Chapter A.6, Problem 143AYU

To determine

To find: The equation that has no solution and give it to a fellow student to solve. Ask the fellow student to write a critique of your equation.

Answer to Problem 143AYU

$\frac{1}{x+3}+\frac{1}{x-3}=\frac{6}{x^2-9}$

With rational equations we must first note the domain, which is all real numbers except $x=-3$ and $x=3$. That is, these are the values of $x$ that will cause the equation to be undefined. Since the least common denominator of $x+3$, $x-3$, and $x^2-9$ is $\left(x+3)(x-3\right)$, we can mulitply each term by the LCD to cancel out the denominators and reduce the equation to $\left(m-3\right)+\left(m+3\right)=6$. Combining like terms, we end up with $2m=6$. Dividing both sides of the equation by the constant, we obtain an answer of $m=3$. However, this solution is NOT in the domain. Thus, there is NO SOLUTION because $m=3$ is an extraneous answer.

$9x+12=3\left(3x+4\right)$

When finding how many solutions an equation has you need to look at the constants and coefficients.

The coefficients are the numbers alongside the variables.

The constants are the numbers alone with no variables.

If the coefficients are the same on both sides then the sides will not equal, therefore no solutions will occur.

$9x+12=3\left(3x+4\right)$

Use distributive property on the right side first.

$9x+4=9x+12$

$4\ne 12$

No solutions.

Explanation of Solution

$\frac{1}{x+3}+\frac{1}{x-3}=\frac{6}{x^2-9}$

With rational equations we must first note the domain, which is all real numbers except $x=-3$ and $x=3$. That is, these are the values of $x$ that will cause the equation to be undefined. Since the least common denominator of $x+3$, $x-3$, and $x^2-9$ is $\left(x+3)(x-3\right)$, we can mulitply each term by the LCD to cancel out the denominators and reduce the equation to $\left(m-3\right)+\left(m+3\right)=6$. Combining like terms, we end up with $2m=6$. Dividing both sides of the equation by the constant, we obtain an answer of $m=3$. However, this solution is NOT in the domain. Thus, there is NO SOLUTION because $m=3$ is an extraneous answer.

$9x+12=3\left(3x+4\right)$

When finding how many solutions an equation has you need to look at the constants and coefficients.

The coefficients are the numbers alongside the variables.

The constants are the numbers alone with no variables.

If the coefficients are the same on both sides then the sides will not equal, therefore no solutions will occur.

$9x+12=3\left(3x+4\right)$

Use distributive property on the right side first.

$9x+4=9x+12$

$4\ne 12$

No solutions.