• Dec 6th 2009, 09:12 AM
Candy101
ok so i got this question and i know how to slove it and everything but im lost as of where the teacher got this one number.. so here it is

(x+5)/(x-2) = 5/(x+2) + 28/(x^2-4)

so from that i got ..

x^2+10 = 5x-10+28

but.. the teacher got

x^2 + 7x +10 = 5x-10+28

where did he get 7x from???
• Dec 6th 2009, 09:36 AM
Massachusetts Cowboys
Quote:

Originally Posted by Candy101
ok so i got this question and i know how to slove it and everything but im lost as of where the teacher got this one number.. so here it is

(x+5)/(x-2) = 5/(x+2) + 28/(x^2-4)

x^2 + 7x +10 = 5x-10+28

where did he get 7x from???

Hmm, not sure myself. If i do the calculation for the left part

1. (x+5) multiplied by (x-2) since in order to get rid of the denumerator
2. gettin rid of brackets
3. x^2 -2X + 5X -10

I get +3x .. lol (Surprised)
• Dec 6th 2009, 09:45 AM
e^(i*pi)
Quote:

Originally Posted by Candy101
ok so i got this question and i know how to slove it and everything but im lost as of where the teacher got this one number.. so here it is

(x+5)/(x-2) = 5/(x+2) + 28/(x^2-4)

so from that i got ..

x^2+10 = 5x-10+28

but.. the teacher got

x^2 + 7x +10 = 5x-10+28

where did he get 7x from???

Note that due to the difference of two squares: $(x^2-4)=(x-2)(x+2)$. Also note that the best method is to get the same denominator for different terms, in this case it makes sense to use $x^2-4$ as it is the LCM of $x-2$ and $x+2$

Multiply the first term by $\frac{x+2}{x+2}$

Multiply the second term by $\frac{x-2}{x-2}$

This should now give each term a denominator of $x^2-4$ which cancels to

$(x+5)(x+2) = 5(x-2)+28$

$x^2+7x+10 = 5x+18
$

For this to work $x \neq \pm 2$

edit: see spoiler for detailed instructions

Spoiler:
$\frac{x+5}{x-2} = \frac{5}{x+2} + \frac{28}{x^2-4} = \frac{x+5}{x-2} = \frac{5}{x+2} + \frac{28}{(x-2)(x+2)}$

$\frac{x+5}{x-2} \cdot \frac{x+2}{x+2} = \frac{(x+5)(x+2)}{(x-2)(x+2)}$

$\frac{5}{x+2} \cdot \frac{x-2}{x-2} = \frac{5(x-2)}{(x-2)(x+2)}$

Combine all these now

$\frac{(x+5)(x+2)}{(x-2)(x+2)} = \frac{5(x-2)}{(x-2)(x+2)} + \frac{28}{x^2-4}$

$\frac{(x+5)(x+2)}{(x-2)(x+2)} = \frac{5(x-2)}{(x-2)(x+2)} + \frac{28}{(x-2)(x+2)}$

$(x-2)(x+2)$ will cancel

$(x+5)(x+2) = 5(x-2) + 28$

Expand using FOIL

$x^2+5x+2x+10 = 5x-10+28$

Simplify

$x^2+7x+10=5x+18$
• Dec 6th 2009, 09:56 AM
masters
Quote:

Originally Posted by Candy101
ok so i got this question and i know how to slove it and everything but im lost as of where the teacher got this one number.. so here it is

(x+5)/(x-2) = 5/(x+2) + 28/(x^2-4)

so from that i got ..

x^2+10 = 5x-10+28

but.. the teacher got

x^2 + 7x +10 = 5x-10+28

where did he get 7x from???

Hi Candy101,

$\frac{x+5}{x-2}=\frac{5}{x+2}+\frac{28}{x^2-4}$

First, multiply through by the common denominator, which is (x - 2)(x + 2).

$(x+5)(x+2)=5(x-2)+28$

Expand and set everything equal to zero. This is where you went wrong. You multiplied the first two binomials incorrectly.

$x^2+7x+10=5x-10+28$

$x^2+7x+10-5x+10-28=0$

$x^2+2x-8=0$

Factor.

$(x+4)(x-2)=0$

$x=-4$

$x \ne 2$ because it would cause the denominator to be zero in the first fraction.
• Dec 7th 2009, 11:38 AM
Candy101
hey thanks guys for the help!!!!!!!!!!!!!!!!!!!!!!!!!!