Question

(2x+3)/(x-4) - (2x-8)/(2x+1) = 1
find x

Answer 1

It is really long...but here you go.
Work on the left side of the equation first. Since they are fraction being subtracted you want to find a common denominator. So on each side you need to multiply by the other denominator......[[(2x+3)/(x-4)] *(2x+1)] - [[(2x-8)/(2x+1)] * (x-4)]
Writing it another way for you to easily solve.....
[(2x+3)*(2x+1) / (x-4)*(2x+1)] - [(2x-8)*(x-4) / (x-4)*(2x+1)]

You need to factor out the numerator and denominator of both of these giving you...

[(4x^2+9x+3) / (2x^2-7x-4)] - [(2x^2-16x+32) / (2x^2-7x-4)]
See how both fractions now have the same denominator. You can then combine them into one fraction.
[(4x^2+9x+3) - (2x^2-16x+32)] / (2x^2-7x-4) but you still need to keep the top part of this fraction separate to take into account the negative sign will distribute over giving you.....
(4x^2+9x+3-2x^2+16x-32) / (2x^2-7x-4) then you will need to combine like terms on your numerator....
(2x^2+25x-29) / (2x^2-7x-4) = 1 now that you have just one simplified fraction on the left side you can now start to solve for x....kind of.
Multiply the denominator to both sides of the equation.
(2x^2-7x-4) * [(2x^2+25x-29) / (2x^2-7x-4)] = (2x^2-7x-4) *1
This cancels out your denominator on the left side leaving you...
(2x^2+25x-29) = (2x^2-7x-4) *1 multiplying by 1 leaves
(2x^2+25x-29) = (2x^2-7x-4) Then you can subtract the like terms from each side cancelling out the 2x^2 and leaving you

25x-29 = -7x-4 adding 7x to each side

32x-29 = -4 and adding 29 to each side

32x = 25 and solving for x

x=25/32 yay!! that didn't take ten hundred steps at all haha....I hope I got the right answer for you and all of my math was right I would definitely suggest double checking the math to make sure. hope this helps :)