Motorboat and Current

**sologuitar** · December 10th 2009, 08:38 PM

A motorboat traveling with the current can go 30 miles in 2 hrs. Against the current , it takes 3 hrs. to go the same distance. Find the rate of the motorboat in calm water and the rate of the current!

My set up:

Let x = speed of motorboat in still water

Let y = speed of current

See attachment for table I created.

From the table, I created the following system of equations but got the wrong answers.

2x + 2y = 30
3x - 3y = 30

What did I do wrong?

**Soroban** · December 11th 2009, 03:30 AM

Hello, sologuitar!

Your table is correct and your equations are correct.
It must be your algebra.

We have: . $\begin{array}{ccccccc}2x + 2y \;=\; 30 & \Rightarrow & x + y \;=\; 15 & [1] \\ 3x - 3y \;=\;30 & \Rightarrow & x - y \;=\; 10 & [2] \end{array}$

Add [1] and [2]: . $2x \:=\:25 \quad\Rightarrow\quad x \:=\:\frac{25}{2}$

Substitute into [1]: . $\frac{25}{2} + y \:=\:15 \quad\Rightarrow\quad y \:=\:\frac{5}{2}$

Therefore: . $\begin{array}{cc}\text{Motorboat} & 12\frac{1}{2}\text{ mph} \\ \\[-4mm]\text{Current} & 2\frac{1}{2}\text{ mph} \end{array}$

**sologuitar** · December 11th 2009, 04:54 AM

Originally Posted by Soroban

Hello, sologuitar!

Your table is correct and your equations are correct.
It must be your algebra.

We have: . $\begin{array}{ccccccc}2x + 2y \;=\; 30 & \Rightarrow & x + y \;=\; 15 & [1] \\ 3x - 3y \;=\;30 & \Rightarrow & x - y \;=\; 10 & [2] \end{array}$

Add [1] and [2]: . $2x \:=\:25 \quad\Rightarrow\quad x \:=\:\frac{25}{2}$

Substitute into [1]: . $\frac{25}{2} + y \:=\:15 \quad\Rightarrow\quad y \:=\:\frac{5}{2}$

Therefore: . $\begin{array}{cc}\text{Motorboat} & 12\frac{1}{2}\text{ mph} \\ \\[-4mm]\text{Current} & 2\frac{1}{2}\text{ mph} \end{array}$

Believe it or not, I got the same answers but decided to reject the fractions. How about that?

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