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Help with fractions that contain exponents, factoring and LCD?

I have three problems that I can't completely figure.  I appreciate any help and explanations!

17        -    6

x2-12x+35      x-7

x+7          +    x +5

x2+7x-18            x2+3x-54

I know the second fraction needs to be flipped so it can be multiplied.

7c+49   Divided by    c2-49          So it would become:   7c+49   X    7c              Then:  (7c+49) X 7

3                           7c                                                    3             c2-49                         3(c2-49)

Kevin S. |
5.0 5.0 (4 lesson ratings) (4)
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When finding the LCD, it is often helpful to factor the denominators, then see what you have:

In the first problem, our denominators are

x2 - 12x + 35     and x - 7

x2 - 12x + 35  factors into (x - 5)(x - 7).

So your LCD is (x - 5)(x - 7)

So, multiply the second term by (x-5)/(x-5) to get your new fractions:

17/[(x - 5)(x - 7)] - [6(x - 5)]/[(x - 5)(x - 7)]

simplify the second numerator and rewrite:

[17 - 6x +30]/[(x - 5)(x - 7)] = [ -6x + 47]/[(x - 5)(x - 7)]

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For the second one, approach the LCD the same way. You'll just wind up with three factors in the denominator as opposed to two

For the third, you've got a good start. Now factor 7 from (7c + 49) and factor (c2 - 49) and I think you'll see terms that you can reduce.

Asok B. | Ph. D in Biochemistry, willing to teach Chemistry, Biochemistry, MathPh. D in Biochemistry, willing to teach ...
4.7 4.7 (18 lesson ratings) (18)
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Problem#1

Let's factorize the denominator x2-12x +35.

x2-12x +35 = x2-7x- 5x+ 35 = x(x-7)-5(x-7) = (x-7)(x-5)

now, 17/ (x2-12x+35) -6/(x-7) = 17/ (x-7)(x-5)-6/(x-7) =    { 17- 6(x-5)}  /(x-7)(x-5)

= (17-6x+30)/ (x-7)(x-5) = (47-6x)/(x-7)(x-5

Problem#2

x2+7x-18 = x2+9x-2x-18= x(x+9)-2(x+9)= (x+9)(x-2)

x2+3x-54 = x2+9x-6x-54= x(x+9)-6(x+9) = (x+9)(x-6)

So, (x+7) /(x2+7x-18)   +  (x+5)/(x2+3x-54) =  (x+7)/(x+9)(x-2) + (x+5)/(x+9)(x-6)

= { (x+7)(x-6) + (x+5)(x-2)} / (x+9)(x-2)(x-6)

={x2+7x-6x-42 + x2+5x-2x-10}/ (x+9)(x-2)(x-6)

=  (2x2 +4x-52)/(x+9)(x-2)(x-6)

=2(x2+2x-26)/(x+9)(x-2)(x-6)

Problem#3

The numerator is (7c+49)/3 = 7(c+7)/3

The denominator is (C2-49)/7c = (c+7) (c-7)/7C

So the fraction= 7(c+7)/3 divided by (c+7) (c-7)/7C = 49c/3(c-7)= 49c/3c-21