5To solve an equation with variables on both sides, use inverse operations to "collect" variable terms on one side of the equation.Helpful HintEquations are often easier to solve when the variable has a positive coefficient. Keep this in mind when deciding on which side to "collect" variable terms.
6Example 1: Solving Equations with Variables on Both Sides Solve 7n – 2 = 5n + 6.7n – 2 = 5n + 6To collect the variable terms on one side, subtract 5n from both sides.–5n –5n2n – 2 =2n =Since n is multiplied by 2, divide both sides by 2 to undo the multiplication.n = 4
7Check It Out! Example 1aSolve 4b + 2 = 3b.4b + 2 = 3bTo collect the variable terms on one side, subtract 3b from both sides.–3b –3bb + 2 = 0– 2 – 2b = –2
8Check It Out! Example 1bSolve y = 0.7y – 0.3.To collect the variable terms on one side, subtract 0.3y from both sides.y = 0.7y – 0.3–0.3y –0.3y= 0.4y – 0.3Since 0.3 is subtracted from 0.4y, add 0.3 to both sides to undo the subtraction.= 0.4ySince y is multiplied by 0.4, divide both sides by 0.4 to undo the multiplication.2 = y
9To solve more complicated equations, you may need to first simplify by using the Distributive Property or combining like terms.
10Example 2: Simplifying Each Side Before Solving Equations Solve 4 – 6a + 4a = –1 – 5(7 – 2a).4 – 6a + 4a = –1 –5(7 – 2a)Distribute –5 to the expression in parentheses.4 – 6a + 4a = –1 –5(7) –5(–2a)4 – 6a + 4a = –1 – a4 – 2a = – aCombine like terms.Since –36 is added to 10a, add 36 to both sides.40 – 2a = a+ 2a aTo collect the variable terms on one side, add 2a to both sides.= a
11Example 2 ContinuedSolve 4 – 6a + 4a = –1 – 5(7 – 2a).40 = 12aSince a is multiplied by 12, divide both sides by 12.
12Check It Out! Example 2aSolveDistribute to the expression in parentheses.12To collect the variable terms on one side, subtract b from both sides.123 = b – 1Since 1 is subtracted from b, add 1 to both sides.4 = b
13Check It Out! Example 2bSolve 3x + 15 – 9 = 2(x + 2).3x + 15 – 9 = 2(x + 2)Distribute 2 to the expression in parentheses.3x + 15 – 9 = 2(x) + 2(2)3x + 15 – 9 = 2x + 43x + 6 = 2x + 4Combine like terms.–2x –2xTo collect the variable terms on one side, subtract 2x from both sides.x + 6 =– – 6Since 6 is added to x, subtract 6 from both sides to undo the addition.x = –2
14An identity is an equation that is true for all values of the variable An identity is an equation that is true for all values of the variable. An equation that is an identity has infinitely many solutions.Some equations are always false. These equations have no solutions.
15Identities and False Equations WORDSIdentityWhen solving an equation, if you get an equation that is always true, the original equation is an identity, and it has infinitely many solutions.NUMBERS2 + 1 = 2 + 13 = 3 ALGEBRA2 + x = 2 + x–x –x2 = 2
16Identities and False Equations When solving an equation, if you get a false equation, the original equation has no solutions.WORDSx = x + 3–x –x0 = 3 1 = 1 + 21 = 3 ALGEBRANUMBERSIdentities and False Equations
17Example 3A: Infinitely Many Solutions or No Solutions Solve 10 – 5x + 1 = 7x + 11 – 12x.10 – 5x + 1 = 7x + 11 – 12x10 – 5x + 1 = 7x + 11 – 12xIdentify like terms.11 – 5x = 11 – 5xCombine like terms on the left and the right.+ 5x xAdd 5x to both sides.= 11True statement.The equation 10 – 5x + 1 = 7x + 11 – 12x is an identity. All values of x will make the equation true. All real numbers are solutions.
18Example 3B: Infinitely Many Solutions or No Solutions Solve 12x – 3 + x = 5x – 4 + 8x.12x – 3 + x = 5x – 4 + 8x12x – 3 + x = 5x – 4 + 8xIdentify like terms.13x – 3 = 13x – 4Combine like terms on the left and the right.–13x –13xSubtract 13x from both sides.–3 = –4False statement.The equation 12x – 3 + x = 5x – 4 + 8x is a false equation. There is no value of x that will make the equation true. There are no solutions.
19 Check It Out! Example 3a Solve 4y + 7 – y = 10 + 3y. Identify like terms.3y + 7 = 3y + 10Combine like terms on the left and the right.–3y –3ySubtract 3y from both sides.7 =False statement.The equation 4y + 7 – y = y is a false equation. There is no value of y that will make the equation true. There are no solutions.
20Check It Out! Example 3b Solve 2c + 7 + c = –14 + 3c + 21. Identify like terms.3c + 7 = 3c + 7Combine like terms on the left and the right.–3c –3cSubtract 3c both sides.7 =True statement.The equation 2c c = –14 + 3c + 21 is an identity. All values of c will make the equation true. All real numbers are solutions.
21Example 4: ApplicationJon and Sara are planting tulip bulbs. Jon has planted 60 bulbs and is planting at a rate of 44 bulbs per hour. Sara has planted 96 bulbs and is planting at a rate of 32 bulbs per hour. In how many hours will Jon and Sara have planted the same number of bulbs? How many bulbs will that be?PersonBulbsJon60 bulbs plus 44 bulbs per hourSara96 bulbs plus 32 bulbs per hour
22Example 4: Application Continued Let b represent bulbs, and write expressions for the number of bulbs planted.60 bulbsplus44 bulbs each hourthe same as96 bulbs32 bulbs each hourWhen is?b = bb = bTo collect the variable terms on one side, subtract 32b from both sides.– 32b – 32bb = 96
23Example 4: Application Continued b = 96Since 60 is added to 12b, subtract 60 from both sides.– – 6012b = 36Since b is multiplied by 12, divide both sides by 12 to undo the multiplication.b = 3
24Example 4: Application Continued After 3 hours, Jon and Sara will have planted the same number of bulbs. To find how many bulbs they will have planted in 3 hours, evaluate either expression for b = 3:b = (3) = = 192b = (3) = = 192After 3 hours, Jon and Sara will each have planted 192 bulbs.
25Let g represent Greg's age, and write expressions for his age. Check It Out! Example 4Four times Greg's age, decreased by 3 is equal to 3 times Greg's age increased by 7. How old is Greg?Let g represent Greg's age, and write expressions for his age.three times Greg's agefour times Greg's ageis equal toincreased bydecreased by7.34g – = g
26Check It Out! Example 4 Continued 4g – 3 = 3g + 7To collect the variable terms on one side, subtract 3g from both sides.–3g –3gg – 3 =Since 3 is subtracted from g, add 3 to both sides.g =Greg is 10 years old.
27Lesson Quiz 1. 7x + 2 = 5x + 8 2. 4(2x – 5) = 5x + 4 Solve each equation.1. 7x + 2 = 5x (2x – 5) = 5x + 43. 6 – 7(a + 1) = –3(2 – a)4. 4(3x + 1) – 7x = 6 + 5x – 25.6. A painting company charges $250 base plus $16 per hour. Another painting company charges $210 base plus $18 per hour. How long is a job for which the two companies costs are the same?38all real numbers120 hours