CHAPTER 7 Systems of Equations Slide 2Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc....

CHAPTER

7Systems of Equations

Slide 2Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

7.1 Systems of Equations in Two Variables

7.2 The Substitution Method

7.3 The Elimination Method

7.4 Applications and Problem Solving

7.5 Applications with Motion

OBJECTIVES



a Solve applied problems by translating to a system of two equations in two variables.

EXAMPLE

Maya sells concessions at a local sporting event. In one hour, she sells 72 drinks. The drink sizes are small, which sells for $2 each, and large, which sells for $3 each. If her total sales revenue was $190, how many of each size did she sell?



A Applications of Systems of Equations

(continued)


EXAMPLE

1. Familiarize. Suppose that of the 72 drinks, 20 where small and 52 were large. The 72 drinks would then amount to a total of 20($2) + 52($3) = $196. Although our guess is incorrect, checking the guess has familiarized us with the problem. We let s = the number of small drinks and l = the number of large drinks.




(continued)


EXAMPLE2. Translate. Since a total of 72 drinks were sold, we must

have s + l = 72.To find a second equation, we reword some information and focus on the drinks sold.

Rewording and Translating:Income from income from

small drinks plus large drinks totaled 190

2s + 3l = 190





EXAMPLE

We have the following system: 72 (1)

2 3 190 (2)

s l

s l




(continued)


EXAMPLE3. Solve.

Next, we replace l in equation (2) with l = 72 s:

72s l 72 (3)l s

2 3 190s l 2 3( ) 972 1 0 ss

2 216 3 190s s 216 190s

26s 26s




(continued)


EXAMPLE

We find l by substituting 26 for s in equation (3): l = 72 s = 72 26 = 46.





EXAMPLE4. Check. If Maya sold 26 small and 46 large drinks, she

would have sold 72 drinks, for a total of 26($2) + 46($3) = $52 + $138 = $190

5. State. Maya sold 26 small drinks and 46 large drinks.





EXAMPLE

A cookware consultant sells two sizes of pizza stones. The circular stone sells for $26 and the rectangular one sells for $34. In one month she sold 37 stones. If she made a total of $1138 from the sale of the pizza stones, how many of each size did she sell?



B Applications and Problem Solving

(continued)


EXAMPLE

1. Familiarize. When faced with a new problem, it is often useful to compare it to a similar problem that you have already solved. Here instead of $2 and $3 drinks, we are counting pizza stones. We let c = the circular stone and r = the rectangular stone.2. Translate. Since a total of 37 stones were sold, we have c + r = 37.




(continued)


EXAMPLE

c + r = 37

26c + 34r = 1138

Presenting the information in a table can be helpful.

$113834r26cMoney Paid

37rcNumber of pans

$34$26Cost per pan

TotalRectangularCircular




(continued)


EXAMPLEWe have translated to a system of equations:

c + r = 3726c + 34r = 1138




(continued)


EXAMPLE3. Carry out. The system can be solved using elimination:

26c 26r = 962 26c + 34r = 1138

8r = 176 r = 22

Multiply equation (1) by 26Equation (2)AddingDividing both sides by 8




(continued)


EXAMPLE

To solve for c, we substitute 22 for r: c + r = 37c + 22 = 37 c = 15




(continued)


EXAMPLE4. Check. If r = 22 and c = 15, a total of 37 stones were

sold. The amount paid was 22($34) + 15($26) = $1138. The numbers check.

5. State. The consultant sold 22 rectangular pizza stones and 15 circular pizza stones.





EXAMPLE



C Applications and Problem Solving

(continued)


A coffee shop is considering a new mixture of coffee beans. It will be created with Italian Roast beans costing $9.95 per pound and the Venezuelan Blend beans costing $11.25 per pound. The types will be mixed to form a 60-lb batch that sells for $10.50 per pound. How many pounds of each type of bean should go into the blend?

EXAMPLE




(continued)


1. Familiarize. This problem is similar to our previous examples. Instead of pizza stones we have coffee blends. We have two different prices per pound. Instead of knowing the total amount paid, we know the weight and price per pound of the new blend being made.

EXAMPLE




(continued)


Let i represent the number of pounds of Italian roast and v represent the number of pounds of Venezuelan blend.2. Translate. Since a 60-lb batch is being made, we have i + v = 60. Present the information in a table.

EXAMPLE

i + v = 60

9.95i + 11.25v = 630




(continued)


Italian Venezuelan New Blend

Number of pounds

i v 60

Price per pound

$9.95 $11.25 $10.50

Value of beans

9.95i 11.25v 630

EXAMPLEWe have translated to a system of equations:

i + v = 60 (1)9.95i + 11.25v = 630 (2)




(continued)


EXAMPLE3. Carry out. When equation (1) is solved for v, we have v = 60 i. We then substitute for v in equation (2).

9.95i + 11.25(60 i) = 630 9.95i + 675 11.25i = 630 1.30i = 45 i = 34.6If i = 34.6, we see from equation (1) that v = 25.4.




(continued)


EXAMPLE




(continued)


4. Check. If 34.6 lb of Italian Roast and 25.4 lb of Venezuelan Blend are mixed, a 60-lb blend will result. The value of 34.6 lb of Italian beans is 34.6($9.95), or $344.27. The value of 25.4 lb of Venezuelan Blend is 25.4($11.25), or $285.75, so the value of the blend is $630.02. A 60-lb blend priced at $10.50 a pound is also worth $630, so our answer checks.

EXAMPLE





5. State. The new blend should be made from 34.6 pounds of Italian Roast beans and 25.4 pounds of Venezuelan Blend beans.


Problem-Solving Tip


When solving problems, see if they are patterned or modeled after other problems that you have studied.

CHAPTER 7 Systems of Equations Slide 2Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc....

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