Question

Simplify 1 over quantity x minus 3 plus 4 over x all over 4 over x minus 1 over quantity x minus 3

Answer 1

Given expression: $\frac{\frac{1}{x-3}+\frac{4}{x}}{\frac{4}{x}-\frac{1}{x-3}}$.

$\mathrm{Least\:Common\:Multiplier\:of\:}x,\:x-3:\quad x\left(x-3\right)$

$\mathrm{Adjust\:Fractions\:based\:on\:the\:LCM}$

$\frac{4}{x}-\frac{1}{x-3}=\frac{4\left(x-3\right)}{x\left(x-3\right)}-\frac{x}{x\left(x-3\right)}$

$=\frac{4\left(x-3\right)-x}{x\left(x-3\right)}$

$=\frac{3x-12}{x\left(x-3\right)}$

$\frac{4}{x}-\frac{1}{x-3}=\frac{x}{x\left(x-3\right)}+\frac{4\left(x-3\right)}{x\left(x-3\right)}$

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

$=\frac{5x-12}{x\left(x-3\right)}$

$\frac{\frac{1}{x-3}+\frac{4}{x}}{\frac{4}{x}-\frac{1}{x-3}}=\frac{\frac{5x-12}{x\left(x-3\right)}}{\frac{3x-12}{x\left(x-3\right)}}$

$\mathrm{Divide\:fractions}:\quad \frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a\cdot \:d}{b\cdot \:c}$

$=\frac{\left(5x-12\right)x\left(x-3\right)}{x\left(x-3\right)\left(3x-12\right)}$

$\mathrm{Cancel\:the\:common\:factor:}\:x$

$=\frac{\left(5x-12\right)\left(x-3\right)}{\left(x-3\right)\left(3x-12\right)}$

$\mathrm{Cancel\:the\:common\:factor:}\:x-3$

$=\frac{5x-12}{3x-12} \ \ \ : Final Answer.$

Answer 2

We can begin this problem by creating a common denominator. our fraction is:

(1/(x-3)+4/x)/(x-1)/(x-3), giving us the common denominator of x(x-3). We then take each individual fraction in the complex fraction, making it so that each one has the denominator x(x-3). 1/(x-3) becomes x/x(x-3), 4/x becomes 4(x-3)/x(x-3), and (x-1)/(x-3) becomes x(x-1)/x(x-3).

Now, our fraction is (x/x(x-3)+ 4(x-3)/x(x-3))/x(x-1)/x(x-3), or simply ((x+4(x-3)/x(x-3))/x(x-1)/x(x-3)). If we multiply the numerator and denominator by x(x-3), we get the fraction (x+4(x+3))/(x(x-1)). 

Distributing the 4, we get (x+4x+12)/(x(x-1)), or (3x+12)/(x(x-1)). (I don't know if you want the denominator factored or not, but if you want it expanded then it's (3x+12)/(x^2-x).)

I hope this was easy enough to follow! I haven't written anything like this before so I'm sorry if it wasn't very good.