General Strategy for Solving Verbal Problems

1. Read the problem carefully.

2. Use diagrams if you think it will make the information clearer.

3. Find the relationship or formula relevant to the problem.

4. Identify the unknown quantity (or quantities), and label them, using one variable.

5. Write an equation involving the unknown quantity, using the relationship or formula

from step 3.

6. Solve the equation.

7. Answer the question.

8. Check the answer in the original words of the problem.

Example. Meg rowed her boat upstream a distance of 9 miles and then rowed back to the starting point. The total time of the trip was 10 hours. If the rate of the current was 4 mph, how fast would the boat have traveled in still water?

Solution. Let x = the rate of the boat in still water

x-4 = the rate of the boat traveling upstream (against the current)

x+4= the rate of the boat traveling downstream (with the current)

Use the distance formula, d = r · t , solved for t: t =

	distance, d	rate, r	time, t =
Upstream	9	x – 4
Downstream	9	x + 4
Total			10

The sum of the time spent traveling upstream and the time traveling downstream is 10.

	Multiply both sides by LCD, (x-4)(x+4)
	Use distributive property
	Simplify
9x + 36 + 9x – 36 = 10x-160
18x = 10x - 160
0 = 10x - 18x – 160	Subtract 18x from both sides to get 0 on one side
0 = 5x - 9x – 80	Divide both sides by 2
0 = (5x + 16)(x – 5)	Factor
5x + 16 = 0 or x – 5 = 0	Use Zero Factor property
5x = -16 x = 5	Solve each equation
x = or x = 5

Looking back, we see that x represents the rate of the boat in still water. The negative solution makes no sense in this case. Therefore, the only valid solution is 5, which means that the rate of the boat in still water is 5 mph.

Example. Working together, two roommates can paint their apartment in 10 hours. Working alone, one of the roommates can complete the job in 2 hours less time than the other roommate working alone. How long would each person take working alone?

Solution. Let x = time for one roommate to do the job alone

x – 2 = time for the other roommate to do the job alone

= rate of one roommate

= rate of the other roommate

Multiply the time it takes them working together, 10 hours, by each rate to get the fractional part of the job done by each person.

	Rate	Time	Part of job accomplished
One Roommate		10
Other Roommate		10

The sum of the parts of the job done by each roommate must equal 1:

+ = 1
x(x – 2) = 1· x (x – 2)	Multiply both sides by the LCD, x(x – 2)
	Use distributive property
10(x – 2) + 10x = x - 2x 10x – 20 + 10x = x - 2x 20x – 20 = x - 2x	Simplify
0 = x - 22x + 20	Get 0 one one side
x =	Use quadratic formula
x =	Simplify
x =

x represents the time it would take one roommate to do the job alone, and x – 2 represents the time it would take the other. If x = 0.95, then x – 2 = -1.05. This negative time does not make sense. So the only valid solution for x is 21.05. Thus, it would take one roommate 21.05 hours to do the job alone, and it would take the other roommate 19.05 hours to do the job alone.

Example. A rectangular pool 20 feet wide and 60 feet long is surrounded by a walkway of uniform width. If the total area of the walkway is 516 square feet, how wide is the walkway?

Let x = the width of the walkway

The length of the outer rectangle is 60 + x + x = 60 + 2x

The width of the outer rectangle is 20 + x + x = 20 + 2x

The area of the outer rectangle is: A = lw = (60 + 2x)(20 + 2x)

The area of the inner rectangle is: A = lw = 60(20) = 1200

Area of the outer rectangle = Area of inner rectangle + Area of Walkway

(60 + 2x)(20 + 2x) = 1200 + 516
1200 + 120x + 40x + 4x = 1716	Simplify
4x + 160x – 516 = 0	Get 0 on one side of the equation
x + 40x – 129 = 0	Divide both sides by 4
(x – 3)(x + 43) = 0	Factor
x – 3 = 0 or x + 43 = 0	Use Zero Factor property
x = 3 x = -43	Solve

x represents the width of the walkway. The negative solution, -43, makes no sense. So, the solution is 3, which means that the width of the walkway is 3 feet.

Pythagorean Theorem

In a right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.

Example. Find the length of the sides of the right triangle shown in the figure below:

	Pythagorean Theorem
64 + x - 4x + 4 = x + 8x + 16	Square the binomials
x - 4x + 68 = x + 8x + 16	Simplify
52 = 12x	Subtract x from both sides; add 4x to both sides; subtract 16 from both sides
= x	Divide both sides by 12

Looking back at the figure, the unknown sides of the triangle are represented by x – 2 and x + 4. So, the lengths of these sides are:

x – 2 = and x + 4 =