Question 201183
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1)  Let *[tex \Large x] represent the width.  Then the length is *[tex \Large 4x - 1].


The area is the product of the length and the width, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(4x - 1) = 60]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x^2 - x - 60 = 0]


Solve the quadratic (it factors).  Exclude the negative root because you are looking for a positive measure of length.  The positive root will be your width.


2) *[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2x}{x^2 - 1}-\frac{x+2}{x^2 + 3x - 4}]


Factor both denominators:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2x}{(x + 1)(x - 1)}-\frac{x+2}{(x + 4)(x - 1)}]


The LCD is then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x + 1)(x - 1)(x + 4)]


Apply the LCD:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2x(x + 4)}{(x + 1)(x - 1)(x + 4)}-\frac{(x + 2)(x + 1)}{(x + 1)(x - 1)(x + 4)}]


Distribute and apply FOIL to the numerators:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2x^2 + 8x}{(x + 1)(x - 1)(x + 4)}-\frac{x^2 + 3x + 2}{(x + 1)(x - 1)(x + 4)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2x^2 + 8x}{(x + 1)(x - 1)(x + 4)}+\frac{-(x^2 + 3x + 2)}{(x + 1)(x - 1)(x + 4)}]


Add the numerators and collect like terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x^2 + 5x - 2}{(x + 1)(x - 1)(x + 4)}]


Since the numerator does not have rational factors, there are no common factors with the denominator, so expand the denominator and you are done:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x^2 + 5x - 2}{(x^2 - 1)(x + 4)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x^2 + 5x - 2}{x^3 + 4x^2 - x - 4}]




John
*[tex \LARGE e^{i\pi} + 1 = 0]
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