Answer:
[tex]x=-9[/tex] or [tex]x=3[/tex]
Step-by-step explanation:
Cross multiply:
[tex]3/x+6=x/9\\\\\\(3) * (9)=x*(x+6)\\\\\\27=x^{2} +6x[/tex]
Subtract x^2 +6x from both sides:
[tex]27-(x^{2}+6x )=x^{2} +6x-(x^{2} +6x)\\\\\\-x^{2} -6x+27=0[/tex]
Factor left side of the equation:
[tex](-x+3)(x+9)=0[/tex]
Set factors equal to 0:
[tex]-x+3=0[/tex] or [tex]x+9=0[/tex]
Therefore, [tex]x=3[/tex] or [tex]x=-9[/tex]
hope this helps.....
