Question

5x-8/x-2 +6x-44/x-7 -10x-8/x-1 =x-8/x-6
value of x=
I need the calculation also.
not only the answer
pls help!!!​

Answer 1

Answer:

x = 4

Step-by-step explanation:

$\frac{5x-8}{x-2} + \frac{6x-44}{x-7} -\frac{10x-8}{x-1} = \frac{x-8}{x-6}$

$or, \frac{5x-8}{x-2} + \frac{6x-44}{x-7} = \frac{x-8}{x-6} +\frac{10x-8}{x-1}$

$or, \frac{(5x-8)(x-7)+(x-2)(6x-44)}{(x-2)(x-7)} = \frac{(x-1)(x-8)+(x-6)(10x-8)}{(x-6)(x-1)}$ $or, \frac{5x^2-43x+56+ 6x^2-56x+88}{x^2-9x+14} = \frac{x^2-9x+8+10x^2-68x+48}{x^2-7x+6}$

$or, \frac{11x^2-99x+144}{x^2-9x+14} =\frac{11x^2-77x+56}{x^2-7x+6}$

$or, \frac{11x^2-77x+56-22x+88}{x^2-9x+14} =\frac{11x^2-77x+56}{x^2-7x+6}$

$or, \frac{11x^2-77x+56}{x^2-9x+14} -\frac{22x-88}{x^2-9x+14}=\frac{11x^2-77x+56}{x^2-7x+6}$

$or, \frac{11x^2-77x+56}{x^2-9x+14} -\frac{11x^2-77x+56}{x^2-7x+6}=\frac{22x-88}{x^2-9x+14}$

$or, (11x^2-77x+56)(\frac{1}{x^2-9x+14} -\frac{1}{x^2-7x+6})=\frac{22x-88}{x^2-9x+14}$

$or, (11x^2-77x+56)(\frac{x^2-7x+6-x^2+9x-14}{(x^2-9x+14)(x^2-7x+6)})=\frac{22x-88}{x^2-9x+14}$

$or, (11x^2-77x+56)(\frac{2x-8}{x^2-7x+6})=\frac{22x-88}{1}$   (multiple both side with x²-9x+14 )

$or, (\frac{11x^2-77x+56}{x^2-7x+6})(2x-8) - (22x-88)= 0$

$or, (\frac{11x^2-77x+56}{x^2-7x+6})2(x-4) - 22(x-4)= 0$

$or, (2\times\frac{11x^2-77x+56}{x^2-7x+6}-22)(x-4) = 0$

$so,\ \ \frac{11x^2-77x+56}{x^2-7x+6}-11 = 0 \ \ \ \ or, \ x-4 = 0$

$so,\ \ \frac{11x^2-77x+56-11x^2+77x-66}{x^2-7x+6} = 0 \ \ \ \ or, \ x= 4$

Now, x = 4

and , $\frac{11x^2-77x+56-11x^2+77x-66}{x^2-7x+6} = 0$

$or , \frac{-10}{x^2-7x+6} = 0$   (not acceptable)

Final answer x = 4