Cross Multiplying rational equation--Need help!!

[muggsy]

I still cannot get this problem and not sure if I am close, got it right or way off.....Can someone help?

I will show you my work thus far.....

Problem: x-1 4 1
----- + --------- = -------
x+3 xSquared +5x+6 x+2


Solving it as: x-1 4 1
----- ------- = -----
x+3 (x+3)(x+2) x+2



3 1
---------------------- = -------
2x+5x+3+6 x+2




7x+9 1
-------- = --------
x+2



7x+9 = x+2

7x+9-2 = x+2-2

7x-7 = x

7x = 7

x=1

 

[soroban]

Hello, muggsy!

\(\displaystyle \text{Problem: }\;\frac{x-1}{x+3} + \frac{4}{x^2+5x+6} \;=\;\frac{1}{x+2}\)

\(\displaystyle \text{Solving it as: }\;\;\frac{x-1}{x+3} + \frac{4}{(x+3)(x+2)} \;=\;\frac{1}{x+2}\) . . . . This is correct.

. . . . . . . . . . . . . . . . . . . .\(\displaystyle \frac{3}{2x+5x+3+6} \;=\;\frac{1}{x+2}\) .. . . . What?

\(\displaystyle \text{Multiply through by }(x+3)(x+2):\)

. . \(\displaystyle \frac{(x+3)(x+2)}{1}\cdot \frac{x-1}{x+3} \;+\; \frac{(x+3)(x+2)}{1}\cdot\frac{4}{(x+3)(x+2)} \;\;=\;\;\frac{(x+3)(x+2)}{1}\cdot\frac{1}{x+2}\)

. . \(\displaystyle (x+2)(x-1) + 4 \;=\;x+3\)

. . . . \(\displaystyle x^2 + x - 2 + 4 \;=\;x + 3\)

. . . . . . . . . . . .\(\displaystyle x^2 \;=\;1\)

. . . . . . . . . . . . \(\displaystyle x \;=\;\pm1\)