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

\[|\frac{1-2x}{x}|=3x => \frac{1-2x}{x}=3x or \frac{1-2x}{x}=-3x\] \[x \neq 0\] \[1-2x=3x(x) or 1-2x=-3x(x)\] \[1-2x=3x^2 or 1-2x=-3x^2\] \[3x^2+2x-1=0 or 3x^2-2x+1=0\] \[3x^2+3x-x-1=0 or x=\frac{-(-2) \pm \sqrt{(-2)^2-4(3)(1)}}{2(3)}\] \[3x(x+1)-1(x+1)=0 or x=\frac{2 \pm \sqrt{4-4(3)}}{6}\] \[(x+1)(3x-1)=0 or x=\frac{2 \pm \sqrt{-8}}{6}\] 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

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

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

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

Join our real-time social learning platform and learn together with your friends!