Question

absolute value (1-2x/x)= 3x solve for x

Im thinking you divided by x and get 2x-1=3x^2 and then bring all to one side and get
3x^2-2x+1=0 but that doesn't simplify and the answer is x=1/3

Answer 1

\[|\frac{1-2x}{x}|=3x => \frac{1-2x}{x}=3x or \frac{1-2x}{x}=-3x\]
\[x \neq 0\]
\[1-2x=3x(x) or 1-2x=-3x(x)\]
\[1-2x=3x^2 or 1-2x=-3x^2\]
\[3x^2+2x-1=0 or 3x^2-2x+1=0\]
\[3x^2+3x-x-1=0 or x=\frac{-(-2) \pm \sqrt{(-2)^2-4(3)(1)}}{2(3)}\]
\[3x(x+1)-1(x+1)=0 or x=\frac{2 \pm \sqrt{4-4(3)}}{6}\]
\[(x+1)(3x-1)=0 or x=\frac{2 \pm \sqrt{-8}}{6}\]
but that second equation is not real
so we have
\[x+1=0 or 3x-1=0\]
\[x=-1 or x=\frac{1}{3}\]
but we need to check these

Answer 2

\[|\frac{1-2(-1)}{(-1)}|=3(-1) \] => x=-1 does not work
now lets check x=1/3

Answer 3

\[|\frac{1-2\cdot \frac{1}{3}}{\frac{1}{3}}|=|\frac{\frac{1}{3}}{\frac{1}{3}}|=1=3(\frac{1}{3}) \]
so x=1/3 is our only real solution

Answer 4

how did you get 3x^2+2x-x-1?