help

[ibanez1608]

1. . . . . . . x+5. . . 1
---- + ----- . . = ----
X^2+x. . . . 3-3x^2. . . . 3x-3

 

[MarkFL]

I am assuming you have been given to solve:

$\displaystyle \dfrac{1}{x^2+x}+\dfrac{x+5}{3-3x^2}=\dfrac{1}{3x-3}$

My first step would be to completely factor the 3 denominators so that I can see what is the simplest factor I need to multiply both sides of the equation with to clear all 3 denominators.

 

[soroban]

Hello, ibanez1608!

Is there a typo?
Since this is listed in Pre-Algebra
. . and the equation has no real roots . . .

$\displaystyle \displaystyle\frac{1}{x^2+x} + \frac{x+5}{3-3x^2} \:=\:\frac{1}{3x-3}$

Factor the first denominator: .$\displaystyle x^2+x \:=\: x(x+1)$

Factor the second denominator: .$\displaystyle 3-3x^2 \:=\:-3(x^2-1) \:=\:-3(x-1)(x+1)$

Factor the third denominator: .$\displaystyle 3x-3 \:=\:3(x-1)$


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

. . . . . . . . . $\displaystyle \displaystyle\frac{1}{x(x+1)} - \frac{x+5}{3(x-1)(x+1)} \;=\;\frac{1}{3(x-1)}$


Multiply by $\displaystyle 3x(x-1)(x+1)\!:$

. . $\displaystyle 3(x-1) - x(x+5) \:=\: x(x+1)$

. . . .$\displaystyle 3x - 3 - x^2 - 5x \:=\: x^2 + x$

. . . . . . $\displaystyle 2x^2 + 3x + 3 \:=\:0$


This quadratic equation has no real roots.

 

[ibanez1608]

thank u

All u need to do is to simple the equation
Thank u soroban