

Graphs Numbers 
Inequalities : SolveBasic  Intermediate  Advanced  Help Enter a polynomial inequality or system of inequalities, enter the variable to be solved for, set the options and click the Solve button.
Using equations and inequalities to solve algebra story problemsNUMBER RELATION PROBLEMSOBJECTIVESUpon completing this section you should be able to:
Read the following three problems carefully:
At first glance these may seem to be three different problems. However, closer examination shows that they are actually the same. For all three a possible outline is The equation (x + 2) + (x) + (x + 6) = 59 will lead to the solutions. Problems such as these are classified as number relation problems. If you recognize a problem in this class, then the outline for solving it must show a relation between the numbers. The equation will then be based on a statement about the sum, difference, and so on. The problems you solved in the last section were all number relation problems. They all involved comparing two or more numbers.
Problems comparing age, height, weight, and so forth are number relation problems.
The field of business contains number relation problems.
Example 4 A customer purchases a stereo radio for $306.80, including a 4% sales tax. Find the price of the radio and the amount of sales tax. Solution If x represents the price of the radio, then the sales tax is found by multiplying the price of the radio by 4%. Thus, we have
Then the total sale is the sum of the price of the radio and the sales tax. x + .04x = 306.80
Solving the equation, we obtain
Check: The total of $295.00 and $11.80 is $306.80. Some number relation problems require prior knowledge of a formula to set up the equation. Example 5 In a certain rectangle the length is two more than the width. If the length is increased by three and the width by two, the perimeter of the new rectangle will be twice that of the original. Find the length and width of the original rectangle. Solution Notice that to solve the problem, you must know that P = 2l + 2w is the formula for the perimeter of a rectangle. We outline this as
Notice that we used subscripts o and n to distinguish the original perimeter from the new perimeter. The equation comes from the statement that
Since x represents the width, we have Thus, the perimeter of the new rectangle is twice that of the original rectangle, and the answer checks.
Example 6 An individual invests $8,000 for one year. Part is invested at 6% and the rest at 5%. If the total interest at the end of the year is $460, how much was invested at each rate? Solution Here we must use the interest formula I = prt, where I represents the interest, r represents the rate of interest, p represents the principal amount invested, and t is the time in years. We let x = amount invested at 5% and 8,000  x = amount invested at 6%. Then, the interest earned at 5% is
The interest earned at 6% is The equation representing the total interest is .05x + .06(8,000  x) = 460. Solving the equation gives Thus, $2,000 was invested at 5% and $6,000 at 6%. Check: 5% of $2,000 is $100 and 6% of $6,000 is $360. The two interest amounts total $460.
Example 7 The interest due on a twoyear $300 loan amounted to $90. What annual rate of interest was charged? Solution Again, the formula I = prt is useful. Here I = $90, p = $300, and t = 2 years. Therefore, Thus, the rate of interest was 15%. Check: $300 at 15% for 2 years yields $90 interest.
DISTANCERATETIME PROBLEMSOBJECTIVESUpon completing this section you should be able to:
Another formula often found in verbal problems is d = rt (distance = rate X time), which is the distance formula for constant rate. Given the rate at which an object is moving and the time that it moves at this rate, we can find the distance the object moves. Example 1 A car travels at the rate of 55 mph for four hours. How far does it travel? Solution Using the formula d = rt, substitute r = 55 and t = 4, obtaining d = (55)(4) = 220. Thus, the distance traveled is 220 miles.
If the distance and rate are both given, we can find the time. Example 2 How long will it take a plane whose ground speed is 530 mph to travel 2,120 miles? Solution In this case, let d = 2,120 and r = 530 in the formula It will therefore take four hours to travel the distance.
We can also solve for the rate if we are given the distance and time. Example 3 A person walked 21 miles in hours. What was the person's average rate? Solution We substitute d = 21 and t = 3.5 in the formula The person's average rate was six miles per hour. One type of distance problem involves two objects leaving from the same point and traveling in the same direction.
Example 4 A bank robber leaves town heading north at an average speed of 120 kilometers per hour. The sheriff leaves two hours later in a plane that travels at 200 kilometers per hour. How long will it take the sheriff to catch the robber? Solution We outline as follows: In this table x represents the time the sheriff takes to catch the robber. Notice that the distance 120(x + 2) and 200x come from the formula d = rt.
The equation comes from the fact that the sheriff and the robber will have traveled the same distance when the sheriff catches the robber. Thus, The sheriff will catch the robber in three hours.
Check: The bank robber has traveled for five hours at 120 km/ hr for a distance of 5(120) = 600 km. The sheriff has traveled for three hours at 200 km/hr for a distance of 3(200) = 600 km. Another type of distance problem involves two objects leaving from the same point and traveling in opposite directions. Example 5 Pamela and Sue start at the same point and walk in opposite directions. The rate at which they are moving away from one another is 11 mph. At the end of three hours Pamela stops and Sue continues to walk for another hour. At the end of that time Pamela has walked twice as far as Sue. How far apart are they? Solution We use the following table:
The sum of their rates is 11 mph. Solving, we obtain Thus, they are 36 miles apart. Check: Pamela's rate is Sue's rate is Their total rate is 11 mph. Also, Pamela's distance (24) is twice Sue's distance (12).
