Answer
The solution set of the rational equation $\frac{1}{4}+\frac{1}{t}=\frac{1}{3}$ is $\left\{ 12 \right\}$.
Work Step by Step
$\frac{1}{4}+\frac{1}{t}=\frac{1}{3}$
Note that, $t\ne 0$.
The denominators of the rational expressions are $4$, $t$, and $3$.
On solving the LCD is $12t$.
Multiply both sides by $12t$.
$12t\left( \frac{1}{4}+\frac{1}{t} \right)=12t\left( \frac{1}{3} \right)$
Apply the distributive property,
$12t\cdot \left( \frac{1}{4} \right)+12t\cdot \left( \frac{1}{t} \right)=12t\left( \frac{1}{3} \right)$
Simplify the terms,
$\begin{align}
& \frac{12t}{4}+\frac{12t}{t}=\frac{12t}{3} \\
& \frac{3\cdot 4\cdot t}{4}+\frac{12t}{t}=\frac{3\cdot 4\cdot t}{3} \\
& 3t+12=4t \\
\end{align}$
Apply the addition principle, subtract $3t$ from both the sides of the equation:
$\begin{align}
& 3t+12-3t=4t-3t \\
& 12=t \\
& t=12
\end{align}$
Now, check the solution of the equation $\frac{1}{4}+\frac{1}{t}=\frac{1}{3}$:
Substitute $t=12$ in $\frac{1}{4}+\frac{1}{t}=\frac{1}{3}$:
$\begin{align}
\frac{1}{4}+\frac{1}{\left( 12 \right)}\overset{?}{\mathop{=}}\,\frac{1}{3} & \\
\frac{3+1}{12}\overset{?}{\mathop{=}}\,\frac{1}{3} & \\
\frac{4}{12}\overset{?}{\mathop{=}}\,\frac{1}{3} & \\
\frac{1}{3}=\frac{1}{3} & \\
\end{align}$
So, $t=12$ is the solution of $\frac{1}{4}+\frac{1}{t}=\frac{1}{3}$.
