Question

Solving Equations Find all real solutions of the equation.

$\frac{1}{x}+\frac{2}{x-1}=3$

Short answer

The given equation has two real solutions$1+\sqrt{\frac{2}{3}}$ and $1-\sqrt{\frac{2}{3}}$.

Step-by-step solution

Step 1. Definition of equation and quadratic formula.

Equation is a mathematical statement in which two expressions are equal to each other.

If a quadratic equation is $a{x}^2+bx+c=0,a\ne 0$, then the quadratic formula is:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Step 2. Given Information.

The given equation is:

$\frac{1}{x}+\frac{2}{x-1}=3$

Step 3. Simplify the given equation.

LCM of$x$ and$x-1$ is $x\left(x-1\right)$. Multiply both sides by$x\left(x-1\right)$ in the given equation.

$\begin{array}{c}x\left(x-1\right)\left(\frac{1}{x}+\frac{2}{x-1}\right)=3x\left(x-1\right)\\ \frac{x\left(x-1\right)}{x}+\frac{2x\left(x-1\right)}{x-1}=3x\left(x\right)+3x\left(-1\right)\\ \left(x-1\right)+2x=3x^2-3x\\ 3x-1=3x^2-3x\\ 0=3x^2-3x-3x+1\\ 0=3x^2-6x+1\end{array}$

Step 4. Solve the given equation by quadratic formula.

In the quadratic equation $3x^2-6x+1=0$, the required values for quadratic formula are $a=3,b=-6,c=1$.

Substitute$a=3,b=-6,c=1$ in the quadratic formula.

$\begin{array}{c}x=\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4\left(3\right)\left(1\right)}}{2\left(3\right)}\\ =\frac{6\pm \sqrt{36-12}}{6}\\ =\frac{6}{6}\pm \frac{\sqrt{24}}{6}\\ =1\pm \sqrt{\frac{24}{36}}\\ =1\pm \sqrt{\frac{2}{3}}\end{array}$

Thus, the given equation has two real solutions$1+\sqrt{\frac{2}{3}}$ and $1-\sqrt{\frac{2}{3}}$.