Question

What Did I Do Wrong?!?

View Equation and My Work

Answer 1

\[(x^2+2x/12x+54)-(3-x/8x+36)=\]
\[(x^2+2x/6(2x+9))-(3-x/12(2x+9))=\]
\[[(2)(x^2+2x)/(2)(6(2x+9))]-[(3-x)(3)/(4(2x+9))(3)]=\]
\[(2x^2+4x)/(12(2x+9))-(9-3x)/(12(2x+9))=\]
\[(2x^2+4x-9-3x)/(12(2x+9))=\]
\[(2x^2+x-9)/(12(2x+9))\]

Answer 2

\[\large \frac {x^2+2x}{12x+54}-\frac {3-x}{8x+36}=\]
is it this problem?

Answer 3

Yes

Answer 4

\[\large \frac {x^2+2x}{12x+54}-\frac {3-x}{8x+36}\]
\[\large =\frac {x(x+2)}{6(2x+9)}-\frac {3-x}{4(2x+9)}\]
\[\large =\frac {x(x+2)}{6(2x+9)}+\frac {x-3}{4(2x+9)}\]

get common denominators next...

Answer 5

12(2x+9)