m81320 Solve Equations Using the Subtraction and Addition Properties of Equality By the end of this section, you will be able to: Solve equations using the Subtraction and Addition Properties of Equality Solve equations that need to be simplified Translate an equation and solve Translate and solve applications 5a8cc4b8-4470-4e4b-b761-1929108174f2 Before you get started, take this readiness quiz. Solve: n−12=16. If you missed this problem, review . 28 Translate into algebra ‘five less than x.’ If you missed this problem, review . x−5 Is x=2 a solution to 5x−3=7? If you missed this problem, review . yes We are now ready to “get to the good stuff.” You have the basics down and are ready to begin one of the most important topics in algebra: solving equations. The applications are limitless and extend to all careers and fields. Also, the skills and techniques you learn here will help improve your critical thinking and problem-solving skills. This is a great benefit of studying mathematics and will be useful in your life in ways you may not see right now.

We began our work solving equations in previous chapters. It has been a while since we have seen an equation, so we will review some of the key concepts before we go any further. We said that solving an equation is like discovering the answer to a puzzle. The purpose in solving an equation is to find the value or values of the variable that make each side of the equation the same. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle. A solution of an equation is a value of a variable that makes a true statement when substituted into the equation. In the earlier sections, we listed the steps to determine if a value is a solution. We restate them here. Substitute the number for the variable in the equation. Simplify the expressions on both sides of the equation. Determine whether the resulting equation is true. If it is true, the number is a solution. If it is not true, the number is not a solution. Determine whether y=34 is a solution for 4y+3=8y. Multiply. Add. Since y=34 results in a true equation, 34 is a solution to the equation 4y+3=8y. Is y=23 a solution for 9y+2=6y? no Is y=25 a solution for 5y−3=10y? no We introduced the Subtraction and Addition Properties of Equality in Solving Equations Using the Subtraction and Addition Properties of Equality. In that section, we modeled how these properties work and then applied them to solving equations with whole numbers. We used these properties again each time we introduced a new system of numbers. Let’s review those properties here. Subtraction Property of Equality For all real numbers a,b, and c, if a=b, then a−c=b−c. Addition Property of Equality For all real numbers a,b, and c, if a=b, then a+c=b+c. When you add or subtract the same quantity from both sides of an equation, you still have equality. We introduced the Subtraction Property of Equality earlier by modeling equations with envelopes and counters. models the equation x+3=8.

The goal is to isolate the variable on one side of the equation. So we ‘took away’ 3 from both sides of the equation and found the solution x=5. Some people picture a balance scale, as in , when they solve equations.

The quantities on both sides of the equal sign in an equation are equal, or balanced. Just as with the balance scale, whatever you do to one side of the equation you must also do to the other to keep it balanced. Let’s review how to use Subtraction and Addition Properties of Equality to solve equations. We need to isolate the variable on one side of the equation. And we check our solutions by substituting the value into the equation to make sure we have a true statement. Solve: x-11=−3. To isolate x, we undo the addition of 11 by using the Subtraction Property of Equality. We "undo" the subtraction of 11 by adding 11 to each side. Simplify. Check: Substitute x=8. Since x=8 makes x-11=−3 a true statement, we know that it is a solution to the equation. Solve: x+9=−7. x = −16 Solve: x+16=−4. x = −20 In the original equation in the previous example, 11 was added to the x, so we subtracted 11 to ‘undo’ the addition. In the next example, we will need to ‘undo’ subtraction by using the Addition Property of Equality. Solve: m+4=−5. Subtract 4 from each side to "undo" the addition. Simplify. Check: Substitute m=−9. The solution to m−4=−5 is m=−1. Solve: n−6=−7. n = −1 Solve: x−5=−9. x = −4 Now let’s review solving equations with fractions. Solve: n−38=12. Use the Addition Property of Equality. Find the LCD to add the fractions on the right. Simplify Check: Subtract. Simplify. The solution checks. Solve: p−13=56. p=76 Solve: q−12=16. q=23 In Solve Equations with Decimals, we solved equations that contained decimals. We’ll review this next. Solve a−3.7=4.3. Use the Addition Property of Equality. Add. Check: Substitute a=8. Simplify. The solution checks. Solve: b−2.8=3.6. b = 6.4 Solve: c−6.9=7.1. c = 14

