What should be subtracted from \(\frac{7x}{x^2-x-12}\) to get \(\fr

1.	What should be subtracted from \(\frac{7x}{x^2-x-12}\) to get \(\frac{3}{x+3}\) ?A) \(\frac{1}{x-4}\)B) \(\frac{2}{x-4}\)C) \(\frac{4}{x-4}\)D) \(\frac{5}{x+4}\)
Answer» Correct option is (C) \(\frac{4}{x-4}\) Let p(x) be subtracted from \(\frac{7x}{x^2-x-12}\) to get \(\frac{3}{x+3}\) i.e., \(\frac{7x}{x^2-x-12}-p(x)=\) \(\frac{3}{x+3}\) \(\Rightarrow\) \(p(x)=\frac{7x}{(x-4)(x+3)}\) \(-\frac{3}{x+3}\) \(=\frac{7x-3(x-4)}{(x-4)(x+3)}\) \(=\frac{4x+12}{(x-4)(x+3)}\) \(=\frac{4(x+3)}{(x-4)(x+3)}\) \(=\frac{4}{x-4}\) Correct option is C) \(\frac{4}{x-4}\)

What should be subtracted from \(\frac{7x}{x^2-x-12}\) to get \(\frac{3}{x+3}\) ?A) \(\frac{1}{x-4}\)B) \(\frac{2}{x-4}\)C) \(\frac{4}{x-4}\)D) \(\frac{5}{x+4}\)

Answer»

Correct option is (C) \(\frac{4}{x-4}\)

Let p(x) be subtracted from \(\frac{7x}{x^2-x-12}\) to get \(\frac{3}{x+3}\)

i.e., \(\frac{7x}{x^2-x-12}-p(x)=\) \(\frac{3}{x+3}\)

\(\Rightarrow\) \(p(x)=\frac{7x}{(x-4)(x+3)}\) \(-\frac{3}{x+3}\)

\(=\frac{7x-3(x-4)}{(x-4)(x+3)}\) \(=\frac{4x+12}{(x-4)(x+3)}\)

\(=\frac{4(x+3)}{(x-4)(x+3)}\) \(=\frac{4}{x-4}\)

Correct option is C) \(\frac{4}{x-4}\)