Question

x/3 +3/6-x = 2(6+x)/15 ; x is not equal to 6​

Answer 1

Answer:

$\frac{x}{3} + \frac{3}{6-x} = \frac{2(6+x)}{15}$

$\frac{6x-x^2+9}{18-3x} = \frac{12+2x}{15}$

$15(6x-x^2+9)=(18-3x)(12+2x)$

$90x-15x^2+135=216+36x-36x-6x^2$

$90x-15x^2+135=216-6x^2$

$90x-15x^2+6x^2+135-216=0$

$-9x^2+90x-81=0$

$-9(x^2-10x+9)=0$

$-9(x^2-x-9x+9)=0$

$-9[x(x-1)-9(x-1)]=0$

$-9(x-1)(x-9)=0$     [If we shift -9 from LHS to RHS the RHS becomes 0/-9

$(x-9)(x-1)=0$          which becomes 0 because anything divided by 0  

$x=9,1$                                     becomes 0]