Multiply out the brackets on RHS
3ax + 2a + 4bx -5b = 34x - 31
collect terms on RHS
3ax + 4bx + 2a - 5b = 34x - 31
so 3a + 4b = 34
and 2a - 5b = -31
now solve the two simultaneous equations to find values of a and b
Yes, Fibonacci's approach will give you the values of a and b. (Both are positive integers less than 10.)
However that won't give a solution for the value of x
Does the question ask you to solve for x?
If so then you don't need to find a and b.
Just expand the brackets, rearrange to have all the x items on one side and then rearrange to find the value of x. I can give the answer if you want but I assume you don't want it yet
Can I join in as I like questions like this! factor is right as usual, what was the wording of the question? If solve for x then follow factor's middle post and know you can't actually end up with a numerical value for x, just an expression.
To save me checking back, here's my solution in case you eventually want it.
Assuming the question asks you to find the value of x:
3ax +2a + 4bx-5b= 34x-31
3ax +4bx-34x = 5b-2a-31
x(3a+4b-34) = 5b-2a-31
So x = (5b-2a-31)/(3a+4b-34)
However if the original 'equation' was in fact an identity and asked you to find the values of a and b that satisfy the identity, then (using Fibonacci's method) the values are a=2 and b =7
Thanks all of you. I have checked back to the question which asked only to find the values of a and b which I managed OK. I had come to the conclusion that a numerical value for x could not be obtained and was well on the way to the final solution as in factor 30's last message. Quite pleased with myself after a 60 year break from algebra!
Are "a" and "b" part of the equation? If so, you have 3 unknowns in one equation and I don't see how this can be solved. However, it was some time ago that I was at school.