Question

solve for x :
$\frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = 3 \frac{1}{3}$
;x is not equals to 2 and 4​

Answer 1

I've done half, rest you shoukd do ok

Thank you

Answer 2

$\bf\large\red{\underline{\underline{Question}}}:$

• Solve for x : $\dfrac{x-1}{x-2}+\dfrac{x-3}{x-4}=3\dfrac{1}{3}$

$\bf\large\red{\underline{\underline{Solution}}}:$

$\longmapsto \dfrac{x-1}{x-2}+\dfrac{x-3}{x-4}=3\dfrac{1}{3}$

• Converting mixed fraction into proper fraction :

$\dashrightarrow \dfrac{x-1}{x-2}+\dfrac{x-3}{x-4}=\dfrac{10}{3}$

• Taking L.C.M :

$\dashrightarrow \dfrac{(x-1)(x-4)+(x-3)(x-2)}{(x-2)(x-4)} = \dfrac{10}{3}$

• Opening brackets by multiplying :

$\rm\dashrightarrow \dfrac{[x(x-4)-1(x-4)]+ [x(x-2)-3(x-2)]}{x(x-4) -2(x-4)}=\dfrac{10}{3}$

$\rm\dashrightarrow\dfrac{[x^2-4x-x+4]+[x^2-2x-3x+6]}{x^2-4x-2x+8}=\dfrac{10}{3}$

•Simplifying by adding like terms :

$\dashrightarrow\dfrac{[x^2-5x+4]+[x^2-5x+6]}{x^2-6x+8}=\dfrac{10}{3}$

$\dashrightarrow\dfrac{x^2-5x+4+x^2-5x+6}{x^2-6x+8}=\dfrac{10}{3}$

$\dashrightarrow\dfrac{2x^2-10x+10}{x^2-6x+8}=\dfrac{10}{3}$

• Cross multiplying :

$\dashrightarrow{3(2x^2-10x+10)}={10(x^2-6x+8)}$

$\dashrightarrow 6x^2-30x+30=10x^2-60x+80$

$\dashrightarrow 6x^2-30x+30- 10x^2+60x-80=0$

• Again ,Simplifying by adding like terms :

$\dashrightarrow 6x^2-30x+30- 10x^2+60x-80=0$

$\dashrightarrow-4x^2+30x-50=0$

$\dashrightarrow 4x^2-30x+50=0$

• Now solving the quadratic equation by factorisation method :

$\longmapsto 4x^2-30x+50=0$

$\dashrightarrow 4x^2-20x-10x+50=0$

$\dashrightarrow 4x(x-5)-10(x-5)=0$

$\dashrightarrow (x-5)(4x-10)=0$

If (x-5) = 0 ,then x = 5

If (4x-10) = 0 , then 4x = 10  ⇒ x = 5/2 = 2.5

$\bf\large\red{\underline{\underline{Therefore}}},$

• x = 5 or 2.5