Question

solve for m. if there are multiple solutions, enter them as a list of values separated by commas. if there are no solutions, enter none.
\\(\frac{m}{m + 6} = \frac{72}{m^2 - 36} + 7\\)

m =

enter any extraneous solutions. if there are multiple extraneous solutions, enter them as a list separated by a comma. if there are no extraneous solutions, enter none.

m =

Explanation:

Step1: Factor the denominator

Notice that \(m^2 - 36=(m + 6)(m - 6)\) by the difference of squares formula \(a^2-b^2=(a + b)(a - b)\) where \(a = m\) and \(b = 6\). So the equation \(\frac{m}{m + 6}=\frac{72}{m^2-36}+7\) becomes \(\frac{m}{m + 6}=\frac{72}{(m + 6)(m - 6)}+7\).

Step2: Multiply through by the LCD

The least common denominator (LCD) of the fractions is \((m + 6)(m - 6)\). Multiply each term by \((m + 6)(m - 6)\) to eliminate the fractions:
\(m(m - 6)=72+7(m + 6)(m - 6)\)

Step3: Expand the expressions

Expand \(m(m - 6)\) to get \(m^2-6m\), and expand \(7(m + 6)(m - 6)\) using the difference of squares: \(7(m^2-36)=7m^2-252\). So the equation is:
\(m^2-6m=72 + 7m^2-252\)

Step4: Simplify the equation

First, move all terms to one side:
\(m^2-6m-7m^2+252 - 72=0\)
Combine like terms:
\(-6m^2-6m + 180=0\)
Divide through by \(-6\) to simplify:
\(m^2+m - 30=0\)

Step5: Factor the quadratic equation

Factor \(m^2+m - 30\). We need two numbers that multiply to \(-30\) and add to \(1\). The numbers are \(6\) and \(-5\). So:
\((m + 6)(m - 5)=0\)

Step6: Solve for \(m\)

Set each factor equal to zero:
\(m+6 = 0\) or \(m - 5=0\)
So \(m=-6\) or \(m = 5\)

Step7: Check for extraneous solutions

We must check if these solutions make the original denominators zero. The original denominators are \(m + 6\) and \(m^2-36=(m + 6)(m - 6)\).

• For \(m=-6\), the denominators \(m + 6=0\) and \(m^2-36 = 0\), so \(m=-6\) is extraneous.
• For \(m = 5\), \(m+6=11

eq0\) and \(m^2-36=25 - 36=-11
eq0\), so \(m = 5\) is a valid solution.

Answer:

(for \(m\)):
5