Solve for : `x :(x-1)/(x-2)+(x-3)/(x-4)=3 1/3,x!=2,4`

Explanation

The given equation is
`(x-1)/(x-2)+(x-3)/(x-4)=(10)/(3)`
`implies" "((x-1)(x-4)+(x-3)(x-2))/((x-2)(x-4))=(10)/(3)`
`implies" "((x^(2)-5x+4)(x^(2)-5x+6))/((x^(2)-6x+8))=(10)/(3)implies(2x^(2)-10x+10)/(x^(2)-6x+8)=(10)/(3)`
`implies" "3(2x^(2)-10x+10)=10(x^(2)-6x+8)`
`implies" "6x^(2)-30x+30=10x^(2)-60x+80`
`implies" "4x^(2)-30x+50=0implies2x^(2)-15x+25=0." "...(i)`
This equation is of the form `ax^(2)+bx+c=0,` where a=2, b=-15 and c=25.
`:." "D=(b^(2)-4ac)={(-15)^(2)-4xx2xx25}=(225-200)=25gt0.`
So, the given equation has real roots.
Now, `sqrt(D)=sqrt(25)=5.`
`alpha=(-b+sqrt(D))/(2a)=((15+5))/(2xx2)=(20)/(4)=5,`
`beta=(-b-sqrt(D))/(2a)=((15-5))/(2xx2)=(10)/(4)=(5)/(2).`
Hence, the required values of x are 5 and `(5)/(2).`