Question

solve the equation.
\frac{8}{t} + \frac{12}{t+8} = \frac{6}{2}
solutions (separate by commas):
t =

Explanation:

Step1: Simplify the right - hand side

First, simplify \(\frac{6}{2}\) to \(3\). So the equation becomes \(\frac{8}{t}+\frac{12}{t + 8}=3\).

Step2: Find a common denominator

The common denominator of \(t\) and \(t + 8\) is \(t(t + 8)\). Multiply each term by \(t(t + 8)\) to clear the fractions:
\(t(t + 8)\times\frac{8}{t}+t(t + 8)\times\frac{12}{t + 8}=3\times t(t + 8)\)
Simplify each term:
For the first term, \(t\) cancels out, and we get \(8(t + 8)\).
For the second term, \(t + 8\) cancels out, and we get \(12t\).
For the right - hand side, we get \(3t(t + 8)=3t^{2}+24t\).
So the equation is \(8(t + 8)+12t = 3t^{2}+24t\).

Step3: Expand and simplify the left - hand side

Expand \(8(t + 8)\): \(8t+64 + 12t=3t^{2}+24t\).
Combine like terms on the left - hand side: \(20t + 64=3t^{2}+24t\).

Step4: Rearrange the equation to form a quadratic equation

Subtract \(20t+64\) from both sides to get \(3t^{2}+24t-(20t + 64)=0\).
Simplify: \(3t^{2}+4t - 64 = 0\).

Step5: Solve the quadratic equation

We can use the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for a quadratic equation \(ax^{2}+bx + c = 0\). Here, \(a = 3\), \(b = 4\), and \(c=-64\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(4)^{2}-4\times3\times(-64)=16 + 768 = 784\).
Then, \(t=\frac{-4\pm\sqrt{784}}{2\times3}=\frac{-4\pm28}{6}\).
We have two solutions:

• When we take the plus sign: \(t=\frac{-4 + 28}{6}=\frac{24}{6}=4\).
• When we take the minus sign: \(t=\frac{-4-28}{6}=\frac{-32}{6}=-\frac{16}{3}\).

We need to check for extraneous solutions because we multiplied by \(t(t + 8)\) (and \(t
eq0\) and \(t
eq - 8\)).
For \(t = 4\):
Left - hand side: \(\frac{8}{4}+\frac{12}{4 + 8}=2 + 1=3\), right - hand side: \(\frac{6}{2}=3\). So \(t = 4\) is a valid solution.
For \(t=-\frac{16}{3}\):
Left - hand side: \(\frac{8}{-\frac{16}{3}}+\frac{12}{-\frac{16}{3}+8}=\frac{8\times(-3)}{16}+\frac{12}{\frac{8}{3}}=-\frac{3}{2}+\frac{12\times3}{8}=-\frac{3}{2}+\frac{9}{2}=3\), right - hand side: \(\frac{6}{2}=3\). So \(t =-\frac{16}{3}\) is also a valid solution.

Answer:

\(4,-\frac{16}{3}\)