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

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.

Answer 2

Given expression:
`((1)/(x-3)+(4)/(x))/((4)/(x)-(1)/(x-3))`.

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

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

`(4)/(x)-(1)/(x-3)=(4\left(x-3\right))/(x\left(x-3\right))-(x)/(x\left(x-3\right))`

`=(4\left(x-3\right)-x)/(x\left(x-3\right))`

`=(3x-12)/(x\left(x-3\right))`

`(4)/(x)-(1)/(x-3)=(x)/(x\left(x-3\right))+(4\left(x-3\right))/(x\left(x-3\right))`

`=(x+4\left(x-3\right))/(x\left(x-3\right))`

`=(5x-12)/(x\left(x-3\right))`

`((1)/(x-3)+(4)/(x))/((4)/(x)-(1)/(x-3))=((5x-12)/(x\left(x-3\right)))/((3x-12)/(x\left(x-3\right)))`

`\mathrm{Divide\:fractions}:\quad ((a)/(b))/((c)/(d))=(a\cdot \:d)/(b\cdot \:c)`

`=(\left(5x-12\right)x\left(x-3\right))/(x\left(x-3\right)\left(3x-12\right))`

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

`=(\left(5x-12\right)\left(x-3\right))/(\left(x-3\right)\left(3x-12\right))`

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

`=(5x-12)/(3x-12) \ \ \ : Final Answer.`