copyright © 2010 pearson education, inc. all rights reserved. 2.4 – slide 1


Equations, Inequalities, and Applications

Chapter 2


2.4

An Introduction to Applicationsof Linear Equations


Objectives

1. Learn the six steps for solving applied problems.2. Solve problems involving unknown numbers.3. Solve problems involving sums of quantities.4. Solve problems involving supplementary and

complementary angles.

5. Solve problems involving consecutive integers.

2.4 An Introduction to Applications of Linear Equations


Solving an Applied Problem

Step 1 Read the problem, several times if necessary, until you understand what is given and what is to be found.

Step 2 Assign a variable to represent the unknown value, using diagrams or tables as needed. Write down what the

variable represents. Express any other unknown values in terms of the variable.

Step 3 Write an equation using the variable expression(s) .Step 4 Solve the equation. Step 5 State your answer. Does it seem reasonable?Step 6 Check the answer in the words of the original problem.

Solving Applied Problems



Example 1 The product of 3, and a number decreased by 2, is 42. What is the number?

Solving Problems Involving Unknown Numbers

Read the problem carefully. We are asked to find a number.Step 1

Assign a variable to represent the unknown quantity. In this problem, we are asked to find a number, so we write

Let x = the number.

There are no other unknown quantities to find.

Step 2



The equation 3x – 2 = 42 corresponds to the statement “The product of 3 and a number, decreased by 2, is 42.” Be careful when reading these types of problems. The placement of the commas is important.

Example 1 (cont.) The product of 3, and a number decreased by 2, is 42. What is the number?


Write an equation.Step 3

The product of 3, a number decreased by 2,and is 42.

3 • x – = 422 ) (


Notice the placement of the commas!




Step 4 Solve the equation.

3 ( x – 2 ) = 42

3x – 6 = 42

3x – 6 + 6 = 42 + 6

3x = 48

3x 483 3

=

Distribute.

Add 6.

Combine terms.

Divide by 3.

x = 16





Step 5 State the answer. The number is 16.

Step 6 Check. When 16 is decreased by 2, we get 16 – 2 = 14. If 3 is multiplied by 14, we get 42, as required. The answer, 16, is correct.



Example 2 A box contains a combined total of 68 pens and pencils. If the box contains 16 more pens than pencils, how many of each are in the box?

Solving Problems Involving Sums of Quantities

Read the problem. We are given information about the total number of pens and pencils and asked to find the number of each in the box.

Step 1

Assign a variable.

Let x = the number pencils in the box.

Then x + 16 = the number pens in the box.

Step 2



Example 2 (cont.)

A box contains a combined total of 68 pens and pencils. If the box contains 16 more pens than pencils, how many of each are in the box?



The total is the number of pens the number of pencilsplus

68 = ( x + 16 ) + x

Recall: x = # of pencils, x + 16 = # of pens



Example 2 (cont.)

A box contains a combined total of 68 pens and pencils. If the box contains 16 more pens than pencils, how many of each are in the box?


Step 4 Solve the equation.

68 = 2x + 16

68 – 16 = 2x + 16 – 16

52 = 2x

52 2x2 2

=

26 = x

68 = ( x + 16 ) + x

Combine terms.

Subtract 16.

Combine terms.

Divide by 2.

or x = 26



Example 2 (cont.) A box contains a combined total of 68 pens and pencils. If the box contains 16 more pens than pencils, how many of each are in the box?


Step 5 State the answer. The variable x represents the number of pencils, so there are 26 pencils. Then the number of pens is x + 16 = 26 + 16 = 42.

Recall: x = # of pencils, x + 16 = # of pens

Step 6 Check. Since there are 26 pencils and 42 pens, the combined total number of pencils and pens is 26 + 42 = 68. Because 42 – 26 = 16, there are 16 more pens than pencils. This information agrees with what is given in the problem, so the answer checks.



Example 3 A mixture of acid and water must be combined to fill a 286 ml beaker. If the mixture must contain 12 ml of water for each milliliter of acid, how many milliliters of water and how many milliliters of acid does it require to fill the beaker?


Read the problem carefully. We must find how many milliliters of water and how many milliliters of acid are needed to fill the beaker.

Step 1

Assign a variable.

Let x = the number of milliliters of acid required.

Then 12x = the number of milliliters of water required.

Step 2



Example 3 (cont.) A mixture of acid and water must be combined to fill a 286 ml beaker. If the mixture must contain 12 ml of water for each milliliter of acid, how many milliliters of water and how many milliliters of acid does it require to fill the beaker?


Write an equation. A diagram is sometimes helpful.Step 3

Acidx

Water12x

Beaker

= 286

x 12x =+ 286

Recall: x = ml. of acid, 12x = ml. of water.





Step 4 Solve.

13x = 286

13x 28613 13

=

x = 22

x + 12x = 286

Combine terms.

Divide by 13.





Step 5 State the answer. The beaker requires 22 ml of acid and 12(22) = 264 ml of water.

Recall: x = ml of acid, 12x = ml of water

Step 6 Check. Since 22 + 264 = 286, and 264 is 12 times 22, the answer checks.



Example 4 Gerald has a wire 96 inches long that he must cut into three pieces. The longest piece must be three times the length of the middle-sized piece and the shortest piece must be 14 inches shorter than the middle-sized piece. How long must each piece be?


