Problem 48 - Algebra A Combined Function

Question

The current on a portion of the Mississippi River is 3 miles per hour. A barge can go 6 miles upstream in the same amount of time it takes to go 10 miles downstream. Find the speed of the boat in still water. Let \(x\) be the speed of the boat in still water. $$ \begin{array}{|l|ll|} \hline & \boldsymbol{r} & \times & \boldsymbol{t} & = & \boldsymbol{d} \\ \hline \text { upstream } & x-3 & & & 6 \\ \hline \text { downstream } & x+3 & & & 10 \\ \hline \end{array} $$

Step-by-Step Solution

Answer

The speed of the boat in still water is 12 mph.

Step 1: Set Up Equations for Travel Time

We know the relationship between distance \(d\), rate \(r\), and time \(t\) is given by \(d = r \times t\). This means the time it takes to travel a distance \(d\) at speed \(r\) is \(t = \frac{d}{r}\). Now, write down the equations for the time it takes to travel upstream and downstream. - Upstream: \( \text{time} = \frac{6}{x-3} \)- Downstream: \( \text{time} = \frac{10}{x+3} \)

Step 2: Set Up Equation for Equal Times

Since the time to travel upstream is equal to the time to travel downstream, set the equations from Step 1 equal to each other: \[ \frac{6}{x-3} = \frac{10}{x+3} \]

Step 3: Solve for \(x\) - Cross Multiply

Cross-multiplying the equation from Step 2 gives: \[ 6(x+3) = 10(x-3) \] This helps eliminate the fractions, simplifying the equation.

Step 4: Expand and Simplify

Expand the equation from Step 3:\[ 6x + 18 = 10x - 30 \]Now, simplify it by bringing the terms involving \(x\) to one side and constant terms to the other side.

Step 5: Isolate \(x\)

Rearrange the equation from Step 4:\[ 18 + 30 = 10x - 6x \]This simplifies to:\[ 48 = 4x \]Now, divide both sides by 4 to solve for \(x\).

Step 6: Solve for \(x\)

Divide both sides of the equation by 4:\[ x = \frac{48}{4} \]\[ x = 12 \]The speed of the barge in still water is 12 miles per hour.

Key Concepts

Distance-rate-time problems

When you're tackling word problems related to distance, rate, and time, it's essential to remember the fundamental relationship:

• Distance (\(d\)) is equal to Rate (\(r\)) multiplied by Time (\(t\)): \[d = r \times t\].

This principle can help you solve various problems by rearranging it to find the unknown quantity:

• Rate: \(r = \frac{d}{t}\)
• Time: \(t = \frac{d}{r}\)

In this particular problem, the boat moves upstream against the current and downstream with the current. This changes its effective speed:

• Upstream, the boat's speed is reduced: \(x - 3\)
• Downstream, the boat's speed is increased: \(x + 3\)

Both routes take the same time, setting up the foundation for solving the problem using these equations.

Solving equations

To find the speed of the boat in still water, we need to solve equations where time is a key element. Given that the times to travel upstream and downstream are equal, you have:

• Upstream time: \( \frac{6}{x-3} \)
• Downstream time: \( \frac{10}{x+3} \)

Setting these equal allows us to formulate an equation:

• \[ \frac{6}{x-3} = \frac{10}{x+3} \]

This relationship can seem confusing at first, but the solution lies in manipulating the equation correctly. You can cross-multiply to eliminate the fractions, bringing you closer to the solution without delving into too many steps or getting lost in the process.

Cross multiplication

Cross multiplication is a handy technique for solving equations involving fractions. It helps you simplify equations by eliminating the fractions:

• From: \[ \frac{6}{x-3} = \frac{10}{x+3} \]
• To: \[ 6(x+3) = 10(x-3) \]

This step is crucial as it removes the fractions and provides a simplified linear equation. By expanding:

• \(6(x+3) = 6x + 18\)
• \(10(x-3) = 10x - 30\)

The next step involves equating the expanded forms and rearranging them:

• \(6x + 18 = 10x - 30\)

Simplifying further by bringing like terms together, we solve for \(x\) by isolating it, resulting in:

• \[x = 12\]

Cross multiplication turns a seemingly complex problem into a straightforward calculation, highlighting its power and utility in algebra.