New Notes Blog: `18/(x^2-3x)-6/(x-3)=5/x` Solve the equation by using the LCD. Check for extraneous solutions.

Tuesday, 14 January 2014

$\displaystyle\frac{{18}}{{{{x}}^{{2}}-{3}{x}}}-\frac{{6}}{{{x}-{3}}}=\frac{{5}}{{x}}$ Solve the equation by using the LCD. Check for extraneous solutions.

LCD is an acronym for least common denominator. It is the product of distinct factors on the denominator side. Basically, find LCD is the same as finding the LCM (least common multiple) of the denominators.

For the given equation $\displaystyle\frac{{18}}{{{{x}}^{{2}}-{3}{x}}}-\frac{{6}}{{{x}-{3}}}=\frac{{5}}{{x}}$ , the denominators are $\displaystyle{\left({{x}}^{{2}}-{3}{x}\right)}$ , $\displaystyle{\left({x}-{3}\right)}$ , and $\displaystyle{x}$ . Note: The factored form of the denominator $\displaystyle{{x}}^{{2}}-{3}{x}$ is $\displaystyle{x}{\left({x}-{3}\right)}$ .

Based on the list of factors, The distinct factors are $\displaystyle{x}$ and...

Based on the list of factors, The distinct factors are $\displaystyle{x}$ and $\displaystyle{\left({x}-{3}\right)}.$

Thus, $\displaystyle{L}{C}{D}={x}\cdot{\left({x}-{3}\right)}$ or $\displaystyle{\left({{x}}^{{2}}-{3}{x}\right)}$ .

To simplify the equation, we multiply each term by the LCD.

$\displaystyle\frac{{18}}{{{{x}}^{{2}}-{3}{x}}}\cdot{\left({{x}}^{{2}}-{3}{x}\right)}-\frac{{6}}{{{x}-{3}}}\cdot{x}\cdot{\left({x}-{3}\right)}=\frac{{5}}{{x}}\cdot{x}\cdot{\left({x}-{3}\right)}$

Cancel out common factors to get rid of the denominators.

$\displaystyle{18}-{6}\cdot{x}={5}\cdot{\left({x}-{3}\right)}$

Apply distribution property.

$\displaystyle{18}-{6}{x}={5}{x}-{15}$

Subtract 18 from both sides.

$\displaystyle{18}-{6}{x}-{18}={5}{x}-{15}-{18}$

$\displaystyle-{6}{x}={5}{x}-{33}$

Subtract $\displaystyle{5}{x}$ from both sides of the equation.

$\displaystyle-{6}{x}-{5}{x}={5}{x}-{33}-{5}{x}$

$\displaystyle-{11}{x}=-{33}$

Divide both sides by $\displaystyle-{11}$ .

$\displaystyle\frac{{-{11}{x}}}{{-{11}}}=\frac{{-{33}}}{{-{11}}}$

$\displaystyle{x}={3}$

To check for extraneous solution, plug-in $\displaystyle{x}={3}$ on $\displaystyle\frac{{18}}{{{{x}}^{{2}}-{3}{x}}}-\frac{{6}}{{{x}-{3}}}=\frac{{5}}{{x}}$ .

$\displaystyle\frac{{18}}{{{{3}}^{{2}}-{3}\cdot{3}}}-\frac{{6}}{{{3}-{3}}}=?\frac{{5}}{{3}}$

$\displaystyle\frac{{18}}{{{9}-{9}}}-\frac{{6}}{{0}}=?\frac{{5}}{{3}}$

$\displaystyle\frac{{18}}{{0}}-\frac{{6}}{{0}}=?\frac{{5}}{{3}}$

undefined -undefined=? $\displaystyle\frac{{5}}{{3}}$ FALSE.

Note: Any value divided by $\displaystyle{0}$ results to undefined value.

An undefined result implies the x value is an extraneous solution.

Therefore, the $\displaystyle{x}={3}$ is an extraneous solution.

There is no real solution for the given equation $\displaystyle\frac{{18}}{{{{x}}^{{2}}-{3}{x}}}-\frac{{6}}{{{x}-{3}}}=\frac{{5}}{{x}}$ .

New Notes Blog

Tuesday, 14 January 2014

$\displaystyle\frac{{18}}{{{{x}}^{{2}}-{3}{x}}}-\frac{{6}}{{{x}-{3}}}=\frac{{5}}{{x}}$ Solve the equation by using the LCD. Check for extraneous solutions.

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Tuesday, 14 January 2014

Solve the equation by using the LCD. Check for extraneous solutions.

No comments:

Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle\frac{{18}}{{{{x}}^{{2}}-{3}{x}}}-\frac{{6}}{{{x}-{3}}}=\frac{{5}}{{x}}$ Solve the equation by using the LCD. Check for extraneous solutions.

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?