, I guessed that we would have:
Y =e – fb + bcd
Z = f – cd + ace

This is indeed the correct solution. Unfortunately, I have no idea of how knowledge of symmetry can simplify finding the value for X.

One could argue that Cramer’s rule demonstrates a certain symmetry, since the solution for a variable is the the determinant of the coefficient matrix with the the variable coefficients replaced by the solution values and then dividing by the determinant of the original coefficient matrix. The present problem is simple enough to easily see the symmetry.
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One final note. There is additional symmetry in the current problem as evidenced by the symmetry of the determinant of the coefficient matrix, which is 1 + abc.

Last note (I promise). The solutions I gave for X, Y and Z are incorrect. They all have to be divided by the coefficient determinant of (1+abc).

Scientific calculator!
Why not think of using technology when fundamentals are understood and practised.

You can try this online calculator to solve the equations.
Here is an excel file I found on google to solve equations. Also, this has the procedure to solve in excel.

The equations can be arranged as below.

1*X + a*Y + 0*Z = b
0*X + 1*Y + c*Z = d
e*X + 0*Y + 1*Z = f

I assumed that you are going to put values for constants – a, b, c, d, f. If it is required to find in its given general form, then we need paper work.

Personally, I would have tried elimination. But, things get messed up if a mistake is made in calculations. It takes much time to go back and find the mistake, and then correct.
Using the matrix to solve equations, as @cockswain and @28lorelei have pointed out, makes it easier to find mistakes in calculations.

Here, at last, is the most succinct way of expressing what I wanted to say.
After finding the solution of X, we can express it as
X = f(a,d,b,e,c,f)

Then do circular rotations on the variables to get:
Y = f(b,e,c,f,a,d) and
Z = f(c,f,a,d,b,e)

X is to Y as Y is to Z as Z is to X

Here’s how I did the Guass-Jordan, though obviously it would be far quicker with actual numbers.

Inspired by your equations for 3 variables, I wondered if there would be the same symmetry for equations with 2 variables. Any pair of equations in 2 variables where none of the coefficients are 0, can be expressed as:
x+ay=b
cx+y=d

If you express x as x=f(a,b,c,d) then
y=f(c,d,a,b)

Not very useful, but pretty

@LostInParadise With two variables, it is very easy to solve. You can divide the second equation by c and get-
x + ay = b
x + y/c = d/c
Subtract the second from the first, and you get-
ay – y/c = b – d/c
Then factor out stuff to get.
y(a – 1/c) = b – d/c
y = (b – d/c)/(a – 1/c)
y = (b – d/c)/(a – 1/c) * (c/c)

y = (bc – d)/(ac – 1)

I wonder if a similar approach works for three variables.

I assume you are familiar with Gaussian elimination, which is considered the most efficient way of solving linear equations. I was looking for symmetric relationships on purely aesthetic grounds.