Question

question 3. one fish aquarium has 50 gallons of water and is being filled at a rate of $\frac{1}{5}$ gallon each second. a second fish aquarium has 130 gallons of water and is being drained at a rate of $\frac{1}{3}$ gallon each second. solve the following equation, where $x$ is the number of seconds, to determine how many seconds pass until the two containers have the same amount: $50+\frac{1}{5}x = 130-\frac{1}{3}x$

Explanation:

Step1: Combine like - terms

First, combine the terms with $x$ on the right - hand side of the equation $50+\frac{1}{5}x = 130-\frac{1}{3}x$. Add $\frac{1}{3}x$ to both sides of the equation:
$50+\frac{1}{5}x+\frac{1}{3}x=130-\frac{1}{3}x+\frac{1}{3}x$.
$50 + (\frac{1}{5}+\frac{1}{3})x=130$.
Find a common denominator for $\frac{1}{5}$ and $\frac{1}{3}$, which is 15. Then $\frac{1}{5}+\frac{1}{3}=\frac{3 + 5}{15}=\frac{8}{15}$. So the equation becomes $50+\frac{8}{15}x=130$.

Step2: Isolate the term with $x$

Subtract 50 from both sides of the equation:
$50+\frac{8}{15}x - 50=130 - 50$.
$\frac{8}{15}x=80$.

Step3: Solve for $x$

Multiply both sides of the equation by the reciprocal of $\frac{8}{15}$, which is $\frac{15}{8}$:
$x = 80\times\frac{15}{8}$.
$x = 150$.

Answer:

$150$