Question

plz solve......
$\frac{1}{x - 1} + \frac{2}{x - 2} = \frac{3}{x - 3}$
I will mark as brainiest if answered correctly with steps

Answer 1

1/x-1 + 2/x-2 = 3/x-3
= 1(x-2) + 2(x-1) ÷ (x-1)(x-2)=3/x-3
=x-2+2x-2/x(x-2)-1(x-2)=3/x-3
=x-4/x2-2x-x+2=3/x-3
=x-4/x2-3x+2=3/x-3
simplifying x2-3x+2
=x2-1x-2x+2=0
=x(x-1)-2(x-1)=0
=(x-2)(x-1)=0
=x=2 and x=1
x-4/2=3/x-3
=x-4(x-3)=2(3x)
=x2-3x-4x+12=6x
x2-3x-4x-6x+12=o
x2-13x+12=o
and x-4/1=3/x-3
x-4(x-3) =3
=x2-3x-4x+12=3
=x2-7x+9 =0
Therefore, x2-13+12=0 and x2-7x+9=0

Answer 2

Solution:-

given by:-
$\frac{1}{x - 1} + \frac{2}{x - 2} = \frac{3}{x - 3} \\ \frac{1}{x - 1} + \frac{2}{x - 2} - \frac{3}{x - 3} = 0 \\ \frac{(x - 2)(x - 3) + 2(x - 1)(x - 3) - 3(x - 1)(x - 2)}{(x - 1)(x - 2)(x - 3)} = 0 \\ = {x}^{2} - 3x - 2x + 6 + 2 {x}^{2} - 6x - 2x + 6 - 3 {x}^{2} + 6x + 3x - 6 = 0 \\ = - 5x - 8x + 9x + 6 = 0 \\ = - 4x = - 6 \\ = x = \frac{6}{4} \\ =( x = \frac{3}{2} )$
☆i hope its help☆