In the examples up to this point, we have been able to isolate the variable with just one operation. Many of the equations we encounter in algebra will take more steps to solve. Usually, we will need to simplify one or both sides of an equation before using the Subtraction or Addition Properties of Equality. You should always simplify as much as possible before trying to isolate the variable. Solve: 3x−7−2x−4=1. The left side of the equation has an expression that we should simplify before trying to isolate the variable. Rearrange the terms, using the Commutative Property of Addition. Combine like terms. Add 11 to both sides to isolate x. Simplify. Check. Substitute x=12 into the original equation. The solution checks. Solve: 8y−4−7y−7=4. y = 15 Solve: 6z+5−5z−4=3. z = 2 Solve: 3(n−4)−2n=−3. The left side of the equation has an expression that we should simplify. Distribute on the left. Use the Commutative Property to rearrange terms. Combine like terms. Isolate n using the Addition Property of Equality. Simplify. Check. Substitute n=9 into the original equation. The solution checks. Solve: 5(p−3)−4p=−10. p = 5 Solve: 4(q+2)−3q=−8. q = −16 Solve: 2(3k−1)−5k=−2−7. Both sides of the equation have expressions that we should simplify before we isolate the variable. Distribute on the left, subtract on the right. Use the Commutative Property of Addition. Combine like terms. Undo subtraction by using the Addition Property of Equality. Simplify. Check.Let k=−7. The solution checks. Solve: 4(2h−3)−7h=−6−7. h = −1 Solve: 2(5x+2)−9x=−2+7. x = 1

In previous chapters, we translated word sentences into equations. The first step is to look for the word (or words) that translate(s) to the equal sign. reminds us of some of the words that translate to the equal sign. Equals (=) is is equal to is the same as the result is gives was will be Let’s review the steps we used to translate a sentence into an equation. Locate the "equals" word(s). Translate to an equal sign. Translate the words to the left of the "equals" word(s) into an algebraic expression. Translate the words to the right of the "equals" word(s) into an algebraic expression. Now we are ready to try an example. Translate and solve: five more than x is equal to 26. Translate. Subtract 5 from both sides. Simplify. Check:Is 26 five more than 21?The solution checks. Translate and solve: Eleven more than x is equal to 41. x + 11 = 41; x = 30 Translate and solve: Twelve less than y is equal to 51. y − 12 = 51; y = 63 Translate and solve: The difference of 5p and 4p is 23. Translate. Simplify. Check: The solution checks. Translate and solve: The difference of 4x and 3x is 14. 4x − 3x = 14; x = 14 Translate and solve: The difference of 7a and 6a is −8. 7a − 6a = −8; a = −8

