Question

solve for x: 1/x-3=1/x-9+2/x-6,where x not equal to 3,6,9​

Answer 1

Answer:$x =\frac{9-3\sqrt5}{2} \ or \ x=\frac{9+3\sqrt5}{2}$

Step-by-step explanation:

$\implies \frac{1}{x-3}=\frac{1}{x-9}+\frac{2}{x-6}\\\therefore \ \frac{1}{x-3}=\frac{1\times(x-6)}{(x-9)\times(x-6)}+\frac{2\times(x-9)}{(x-9)\times(x-6)}\\\therefore \ \frac{1}{x-3}=\frac{(x-6)}{(x-9)(x-6)}+\frac{2(x-9)}{(x-9)(x-6)}\\\therefore \ \frac{1}{x-3}=\frac{(x-6)+(2x-18)}{(x-6)(x-9)}\\\therefore \ (x-6)(x-9)=(x-3)(3x-24)\\\therefore \ (x-6)(x-9)=3(x-3)(x-8)\\\\\therefore \ x^2-(9+6)x+54=3[x^2-(3+8)x+24]\\\therefore \ x^2-15x+54=3[x^2-11x+24]\\\therefore \ x^2-15x+54=3x^2-33x+72$

$\therefore \ 2x^2-18x+18=0\\\therefore \ x^2-9x+9=0\\\therefore \ x^2-9x+\frac{81}{4}+9-\frac{81}{4}=0\\\therefore \ x^2-9x+\frac{81}{4}-\frac{45}{4}=0\\\therefore \ (x-\frac{9}{2})^2-(\frac{3\sqrt5}{2})^2=0\\\therefore \ (x-\frac{9}{2}+\frac{3\sqrt5}{2})(x-\frac{9}{2}-\frac{3\sqrt5}{2})=0\\\therefore \ (x-\frac{9-3\sqrt5}{2})(x-\frac{9+3\sqrt5}{2})=0\\\therefore \ x =\frac{9-3\sqrt5}{2} \ or \ x=\frac{9+3\sqrt5}{2}$

Answer 2

Answer:

Step-by-step explanation:

$\frac{1}{x-3}$ = $\frac{1}{x-9}$  + $\frac{2}{x-6}$

RHS

= x-6+2x-18\$x^{2}$ - 9x-6x+54

= 3x-24\$x^{2}$ - 15x+54

RHS=LHS

1\x-3=3x-24\$x^{2}$ - 15x+54

$x^{2}$ - 15x+54 = (3x-24)(x-3)

$x^{2}$ - 15x+54=$3x^{2}$ -24x-9x-72

$x^{2}$ - 15x+54=$3x^{2}$ - 33x-72

$3x^{2}$ -$x^{2}$ -33x+15x -72-54= 0

2x^{2} - 18x-126=0

it has no real roots