# 2.7 Further Applications of Linear Equations

## Presentation on theme: "2.7 Further Applications of Linear Equations"— Presentation transcript:

2.7 Further Applications of Linear Equations

Use percent in solving problems involving rates.
Objective 1 Use percent in solving problems involving rates. Slide 2.7-3

Use percent in solving problems involving rates.
Recall that percent means “per hundred.” Thus, percents are ratios in which the second number is always 100. For example, 50% represents the ratio 50 to 100 and 27% represents the ratio 27 to 100. PROBLEM-SOLVING HINT Mixing different concentrations of a substance or different interest rates involves percents. To get the amount of pure substance or the interest, we multiply. Mixture Problems base × rate (%) = percentage b × r = p Interest Problems (annual) principle × rate (%) = interest p × r = I In an equation, percent is always written as a decimal. Slide 2.7-4

Using Percents to Find Percentages
CLASSROOM EXAMPLE 1 Using Percents to Find Percentages How much pure acid is in 40 L of a 16% acid solution? Find the annual simple interest if \$5000 is invested at 4%. Solution: Solution: Slide 2.7-5

Solve problems involving mixtures. Solve problems involving mixtures.
Objective 2 Objective 2 Solve problems involving mixtures. Solve problems involving mixtures. Slide 2.7-6

Use percent in solving problems involving rates. (cont’d)
PROBLEM-SOLVING HINT Using a table helps organize the information in a problem and more easily set up an equation, which is usually the most difficult step. In a mixture problem, the concentration of the final mixture must be between the concentrations of the two solutions making up the mixture. Slide 2.7-7

Solving a Mixture Problem
CLASSROOM EXAMPLE 2 Solving a Mixture Problem A certain metal is 40% copper. How many kilograms of this metal must be mixed with 80 kg of a metal that is 70% copper to get a metal that is 50% copper? Solution: Let x = kg of 40% copper metal. Kg of Percentage Metal (as a decimal) Copper x 0.4 0.4x 80 0.7 80(0.7)=56 x+80 0.5 0.5(x + 80) 160 kg of the 40% copper metal is needed. Slide 2.7-8

Solve problems involving simple interest.
Objective 3 Solve problems involving simple interest. Slide 2.7-9

Solving a Simple Interest Problem
CLASSROOM EXAMPLE 3 Solving a Simple Interest Problem With income earned by selling a patent, an engineer invests some money at 5% and \$3000 more than twice as much at 8%. The total annual income from the investments is \$1710. Find the amount invested at 5%. Amount Invested Rate of Interest for in Dollars Interest One Year x 0.05 0.05x 2x 0.08 0.08(2x+3000) Solution: Let x = amount invested at 5%. \$7000 was invested at 5% interest. Slide

Solve problems involving denominations of money.
Objective 4 Solve problems involving denominations of money. Slide

Money Denominations Problems number × value of one item = total value
Solve problems involving denominations of money. PROBLEM-SOLVING HINT To get the total value in problems that involve different denominations of money or items with different monetary values, we multiply. Money Denominations Problems number × value of one item = total value A table is also helpful for these problems. Slide

Solving a Money Denominations Problem
CLASSROOM EXAMPLE 4 Solving a Money Denominations Problem A man has \$2.55 in quarters and nickels. He has 9 more nickels than quarters. How many nickels and how many quarters does he have? Solution: Let x = amount of quarters. Number of Denomination Total Coins (as a decimal) Value x 0.25 0.25x x + 9 0.05 0.05(x + 9) The man has 7 quarters and 16 nickels. Slide

Solve problems involving distance, rate, and time.
Objective 5 Solve problems involving distance, rate, and time. Slide

Distance, Rate, and Time Relationship
Solve problems involving distance, rate, and time. If your car travels at an average rate of 50 mph for 2 hr, then it travels 50 × 2 = 100 mi. This is an example of the basic relationship between distance, rate, and time, given by the formula d = rt. By solving, in turn, for r and t in the formula, we obtain two other equivalent forms of the formula. The three forms are given here. distance = rate × time, Distance, Rate, and Time Relationship Slide

Finding Distance, Rate, or Time
CLASSROOM EXAMPLE 5 Finding Distance, Rate, or Time A new world record in the men’s 100-m dash was set in 2008 by Usain Bolt of Jamaica, who ran it in 9.69 sec. What was his rate (to two decimal places)? (Source: World Almanac and Book of Facts.) Solution: Usain Bolt’s rate was m per sec. Slide

Solving a Motion Problem
CLASSROOM EXAMPLE 6 Solving a Motion Problem Two airplanes leave Boston at 12:00 noon and fly in opposite directions. If one flies at 410 mph and the other flies at 530 mph, how long will it take them to be 3290 mi apart? Solution: Let t = time. Rate × Time = Distance Faster plane 410 t 410t Slower plane 530 530t It will take the planes 3.5 hr to be 3290 mi apart. Slide

Solve problems involving distance, rate, and time. (cont’d)
PROBLEM-SOLVING HINT In motion problems, once you have filled in the first two pieces of information in each row of your table, we can automatically fill in the third piece of information, using the appropriate form of the distance formula. Then we set up the equation on the basis of our sketch and the information in the table. Slide

Solving a Motion Problem
CLASSROOM EXAMPLE 7 Solving a Motion Problem Two buses left the downtown terminal, traveling in opposite directions. One had an average speed of 10 mph more than the other. Twelve minutes (1/5 hr) later, they were 12 mi apart. What were their speeds? Solution: Let x = rate of the slower bus. Rate × Time = Distance Faster bus x + 10 1/5 (1/5)(x + 10) Slower bus x (1/5)x The slower bus was traveling at 25 mph and the faster bus at 35 mph. Slide