I have tried to work out the following questions, but I am not sure if my answer is correct The question says to simplify.. enter image description here

What I basically did for the first one is combine the denominators and numerators, and multiplied them together to find the lowest common factor.

For the second question, I inverted the second part of the equation to make it a multiplication sign, then I just factored it out and cancelled anything I could.

Thanks, regards.

  • $\begingroup$ If you just want to check your answer, just use Wolfram Alpha... $\endgroup$ – fgp Mar 31 '14 at 11:40
  • $\begingroup$ Asking Wolfram Alpha about the first expression yields this $\endgroup$ – fgp Mar 31 '14 at 11:42
  • 1
    $\begingroup$ WA won't help many during an exam...and many times WA yields wrong/incomplete solutions. $\endgroup$ – DonAntonio Mar 31 '14 at 11:54

First recommended step: simplify as much as possible each numerator and denominator in each summand



Now, since the above happens to be the sum of two fractions with a common denominator, just...sum them:


For the first one directly do the fractions' (check carefully the minimal common denominator):

$$\frac y{x(x-y)}+\frac1x=\frac{y+x-y}{x(x-y)}=\frac x{x(x-y)}=\frac1{x-y}$$

In case the second equation has a division symbol instead of a plus one:



And we now use the basic rule $\;\frac ab\div\frac cd=\frac ab\cdot \frac dc\;$ :

$$(**)=\frac{\color{green}xy}{\color{red}{x-y}}\cdot\frac{\color{red}{x-y}}{\color{green}x^2}=\frac yx$$

  • $\begingroup$ I think OP's second expression has a division sign not addition. $\endgroup$ – user103816 Mar 31 '14 at 11:58
  • $\begingroup$ Hmmm...perhaps you're right, @Anupam . Let us wait until the OP addresses this matter. Thanks. $\endgroup$ – DonAntonio Mar 31 '14 at 12:01

For first expression you can try taking $\dfrac{1}{x}$ common in the denominator. $$\dfrac{y}{x^2-xy}+\dfrac{1}{x}=\dfrac{1}{x}(\dfrac{y}{x-y})+\dfrac{1}{x}=\dfrac{1}{x}(\dfrac{y}{x-y}+1)=\dfrac{1}{x}(\dfrac{y+x-y}{x-y})=\dfrac{1}{x-y}$$

For the second expression first take out common things apart, that is
$$\dfrac{{(xy)}^2}{x^2y-xy^2}=\dfrac{{(xy)}^2}{xy(x-y)}$$ And
And then proceed. The result of the first expression should come out $\dfrac{1}{x-y}$ and the result of the second should come out $\dfrac{y}{x}$.


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