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simplifying complex fractions
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\[[\frac{1*x+(4*(x-3))}{x*(x-3)}]: \frac{(4*(x-3)-1*x)}{x(x-3)}\]
\[\frac{x+4x-12}{x^2-3x}: \frac{4x-12-x}{x^2-3x}\]
so it would be D?
\[\frac{5x-12}{x^2-3x}:\frac{3x-12}{x^2-3x}\]
I don't know yet because I not done
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\[\frac{5x-12}{x^2-3x}\times\frac{x^2-3x}{3x-12}\]x^2-3x cancle\[\frac{5x-12}{3x-12}=\frac{5x-12}{3(x-4)}\]here your answer
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