Still another type of distance problem involves two objects that leave from two different points and travel toward each other. Example 6 Juan and Steven started 36 miles apart and walked toward each other, meeting in three hours. If Juan walked two miles per hour faster than Steven, find the rate of each. Solution
First set up the following table:
The total distance they have traveled is 36 miles. Thus, 3(x + 2) + 3x = 36. Solving, we obtain x = 5 The answers are Juan's rate = 7 mph Check: Juan's rate (7) is two miles per hour faster than Steven's (5). Also 3(7) + 3(5) = 21 + 15 = 36. Within the class of problems using d = rt is a subclass of problems concerned with parallel and opposing forces.
Example 7 A plane, whose speed in still air is 550 mph, flies against a headwind of 50 mph. How long will it take to travel 1,500 miles? Solution We use the formula d = rt. The distance is 1,500 miles and the rate of the plane against the wind will be its still air speed (550 mph) reduced by the headwind (50 mph). Thus, it will take three hours to travel the distance. Example 8 Mike can row his boat from the hunting lodge upstream to the park in five hours. He can row back from the park to the lodge downstream in three hours. If Mike can row x kilometers per hour in still water, and if the stream is flowing at the rate of two kilometers per hour, how far is it from the lodge to the park? Solution In working the problem we assume that the rate of the stream will increase or decrease the rate of the boat by two kilometers per hour. Since x represents Mike's speed in still water, we obtain Setting the distance upstream equal to the distance downstream, we obtain Notice that x is not the solution to the problem but is Mike's rate of rowing in still water. However, the question asked is "What is the distance from the lodge to the park?" To answer this use either the distance upstream or downstream since they are the same. Using the upstream distance, we have
MIXTURE PROBLEMSOBJECTIVESUpon completing this section you should be able to:
The final type of problem we will discuss in this chapter is the mixture problem. Mixture problems come in various settings. They may involve mixing candy, coffee, coins, liquids, and so on. They all have a central concept however, which helps in solving them. Example 1 A merchant has one brand of coffee that sells for $1.80 per pound and a second brand that sells for $2.20 per pound. The merchant wishes to make a blend of the two coffees. How many pounds of each price coffee must be used to make 20 pounds of a mixture that would sell for $ 1.90 per pound? Solution We set up the following table from the information given in the problem. Looking at the "total value" column of the table, notice that the sum of the total values of the two brands must equal the total value of the mixture. This fact gives us the equation 1.80x + 2.20(20  x) = 1.90(20).
Solving, we obtain Thus, the merchant must use 15 pounds of the first brand and 5 pounds of the second brand.
Example 2 A total of 100 coins in nickels and quarters has a value of $14.40. How many of each type of coin are there? Solution From the given information we set up the following table: Looking at the "total value" column of the table, notice that theAgain we see that the sum of the total value of nickels and the total value of quarters is equal to the total value of the mixture ($14.40). We therefore write the equation .05x + .25(100  x) = 14.40.
Solving, we obtain Thus, there are 53 nickels and 47 quarters. Check: .05(53) + .25(47) = 2.65 + 11.75 = 14.40 Example 3 A merchant has three types of hard candy. The first sells for $1.10 per kilogram, the second for $1.50 per kilogram, and the third for $1.60 per kilogram. How many kilograms of each candy would be used to make 100 kilograms that would sell for $1.36 per kilogram, if twice as much of the second type must be used as the third? Solution We may outline as follows:
Notice that the equation must come from the value column. Any equation must have like quantities on each side. You can never equate pounds with value, number of coins with the dollar value of the coins, and so forth.
The equation is
Example 4 How much pure alcohol must be added to nine quarts of water to obtain a mixture of 40% alcohol? Solution We set up the table.
Notice that the amount of alcohol in the mixture will be equal to the amount added. We can therefore write the equation x = .40(x + 9). Solving, we obtain Therefore, six quarts of alcohol must be added. Check: If we add 6 quarts to 9 quarts, there will be a total of 15 quarts in the mixture, of which 6 quarts, or 40%, is alcohol. Example 5 The capacity of a car's radiator is nine liters. The mixture of antifreeze and water is 30% antifreeze. How much of the mixture in the radiator must be drawn off and replaced with pure antifreeze to raise the percentage of the mixture to 65% antifreeze? Solution
In outlining this problem again keep in mind that the equation must equate like quantities. We can choose to make an equation equating antifreeze with antifreeze, or water with water, but not water with antifreeze. If we choose to equate antifreeze with antifreeze, we proceed as follows. If x represents the number of liters of the mixture to be drawn off, the outline would be:
Note that the original amount of antifreeze, less the amount of antifreeze drained off, plus the amount of antifreeze added back will equal the final amount of antifreeze. Thus,
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