# Show steps how to solve: (x - 2)/(x) + (1/2) = (x + 1)/(x - 2).

## Question:

Show steps how to solve: {eq}\frac{(x - 2)}{x} + \frac{1}{2} = \frac{(x + 1)}{(x - 2)} {/eq}.

The quadratic equation can be used to solve problems that follow this form, {eq}0 = ax^{2} + bx + c {/eq}. {eq}a {/eq}, {eq}b {/eq}, and {eq}c {/eq} are pulled from the equation and plugged into the quadratic equation to get an answer to the problem.

{eq}\displaystyle\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} {/eq}

Solve:

{eq}\displaystyle \frac{(x-2)}{x}+\frac{1}{2}=\frac{(x+1)}{(x-2)} {/eq}

We need to get rid of the denominators, by multiplying by {eq}2x(x-2) {/eq},

{eq}\displaystyle2x(x-2)(\frac{(x-2)}{x}+\frac{1}{2}=\frac{(x+1)}{(x-2)}) {/eq}

The result is,

{eq}\displaystyle 2(x-2)^{2}+x(x-2)=2x(x+1) {/eq}

Now we can distribute and combine like terms.

{eq}\begin{align} 2(x^{2}-4x+4)+x^{2}-2x&=2x^{2}+2x\\x^{2}-18x+8&=0\end{align} {/eq}.

Now we need to solve for x. We know that there will be two solutions because of the degree of x. For this example, we will use the quadratic equation:{eq}\displaystyle\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} {/eq}

We need to use the equation {eq}1x^{2}-12x+8=0 {/eq} to substitute {eq}a=1 {/eq}, {eq}b=-12 {/eq}, and {eq}c=8 {/eq} into the quadratic equation.

The steps are below:

{eq}\begin{align}\displaystyle\frac{12\pm \sqrt {(-12)^{2}-4(1)(8)}}{2(1)}\\\frac{12\pm \sqrt {144-32}}{2}\\\frac{12\pm \sqrt {112}}{2}\\\frac{12\pm 4\sqrt {7}}{2}\\\boxed{6\pm 2\sqrt {7}} \end{align} {/eq}