##### write the value or values of the variable that make a denominator 0

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write the value or values of the variable that make a denominator 0

3/(x+3)-4/(x-3)=2x/(x^2-9)

May 26th, 2015

$\frac{3}{x+3}-\frac{4}{x-3}=\frac{2x}{x^2-9}$

To find the value for which denominator becomes 0, equate the denominator of each term to 0.

x + 3 = 0

x = -3

x - 3 = 0

x = 3

$x^2-9=0\\ x^2-3^2=0\\ (x+3)(x-3)=0\hspace{5}\big(because\hspace{5}a^2-b^2=(a+b)(a-b)\big)\\$

x + 3 = 0 or x - 3 = 0

x = -3 or x = 3

ANSWER:  For x = 3 and x = -3 the denominator becomes 0 for the given equation.

May 26th, 2015

what value would make the denominator equal to 0...that is the first part of the question and I don't understand... thx

May 26th, 2015

If you put x = 3 in the equation, it becomes

$\frac{3}{3+3}-\frac{4}{3-3}=\frac{2*3}{3^2-9}\\ \\ \frac{3}{6}-\frac{4}{0}=\frac{6}{9-9}\\ \\ \frac{3}{6}-\frac{4}{0}=\frac{6}{0}\\$

Here denominator of the 4/x-3  term and 2x/x^2-9 terms became 0.

If you put x = -3 in the equation it becomes

$\frac{3}{-3+3}-\frac{4}{-3-3}=\frac{2*(-3)}{(-3)^2-9}\\ \\ \frac{3}{0}-\frac{4}{-6}=\frac{6}{9-9}\\ \\ \frac{3}{0}-\frac{4}{-6}=\frac{6}{0}\\$

Here denominator of the 3/x+3  term and 2x/x^2-9 term became 0.

So, the values which make the denominator 0 are x = 3 and x = -3.

May 26th, 2015

May 26th, 2015

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May 26th, 2015
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May 26th, 2015
Oct 18th, 2017
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