I think it can be done with algebra but there are lots of equations to do! Compared to T&E which is relatively quick
You sure? I make it 9 equations if you think about it ...............
And there is an obvious 10th equation sat there as well.
Ah trial and error
I didn't even use/look at the grid. I just wrote the equations out and crossed out the duplicate. 1st column, 1st row and 3rd column gave me three letters right off. After that it was easy.

I think I have the obvious 10th equation that you speak of, but I don't see how you get nine equations from the table. Four rows and four columns make eight, but one of the rows is the same as one of the columns, so you're left with just seven.

Ah sorry my mistake. 7 equations from the grid plus another obvious one made up from the fact that each letter is an unique number from 1 to 9.![]()


Pretty clever puzzle. Had to assume to values of two symbols at one stage, but that assumption proved to be correct as all other numbers then fell perfectly into place.![]()

I came upon this and there is a solution which means it no longer becomes an assumption but fact...
You probably did it a lot quicker than me though just going on an assumption![]()
Yeah I got to the same bit and I thought it could be one of three numbers but saw that in fact it couldn't if you thought about it and could only be one.
Would have been quicker to try all 3 than work it out.