2 Equations, Inequalities, and Applications

2.4 An Introduction to Applications of 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 consecutive integers. 5. Solve problems involving complementary and supplementary angles.

Solving Applied Problems 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.

Finding the Value of an Unknown Number Example 1 The product of 3, and a number decreased by 2, is 42. What is the number? 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

Finding the Value of an Unknown Number Example 1 (cont.) The product of 3, and a number decreased by 2, is 42. What is the number? Notice the placement of the commas! Step 3 Write an equation. The product of 3, and a number decreased by 2, is 42. 3 • ( x – 2 ) = 42 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.

Finding the Value of an Unknown Number Example 1 (cont.) The product of 3, and a number decreased by 2, is 42. What is the number? Solve the equation. Step 4 3 ( x – 2 ) = 42 3x – 6 = 42 Distribute. 3x – 6 + 6 = 42 + 6 Add 6. 3x = 48 Combine terms. 3x 48 3 3 = Divide by 3. x = 16

Finding the Value of an Unknown Number Example 1 (cont.) The product of 3, and a number decreased by 2, is 42. What is the number? State the answer. The number is 16. Step 5 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.

Solve Problems Involving Sums of Quantities 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? Step 1 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 2 Assign a variable. Let x = the number of pencils in the box. Then x + 16 = the number of pens in the box.

Solving Problems Involving Sums of Quantities 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? Recall: x = # of pencils, x + 16 = # of pens Write an equation. Step 3 The total is the number of pens plus the number of pencils 68 = ( x + 16 ) + x

Solve Problems Involving Sums of Quantities 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 = ( x + 16 ) + x 68 = 2x + 16 Combine terms. 68 – 16 = 2x + 16 – 16 Subtract 16. 52 = 2x Combine terms. 52 2x 2 2 = Divide by 2. 26 = x or x = 26

Solve Problems Involving Sums of Quantities 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? 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. Step 5 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. Step 6

Solve Problems Involving Sums of Quantities 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 Step 2 Assign a variable. Let x = the number of milliliters of acid required. Then 12x = the number of milliliters of water required.

Solve Problems Involving Sums of Quantities 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? Recall: x = ml. of acid, 12x = ml. of water. Step 3 Write an equation. A diagram is sometimes helpful. Beaker Acid x Water 12x = 286 x + 12x = 286

Solve Problems Involving Sums of Quantities 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? Step 4 Solve. x + 12x = 286 13x = 286 Combine terms. 13x 286 13 13 = Divide by 13. x = 22

Solve Problems Involving Sums of Quantities 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? Recall: x = ml of acid, 12x = ml of water State the answer. The beaker requires 22 ml of acid and 12(22) = 264 ml of water. Step 5 Check. Since 22 + 264 = 286, and 264 is 12 times 22, the answer checks. Step 6

Solve Problems Involving Sums of Quantities 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

Solve Problems Involving Sums of Quantities 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? Write an equation. Step 3 96 inches 3x x x – 14 Longest Middle-sized Shortest is Total length 3x + x + x – 14 = 96

Solve Problems Involving Sums of Quantities 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? Solve. 96 = 3x + x + x – 14 Step 4 96 = 5x – 14 Combine terms. 96 + 14 = 5x – 14 + 14 Add 14. 110 = 5x Combine terms. 110 5x 5 5 = Divide by 5. 22 = x or x = 22

Solve Problems Involving Sums of Quantities 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? Recall: x = length of middle-sized piece, 3x = length of longest piece, and x – 14 = length of shortest piece. 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. Step 5 Check. The sum of the lengths is 96 in. All conditions of the problem are satisfied. Step 6

Solving Problems with Consecutive Integers 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. x, x + 2; x, x + 2.

Solve Problems with Consecutive Integers Example 5 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

Solve Problems with Consecutive Integers 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. 2x + 1 = 893 Combine terms. Step 4 2x = 892 Subtract 1. x = 446 Divide by 2.

Solve Problems with Consecutive Integers Example 6 (cont.) The sum of two consecutive checkbook check numbers is 893. Find the numbers. Step 5 State the answer. The lesser check number is 446 and the greater check number is 447. Step 6 Check. The sum of 446 and 447 is 893. The answer is correct.

Solve Problems with Consecutive Integers Example 6 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 consecutive odd integer. Then x + 2 = the greater consecutive odd integer. Step 2 Write an equation. Step 3 4 times the smaller is added to 3 times the larger, the result is 125 4 • x + 3 • ( x + 2 ) = 125

Solve Problems with Consecutive Integers Example 6 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. 4x + 3 ( x + 2 ) = 125 Step 4 4x + 3x + 6 = 125 Distribute. 7x + 6 = 125 Combine terms. 7x + 6 – 6 = 125 – 6 Subtract 6. 7x = 119 Combine terms. x = 17 Divide by 7. 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

Solve Problems with Supplementary and Complementary Angles 1 2 3 4 180 (degrees) 1 2 Angles and are complementary. They form a right angle, indicated by . 3 4 Angles and are supplementary. They form a straight angle. Straight angle

Solve Problems with Supplementary and Complementary Angles Measure of x: Complement of x: Supplement of x: 90 – x 180 – x x x x x x 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.

Solve Problems with Supplementary and Complementary Angles Example 8 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 Step 2 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. Write an equation. Step 3 Supplement is 20 more than three times its complement. 180 – x = 20 + 3 • ( 90 – x )

Solve Problems with Supplementary and Complementary Angles Example 8 (cont.) Find the measure of an angle whose supplement is 20° more than three times its complement. Solve. 180 – x = 20 + 3 ( 90 – x ) Step 4 180 – x = 20 + 270 – 3x Distribute. 180 – x = 290 – 3x Combine terms. 180 – x + 3x = 290 – 3x + 3x Add 3x. 180 + 2x = 290 Combine terms. 180 + 2x – 180 = 290 – 180 Subtract 180. 2x = 110 Combine terms. 2x 110 2 2 = Divide by 2. x = 55

Solve Problems with Supplementary and Complementary Angles Example 8 (cont.) Find the measure of an angle whose supplement is 20° more than three times its complement. Step 5 State the answer. The measure of the angle is 55°. Step 6 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.