Question

How do you solve `1/ (2x+ 3) + 1/ (4x-3) = 5`?

Answer 1

`x=(-3+-3sqrt(6))/10`

Explanation:

`1/(2x+3)+1/(4x-3)=5`
Take LCM of denominators `(2x+3)(4x-3)`
Multiply both sides by `(2x+3)(4x-3)`
`(2x+3)(4x-3)(1/(2x+3)+1/(4x-3))=5(2x+3)(4x-3)`
`((2x+3)(4x-3))/(2x+3)+((2x+3)(4x-3))/(4x-3)=5(2x+3)(4x-3)`

Cancel Equal like terms
`(4x-3)+(2x+3)=5(2x+3)(4x-3)`
`6x=5(8x^2-6x+12x-9)`
`6x=5(8x^2+6x-9)`
`6x=40x^2+30x-45`
Subtract `6x` from both side
`6x-6x=40x^2+30x-45-6x`
`0=40x^2+24x-45`

`40x^2+24x-45=0`
comparing with `ax^2+bx+c=0`
`a=40, b=24, c=-45`
Solution `x=(-b+-sqrt(b^2-4ac))/(2a)`
`x=(-24+-sqrt((24)^2-4(40)(-45)))/(2(40))`
`x=(-24+-sqrt(576+7200))/80`
`x=(-24+-sqrt(7776))/80`
`x=(-24+-sqrt(1296xx6))/80`
`x=(-24+-36sqrt(6))/80`
`x=(-3+-3sqrt(6))/10`