In most of the application problems we solved earlier, we were able to find the quantity we were looking for by simplifying an algebraic expression. Now we will be using equations to solve application problems. We’ll start by restating the problem in just one sentence, assign a variable, and then translate the sentence into an equation to solve. When assigning a variable, choose a letter that reminds you of what you are looking for. The Robles family has two dogs, Buster and Chandler. Together, they weigh 71 pounds. Chandler weighs 28 pounds. How much does Buster weigh? Read the problem carefully. Identify what you are asked to find, and choose a variable to represent it. How much does Buster weigh? Let b= Buster's weight Write a sentence that gives the information to find it. Buster's weight plus Chandler's weight equals 71 pounds. We will restate the problem, and then include the given information. Buster's weight plus 28 equals 71. Translate the sentence into an equation, using the variable b. Solve the equation using good algebraic techniques. Check the answer in the problem and make sure it makes sense. Is 43 pounds a reasonable weight for a dog? Yes. Does Buster's weight plus Chandler's weight equal 71 pounds? 43+28=?71 71=71✓ Write a complete sentence that answers the question, "How much does Buster weigh?" Buster weighs 43 pounds Translate into an algebraic equation and solve: The Pappas family has two cats, Zeus and Athena. Together, they weigh 13 pounds. Zeus weighs 6 pounds. How much does Athena weigh? a + 6 = 13; Athena weighs 7 pounds. Translate into an algebraic equation and solve: Sam and Henry are roommates. Together, they have 68 books. Sam has 26 books. How many books does Henry have? 26 + h = 68; Henry has 42 books. Read the problem. Make sure you understand all the words and ideas. Identify what you are looking for. Name what you are looking for. Choose a variable to represent that quantity. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation. Solve the equation using good algebra techniques. Check the answer in the problem and make sure it makes sense. Answer the question with a complete sentence. Shayla paid $24,575 for her new car. This was $875 less than the sticker price. What was the sticker price of the car? What are you asked to find? "What was the sticker price of the car?" Assign a variable. Let s= the sticker price of the car. Write a sentence that gives the information to find it. $24,575 is $875 less than the sticker price $24,575 is $875 less than s Translate into an equation. Solve. Check: Is $875 less than $25,450 equal to $24,575? 25,450−875=?24,575 24,575=24,575✓ Write a sentence that answers the question. The sticker price was $25,450. Translate into an algebraic equation and solve: Eddie paid $19,875 for his new car. This was $1,025 less than the sticker price. What was the sticker price of the car? 19,875 = s − 1025; the sticker price is $20,900. Translate into an algebraic equation and solve: The admission price for the movies during the day is $7.75. This is $3.25 less than the price at night. How much does the movie cost at night? 7.75 = n − 3.25; the price at night is $11.00. The Links to Literacy activity, "The 100-pound Problem", will provide you with another view of the topics covered in this section. Solving One Step Equations By Addition and Subtraction Solve One Step Equations By Add and Subtract Whole Numbers (Variable on Left) Solve One Step Equations By Add and Subtract Whole Numbers (Variable on Right)

Determine whether a number is a solution to an equation.Substitute the number for the variable in the equation. Simplify the expressions on both sides of the equation. Determine whether the resulting equation is true. If it is true, the number is a solution. If it is not true, the number is not a solution. Subtraction and Addition Properties of EqualitySubtraction Property of EqualityFor all real numbers a, b, and c,if a = b then a-c=b-c. Addition Property of EqualityFor all real numbers a, b, and c,if a = b then a+c=b+c. Translate a word sentence to an algebraic equation.Locate the “equals” word(s). Translate to an equal sign. Translate the words to the left of the “equals” word(s) into an algebraic expression. Translate the words to the right of the “equals” word(s) into an algebraic expression. Problem-solving strategyRead the problem. Make sure you understand all the words and ideas. Identify what you are looking for. Name what you are looking for. Choose a variable to represent that quantity. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation. Solve the equation using good algebra techniques. Check the answer in the problem and make sure it makes sense. Answer the question with a complete sentence.