Read the problem carefully. Three lengths must be found.Step 1

Assign a variable.

x = the length of the middle-sized piece,

3x = the length of the longest piece, and

x – 14 = the length of the shortest piece.

Step 2



Example 4 (cont.) Gerald has a wire 96 inches long that he must cut into three pieces. The longest piece must be three times the length of the middle-sized piece and the shortest piece must be 14 inches shorter than the middle-sized piece. How long must each piece be?


Step 3 Write an equation.

96 inches

3x x x – 14

Longest Middle-sized Shortest is Total length3x x x – 14 = 96+ +





Step 4 Solve.

96 = 5x – 14

96 + 14 = 5x – 14 + 14

110 = 5x

110 5x5 5

=

22 = x

96 = 3x + x + x – 14

Combine terms.

Add 14.

Combine terms.

Divide by 5.

or x = 22





Step 5 State the answer. The middle-sized piece is 22 in. long, the longest piece is 3(22) = 66 in. long, and the shortest piece is 22 – 14 = 8 in. long.

Recall: x = length of middle-sized piece, 3x = length of longest piece, and x – 14 = length of shortest piece.

Step 6 Check. The sum of the lengths is 96 in. All conditions of the problem are satisfied.



3 41

2

3 4Angles and are supplementary.

They form a straight angle.

Straight angle1 2Angles and are complementary. They form a right angle, indicated by .

Solving with Supplementary and Complementary Angles


180 (degrees)


Problem-Solving Hint

If x represents the degree measure of an angle, then

90 – x represents the degree measure of its complement, and

180 – x represents the degree measure of its supplement.

x180 – x

x

90 – x


Complement of x:

x

Supplement of x:


Measure of x:

x x


Example 5 Find the measure of an angle whose supplement is 20° more than three times its complement.


Read the problem. We are to find the measure of an angle, given information about its complement and supplement.

Step 1

Assign a variable.

Let x = the degree measure of the angle.

Then 90 – x = the degree measure of its complement;

180 – x = the degree measure of its supplement.

Step 2


Supplement is 20 more than three times its complement.

180 – x = 20 + 3 • ( 90 – x )



Example 5 (cont.) Find the measure of an angle whose supplement is 20° more than three times its complement.


Solve.Step 4 180 – x = 20 + 3 ( 90 – x )

180 – x = 20 + 270 – 3x

180 – x = 290 – 3x

180 – x + 3x = 290 – 3x + 3x

2x 1102 2

=

Distribute.

Combine terms.

Add 3x.

Divide by 2.

x = 55

180 + 2x = 290

180 + 2x – 180 = 290 – 180

2x = 110

Combine terms.

Subtract 180.

Combine terms.



Example 5 (cont.) Find the measure of an angle whose supplement is 20° more than three times its complement.


State the answer. The measure of the angle is 55°.Step 5

Check. The complement of 55° is 35° and the supplement of 55° is 125°. Also, 125° is equal to 20° more than three times 35° (that is, 125 = 20 + 3(35) is true). Therefore, the answer is correct.

Step 6



Problem-Solving Hint

When solving consecutive integer problems, if x = the lesser integer, then for any

two consecutive integers, use x, x + 1;

two consecutive even integers, use x, x + 2;

two consecutive odd integers, use x, x + 2.

Solving Problems with Consecutive Integers

x, x + 2;

x, x + 2.



Example 6 The sum of two consecutive checkbook check numbers is 893. Find the numbers.


Read the problem. We are to find the check numbers.Step 1

Assign a variable.

Let x = the lesser check number.

Then x + 1 = the greater check number.

Step 2



Example 6 (cont.) The sum of two consecutive checkbook check numbers is 893. Find the numbers.


Recall: Let x = the lesser check number.

Then x + 1 = the greater check number.

Write an equation. The sum of the check numbers is 893, so

Step 3

x + ( x + 1 ) = 893

Solve. Step 4 2x + 1 = 893

2x = 892

x = 446

Subtract 1.

Combine terms.

Divide by 2.



Example 6 (cont.) The sum of two consecutive checkbook checknumbers is 893. Find the numbers.


State the answer. The lesser check number is 446 and the greater check number is 447.

Step 5

Check. The sum of 446 and 447 is 893. The answer is correct.Step 6



4 times the smaller is added to 3 times the larger, the result is 125

Example 7 If four times the smaller of 2 consecutive odd integers is added to three times the larger, the result is 125. Find the integers.


Read the problem. We are to find two consecutive odd integers.

Step 1

Assign a variable.

Let x = the lesser integer.

Then x + 2 = the greater integer.

Step 2

Write an equation. Step 3

4 • x + 3 • ( x + 2 ) = 125



Step 4

Example 7 If four times the smaller of 2 consecutive odd integers is added to three times the larger, the result is 125. Find the integers.


Solve.

State the answer. The lesser integer is 17 and the greater integer is 17 + 2 = 19.

Step 5

Check. The sum of 4(17) and 3(19) is 125. The answer is correct.

Step 6

4x + 3 ( x + 2 ) = 125

4x + 3x + 6 = 125

7x + 6 = 125

7x + 6 – 6 = 125 – 6

7x = 119

x = 17

Combine terms.Distribute.

Subtract 6.

Combine terms.

Divide by 7.


copyright © 2010 pearson education, inc. all rights reserved. 2.4 – slide 1

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