Question

solve for X :
one upon (x minus 1) into( x minus 2 )+ 1 upon x minus 2) into (x minus 3) is equal to 2 upon 3

Answer 1

1/(x - 1)(x - 2) + 1/(x - 2)(x - 3) = 2/3

or, 1/(x - 2) [ 1/(x - 1) + 1/(x - 3) ] = 2/3 ]

or, 1/(x - 2) [ (x - 3 + x - 1)/(x - 1)(x - 3) ] = 2/3

or, 1/(x - 2) [ (2x - 4)/(x² - 4x + 3) ] = 2/3

or, (2x - 4)/(x - 2)(x² - 4x + 3) = 2/3

or, 3(2x - 4) = 2(x - 2)(x² - 4x + 3)

or, 6x- 12 = 2(x³ - 4x² + 3x - 2x² + 8x - 6)

or, 6x - 12 = 2(x³ - 6x² + 11x - 6)

or, 6x - 12 = 2x³ - 12x² + 22x - 12

or, 2x³ - 12x² + 16x = 0

or, 2x(x² - 6x + 8) = 0

or, 2x(x² - 4x - 2x + 8) = 0

or, 2x(x - 4)(x - 2) = 0

x = 0, 2 , 4

but x ≠ 2 because if we take x = 2 then equation will be undefined.

therefore ,x = 0, 4

Answer 2

Answer:

Step-by-step explanation:

$\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}=\frac{2}{3}$

$\frac{(x-3)+(x-1)}{(x-1)(x-2)(x-3)}=\frac{2}{3}$

$\frac{2x-4}{(x-1)(x-2)(x-3)}=\frac{2}{3}$

$\frac{2(x-2)}{(x-1)(x-2)(x-3)}=\frac{2}{3}$

$\frac{(x-2)}{(x-1)(x-2)(x-3)}=\frac{1}{3}$

$\frac{1}{(x-1)(x-3)}=\frac{1}{3}$

$(x-1)(x-3)=3\\\\x^2-4x+3=3\\\\x^2-4x=0\\\\x(x-4)=0\\x=0,4$