Solve Equations Using the Subtraction and Addition Properties of Equality In the following exercises, determine whether the given value is a solution to the equation. Is y=13 a solution of 4y+2=10y? yes Is x=34 a solution of 5x+3=9x? Is u=−12 a solution of 8u−1=6u? no Is v=−13 a solution of 9v−2=3v? In the following exercises, solve each equation. x+7=12 x = 5 y+5=−6 b+14=34 b=12 a+25=45 p+2.4=−9.3 p = −11.7 m+7.9=11.6 a−3=7 a = 10 m−8=−20 x−13=2 x=73 x−15=4 y−3.8=10 y = 13.8 y−7.2=5 x−15=−42 x = −27 z+5.2=−8.5 q+34=12 q=−14 p−25=23 y−34=35 y=2720 Solve Equations that Need to be Simplified In the following exercises, solve each equation. c+3−10=18 m+6−8=15 m = 17 9x+5−8x+14=20 6x+8−5x+16=32 x = 8 −6x−11+7x−5=−16 −8n−17+9n−4=−41 n = −20 3(y−5)−2y=−7 4(y−2)−3y=−6 y = 2 8(u+1.5)−7u=4.9 5(w+2.2)−4w=9.3 w = −1.7 −5(y−2)+6y=−7+4 −8(x−1)+9x=−3+9 x = −2 3(5n−1)−14n+9=1−2 2(8m+3)−15m−4=3−5 m = −4 −(j+2)+2j−1=5 −(k+7)+2k+8=7 k = 6 6a−5(a−2)+9=−11 8c−7(c−3)+4=−16 c = −41 8(4x+5)−5(6x)−x=53 6(9y−1)−10(5y)−3y=22 y = 28 Translate to an Equation and Solve In the following exercises, translate to an equation and then solve. Five more than x is equal to 21. The sum of x and −5 is 33. x + (−5) = 33; x = 38 Ten less than m is −14. Three less than y is −19. y − 3 = −19; y = −16 The sum of y and −3 is 40. Eight more than p is equal to 52. p + 8 = 52; p = 44 The difference of 9x and 8x is 17. The difference of 5c and 4c is 60. 5c − 4c = 60; c = 60 The difference of n and 16 is 12. The difference of f and 13 is 112. f−13=112;f=512 The sum of −4n and 5n is −32. The sum of −9m and 10m is −25. −9m + 10m = −25; m = −25 Translate and Solve Applications In the following exercises, translate into an equation and solve. Pilar drove from home to school and then to her aunt’s house, a total of 18 miles. The distance from Pilar’s house to school is 7 miles. What is the distance from school to her aunt’s house? Jeff read a total of 54 pages in his English and Psychology textbooks. He read 41 pages in his English textbook. How many pages did he read in his Psychology textbook? Let p equal the number of pages read in the Psychology book. 41 + p = 54. Jeff read 13 pages in his Psychology book. Pablo’s father is 3 years older than his mother. Pablo’s mother is 42 years old. How old is his father? Eva’s daughter is 5 years younger than her son. Eva’s son is 12 years old. How old is her daughter? Let d equal the daughter’s age. d = 12 − 5. Eva’s daughter’s age is 7 years old. Allie weighs 8 pounds less than her twin sister Lorrie. Allie weighs 124 pounds. How much does Lorrie weigh? For a family birthday dinner, Celeste bought a turkey that weighed 5 pounds less than the one she bought for Thanksgiving. The birthday dinner turkey weighed 16 pounds. How much did the Thanksgiving turkey weigh? 21 pounds The nurse reported that Tricia’s daughter had gained 4.2 pounds since her last checkup and now weighs 31.6 pounds. How much did Tricia’s daughter weigh at her last checkup? Connor’s temperature was 0.7 degrees higher this morning than it had been last night. His temperature this morning was 101.2 degrees. What was his temperature last night? 100.5 degrees Melissa’s math book cost $22.85 less than her art book cost. Her math book cost $93.75. How much did her art book cost? Ron’s paycheck this week was $17.43 less than his paycheck last week. His paycheck this week was $103.76. How much was Ron’s paycheck last week? $121.19

Baking Kelsey needs 23 cup of sugar for the cookie recipe she wants to make. She only has 14 cup of sugar and will borrow the rest from her neighbor. Let s equal the amount of sugar she will borrow. Solve the equation 14+s=23 to find the amount of sugar she should ask to borrow. Construction Miguel wants to drill a hole for a 58-inch screw. The screw should be 112 inch larger than the hole. Let d equal the size of the hole he should drill. Solve the equation d+112=58 to see what size the hole should be. d=1324

Is −18 a solution to the equation 3x=16−5x? How do you know? Write a word sentence that translates the equation y−18=41 and then make up an application that uses this equation in its solution. Answers will vary.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ If most of your checks were: …confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific. …with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math, every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved? …no—I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

solution of an equationA solution of an equation is a value of a variable that makes a true statement when substituted into the equation.