Ryann and Lydia are taking canoe ride across Smithville lake. They can go 14 miles across the lake in the same time it takes them to go 9 miles back against the current. If their speed in still water is 6 mph , What is the rate of the current? Round your answer to the nearest tenth. Type your answer...

Let's denote the rate of the current as \( c \) mph. When Ryann and Lydia are going across the lake, their effective speed is \( 6 + c \) mph. When they are going back against the current, their effective speed is \( 6 - c \) mph. Given that they can go 14 miles across the lake in the same time it takes them to go 9 miles back against the current, we can set up the following equation based on the time taken for each trip: \[ \frac{14}{6+c} = \frac{9}{6-c} \] Now, we can solve this equation to find the rate of the current \( c \). Solve the equation by following steps: - step0: Solve for \(c\): \(\frac{14}{6+c}=\frac{9}{6-c}\) - step1: Find the domain: \(\frac{14}{6+c}=\frac{9}{6-c},c \in \left(-\infty,-6\right)\cup \left(-6,6\right)\cup \left(6,+\infty\right)\) - step2: Cross multiply: \(14\left(6-c\right)=\left(6+c\right)\times 9\) - step3: Simplify the equation: \(14\left(6-c\right)=9\left(6+c\right)\) - step4: Calculate: \(84-14c=9\left(6+c\right)\) - step5: Calculate: \(84-14c=54+9c\) - step6: Move the expression to the left side: \(84-14c-\left(54+9c\right)=0\) - step7: Calculate: \(30-23c=0\) - step8: Move the constant to the right side: \(-23c=0-30\) - step9: Remove 0: \(-23c=-30\) - step10: Change the signs: \(23c=30\) - step11: Divide both sides: \(\frac{23c}{23}=\frac{30}{23}\) - step12: Divide the numbers: \(c=\frac{30}{23}\) - step13: Check if the solution is in the defined range: \(c=\frac{30}{23},c \in \left(-\infty,-6\right)\cup \left(-6,6\right)\cup \left(6,+\infty\right)\) - step14: Find the intersection: \(c=\frac{30}{23}\) The rate of the current is \( c = \frac{30}{23} \) mph. Rounded to the nearest tenth, the rate of the current is approximately 1.3 mph.