• Nov 8th 2006, 01:23 AM
whoa

1.) 13/x^2-9x+14 = 5/x-3 + 3x-7/x^2+2x-15

2.) 3x/x+1 - 1/3 = 5/x+1

3.) 11/x+5 - 5/x-3 = 3x-7/x^2+2x-15

4.) 17/y + 13/y+2 = 14/y^2-12y

5.) y/y-8 - 48/y^2-64 = y/y+8

thank you for your kind consideraton :-)
• Nov 8th 2006, 03:21 AM
CaptainBlack
Quote:

Originally Posted by whoa

1.) 13/x^2-9x+14 = 5/x-3 + 3x-7/x^2+2x-15

2.) 3x/x+1 - 1/3 = 5/x+1

3.) 11/x+5 - 5/x-3 = 3x-7/x^2+2x-15

4.) 17/y + 13/y+2 = 14/y^2-12y

5.) y/y-8 - 48/y^2-64 = y/y+8

thank you for your kind consideraton :-)

and unambiguous. Otherwise the helpers are going to have to guess
what you mean and may waste their time answering the wrong questions

RonL
• Nov 8th 2006, 04:34 AM
Soroban
Hello, whoa!

Is there a typo in #1?
We end up with an ugly cubic equation.

I'll do #2 . . .

Quote:

$2)\;\frac{3x}{x+1} - \frac{1}{3} \:= \:\frac{5}{x+1}$

Find the LCD and multiply through by it . . .

The LCD is: $3(x+1)$

We have: . $3(x+1)\cdot\frac{3x}{x+1} -\, 3(x+1)\cdot\frac{1}{3} \;= \;3(x+1)\cdot\frac{5}{x+1}$

Cancel and get: . $3\cdot3x - (x + 1) \;=\;3\cdot5$

. . .Solve for $x\!:$ . . . . $9x - x - 1 \;=\;15$

. . . . . . . . . . . . . . . . . . . . $8x \;= \;16$

. . . . . . . . . . . . . . . . . . . . . $\boxed{x\;=\;2}$

• Nov 8th 2006, 05:17 AM
earboth
nr. 3 only
Quote:

Originally Posted by whoa
3.) 11/x+5 - 5/x-3 = 3x-7/x^2+2x-15
...
thank you for your kind consideraton :-)

Hello, whoa,

1. The LCD is $(x+5)(x-3) = x^2 +2x-15$

2. Multiply by LCD:

$11(x-3)-5(x+5) = 3x-7$

3. Multiply the brackets and rearrange:

$11x-33-5x-25=3x-7\ \Longleftrightarrow \ 3x=51$

So x = 17
• Nov 8th 2006, 05:30 AM
earboth
nr. 5 this time
Quote:

Originally Posted by whoa
5.) y/y-8 - 48/y^2-64 = y/y+8
thank you for your kind consideraton :-)

Hello, whoa,

1. The LCD is: $(y+8)(y-8)=y^2-64$

2. Multiply by LCD

$y(y+8)-48=y(y-8)$. Multiply the brackets and rearrange:

$y^2+8y-48=y^2-8y\ \Longleftrightarrow\ 3y=48$

Thus y = 3

EB
• Nov 8th 2006, 06:55 AM
Soroban
Hello, whoa!

Quote:

$1)\;\;\frac{13}{x^2-9x+14} \:= \:\frac{5}{x-3} + \frac{3x-7}{x^2+2x-15}$

We have: . $\frac{13}{(x-2)(x-7)} \;=\;\frac{5}{x-3} + \frac{3x-7}{(x-3)(x+5)}$

The LCD is: $(x-2)(x-3)(x-7)(x+5)$

Multiply through by the LCD:

. . $13(x-3)(x+5) \:=\:5(x-2)(x-7)(x+5) + (3x-7)(x-2)(x-7)$

which simplifies to that ugly cubic I promised you:

. . $8x^3 - 80x^2 + 94x + 447 \;= \;0$ . . . ack!

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I conclude that you copied the first fraction incorrectly.

Problem #1 is expected to be simple and straight-forward.
. . I suspect that its denominator is $x + 5$.

Can you please get back to us?

• Nov 8th 2006, 12:40 PM
whoa
Thank you guys!

oh by the way, i'd copied number 1 right, I think my math instructor made a mistake.;)