Problem 179 - Elementary Algebra

Question

179\. Norris can row 3 miles upstream against the current in 1 hour, the same amount of time it takes him to row 5 miles downstream, with the current. Solve the system. \(\left\\{\begin{array}{l}r-c=3 \\ r+c=5\end{array}\right.\) (a) for \(r,\) his rowing speed in still water. (b) Then solve for \(c,\) the speed of the river current.

Step-by-Step Solution

Answer

Norris's rowing speed is 4 mph, and the speed of the river current is 1 mph.

Step 1: Write Down the System of Equations

The system of equations given is: \[ \begin{array}{l} r - c = 3 \ r + c = 5 \end{array} \]

Step 2: Add the Equations

Add the two equations to eliminate the variable c. \[ (r - c) + (r + c) = 3 + 5 \] This simplifies to: \[ 2r = 8 \]

Step 3: Solve for r

Divide both sides of the equation by 2: \[ r = \frac{8}{2} = 4 \] So, Norris's rowing speed in still water, \( r \), is 4 miles per hour.

Step 4: Substitute r into One of the Original Equations

Substitute \( r = 4 \) into the first original equation: \[ 4 - c = 3 \]

Step 5: Solve for c

Rearrange the equation to solve for \( c \): \[ c = 4 - 3 = 1 \] Thus, the speed of the river current, \( c \), is 1 mile per hour.

Key Concepts

solving systems of equations

When we talk about solving systems of equations, we refer to finding the values of variables that satisfy all equations in the system simultaneously. In our case, Norris's rowing problem can be represented by two linear equations:

• row speed against the current: \(r - c = 3\)
• row speed with the current: \(r + c = 5\)

These equations capture two different but related rowing scenarios. The key to solving such systems is to find a method that isolates one of the variables, making it easier to solve for the other. There are several methods to solve systems of equations, including substitution and graphing. Here, we will use the elimination method for clarity and simplicity.

linear equations

A linear equation is an equation where each term is either a constant or the product of a constant and a single variable. The general form of a linear equation in one or two variables is:

• In one variable: \(ax + b = 0\)
• In two variables: \(ax + by = c\)

Our problem involves two linear equations: \(r - c = 3\) and \(r + c = 5\). These equations are considered linear because each term is either a constant or a variable raised to the power of one. Unlike quadratic or higher-order equations, linear equations graph as straight lines. When solving these two equations simultaneously, the goal is to find the intersection point that satisfies both lines.

speed and distance problems

Speed and distance problems often involve using the relationship:

• Speed = Distance / Time

In our rowing problem, Norris rows 3 miles upstream in 1 hour and 5 miles downstream in 1 hour. Here, the speed upstream and downstream can be expressed as:

• Upstream speed: \(r - c\)
• Downstream speed: \(r + c\)

The system of equations represents these relationships and allows us to solve for the unknowns (Norris's rowing rate and the river current). When combined properly, these equations provide valuable pieces of information that help us determine the speeds.

elimination method

The elimination method involves adding or subtracting equations to eliminate one of the variables. For our system:

• \(r - c = 3\)
• \(r + c = 5\)

By adding these equations, we eliminate \(c\):

• \((r - c) + (r + c) = 3 + 5\)

This simplifies to:

• \(2r = 8\)

Solving for \(r\), we divide both sides by 2:

• \(r = 4\)

Now, knowing \(r\), we substitute \(r = 4\) back into one of the original equations to find \(c\):

• \(4 - c = 3\)

Solving for \(c\), we get:

• \(c = 1\)

This method is efficient and straightforward, providing us with the values of both unknowns.