6.2 Linear equations in one variable  (Page 3/15)

We have three denominators: $x,$ $2,$ and $2x.$ No factoring is required. The product of the first two denominators is equal to the third denominator, so, the LCD is $2x.$ Only one value is excluded from a solution set, 0. $$ Next, multiply the whole equation (both sides of the equal sign) by $2x.$

The proposed solution is −1, $$ which is not an excluded value, so the solution set contains one number, $x=\mathrm{-1},$ or $\left\{\mathrm{-1}\right\}$ written in set notation.

First find the common denominator. The three denominators in factored form are $x,10=2\cdot 5,$ and $4x=2\cdot 2\cdot x.$ The smallest expression that is divisible by each one of the denominators is $20x.$ Only $x=0$ is an excluded value. Multiply the whole equation by $20x.$

The solution is $\frac{35}{2}.$

1. The denominators $x$ and $x-6$ have nothing in common. Therefore, the LCD is the product $x\left(x-6\right).$ However, for this problem, we can cross-multiply.

   $$\begin{array}{cccc}\hfill \frac{3}{x-6}& =& \frac{5}{x}\hfill & \\ \hfill 3x& =& 5(x-6)\hfill & \text{Distribute}\text{.}\hfill \\ \hfill 3x& =& 5x-30\hfill & \\ \hfill \mathrm{-2}x& =& \mathrm{-30}\hfill & \\ \hfill x& =& 15\hfill & \end{array}$$

   The solution is 15. $$ The excluded values are $6$ and $0.$

2. The LCD is $2\left(x-3\right).$ Multiply both sides of the equation by $2\left(x-3\right).$

   $$\begin{array}{ccc}\hfill 2(x-3)\left[\frac{x}{x-3}\right]& =& \left[\frac{5}{x-3}-\frac{1}{2}\right]2(x-3)\hfill \\ \hfill \frac{2\cancel{(x-3)}x}{\cancel{x-3}}& =& \frac{2\cancel{(x-3)}5}{\cancel{x-3}}-\frac{\cancel{2}(x-3)}{\cancel{2}}\hfill \\ \hfill 2x& =& 10-(x-3)\hfill \\ \hfill 2x& =& 10-x+3\hfill \\ \hfill 2x& =& 13-x\hfill \\ \hfill 3x& =& 13\hfill \\ \hfill x& =& \frac{13}{3}\hfill \end{array}$$

   The solution is $\frac{13}{3}.$ The excluded value is $3.$

3. The least common denominator is $2\left(x-2\right).$ Multiply both sides of the equation by $x\left(x-2\right).$

   $$\begin{array}{ccc}\hfill 2(x-2)\left[\frac{x}{x-2}\right]& =& \left[\frac{5}{x-2}-\frac{1}{2}\right]2(x-2)\hfill \\ \hfill 2x& =& 10-(x-2)\hfill \\ \hfill 2x& =& 12-x\hfill \\ \hfill 3x& =& 12\hfill \\ \hfill x& =& 4\hfill \end{array}$$

   The solution is 4. The excluded value is $2.$