Algebra homework

[GlenB]

I have been helping my grandson with algebra homework during the half term but we are struggling to solve the following equation.

a(3x + 2) + b(4x - 5) = 34x - 31

I am not asking for the answer but can someone tell me the first one or two steps please.

I am not sure this is the correct section of AB but couldn't see one more appropriate.

Thanks

[Fibonacci]

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

[factor-fiction]

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

[factor-fiction]

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

[factor-fiction]

Does the question use an equals sign or an identity sign (Ξ) ?
What is the exact wording of the question please?

[Prudie]

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.

[factor-fiction]

I'll check back on the thread tomorrow, GenB.

[factor-fiction]

<<<<SPOILER ALERT>>>>


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

[GlenB]

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!

[factor-fiction]

Happy to help.

[squarebear]

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.

[squarebear]

Ignore me. I didn't read the other responses.