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 :-)

2. 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 :-)
Please add brackets to make the meaning of these expressions clear
and unambiguous. Otherwise the helpers are going to have to guess
what you mean and may waste their time answering the wrong questions

RonL

3. Hello, whoa!

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

I'll do #2 . . .

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

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

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

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

4. ## nr. 3 only

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

5. ## nr. 5 this time

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

6. Hello, whoa!

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

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?

7. Thank you guys!

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