Verified Answer :

Ryan2

\(\frac{1}{x-1}-\frac{3}{x+2}=\frac{1}{4}\\ \\ \frac{x+2-3(x-1)}{(x-1)(x+2)}=\frac{1}{4}\\ \\ \frac{x+2-3x+3}{(x-1)(x+2)}=\frac{1}{4}\\ \\ \frac{5-2x}{x^2+x-2}=\frac{1}{4}\\ \\ 4(5-2x)=x^2+x-2\\ \\ 20-8x=x^2+x-2\\ \\ \boxed{x^2+9x-22=0}\)

Solvin this quadratic equation with Bhaskara formula we find:

x = -11 or x = 2

kate200468

\(\frac{1}{x-1} - \frac{3}{x+2} = \frac{1}{4} \ /\cdot 4(x-1)(x+2)\\\\\frac{4(x-1)(x+2)}{x-1} - \frac{3\cdot 4(x-1)(x+2)}{x+2} = \frac{4(x-1)(x+2)}{4}\\\\4(x+2)-12(x-1)=(x-1)(x+2)\\\\4x+8-12x+12=x^2+2x-x-2\\\\-8x+20=x^2+x-2\\\\-x^2-9x+22=0\ /\cdot(-1)\\\\x^2+9x-22=0\ \ \ \Rightarrow\ \ \ x^2+11x-2x-22=0\\\\x(x+11)-2(x+11)=0\\\\(x+11)(x-2)=0\ \ \ \ \ \Leftrightarrow\ \ \ (x+11=0\ \ \ or\ \ \ x-2=0)\\\\x=-11\ \ \ or\ \ \ x=2\)

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