Need help with simplifying rational expressions

I can't figure out how to do these; can someone run me through the process? And please, pretend you're explaining to someone who's a complete and utter dunce when it comes to math.

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Simplify the rational expression:

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Factor out the coefficients of *x on the numerator and denominator of the rational expression: *

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Simplify the rational expression:

*t*^{2} + 3*t* − 18 |

*t*^{2} − 12*t* + 27 |

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Simplify the rational expression.

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These were the ones I was having the most trouble on. I'm not asking you to do all of them for me (though feel free if you want;)), I just need to have it slowly explained how.

Re: Need help with simplifying rational expressions

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Originally Posted by

**InDrk** I can't figure out how to do these; can someone run me through the process? And please, pretend you're explaining to someone who's a complete and utter dunce when it comes to math.

There would be no point in trying to explain to a "complete and utter dunce". I would prefer to assume an intelligent person who has little background in algebra.

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Simplify the rational expression:

I presume you know that $\displaystyle \frac{a}{a}= 1$ (as long as $\displaystyle a\ne 0$).

95= 5(19) and 35= 5(7) is $\displaystyle \frac{95}{35}= \frac{5}{5}\frac{19}{7}= \frac{19}{7}$

$\displaystyle \frac{x^2}{x}= x\frac{x}{x}= x$

$\displaystyle \frac{y}{y}= 1$

so $\displaystyle \frac{95x^2y}{35xy}= \frac{19x}{7}$

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Factor out the coefficients of *x on the numerator and denominator of the rational expression: *

Okay, the "coefficient of x" in the numerator is -1 and 2= (-1)(-2). 2- x= -2(-1)+ x(-1)= -1(-2+ x)= -1(x- 2). The coefficient of x in the denominator 5: 5x- 10= 5(x-2). So, as long as x is not 2, we cancel the "x- 2" in numerator and denominator. That is $\displaystyle \frac{2- x}{5x- 10}= \frac{-1(x- 2)}{5(x- 2)}= \frac{-1}{5}$ which could also be written as $\displaystyle -\frac{1}{5}$.

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Simplify the rational expression:

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*t*^{2} + 3*t* − 18 |

*t*^{2} − 12*t* + 27 |

Start by trying to factor. 18= 2(3)(3) can be factored with two factors as either 2(9) or 6(3). 9- 2= 7, not 3 but 6- 3= 3. So $\displaystyle t^2+ 3x- 18= (t+ 6)(t- 3). 27= 3(3)(3$) can be factored with two factors are 3(9) and 9+ 3= 12! So $\displaystyle t^2- 12t+ 27= (t- 3)(t- 9)$. So the fraction is $\displaystyle \frac{t^2+ 3t- 18}{t^2- 12t+ 27}= \frac{(t+ 6)(t- 3)}{(t- 3)(t- 9)}$. Now you can cancel the two "t- 3" terms.

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Simplify the rational expression.

The denominator is (x- 2)+ 5= x+ 3 which has no factors in common with x- 2. So we cannot cancel anything and only have

$\displaystyle \frac{x- 2}{x+ 3}$ as the simplest form.

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These were the ones I was having the most trouble on. I'm not asking you to do all of them for me (though feel free if you want;)), I just need to have it slowly explained how.