California Standards 4.0 Students simplify expressions before solving linear equations and inequalities in one variable, such as 3(2x – 5) + 4(x – 2) = 12. 5.0 Students solve multistep problems, including word problems, involving linear equations and linear inequalities in one variable and provide justification for each step. To solve an equation with variables on both sides, use inverse operations to "collect" variable terms on one side of the equation. Helpful Hint Equations 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. Solving Equations with Variables on Both Sides Solve 7n – 2 = 5n + 6. 7n – 2 = 5n + 6 –5n –5n 2n – 2 = +2 2n = n=4 6 +2 8 1: Move variable to one side 2: Add/subtract 3: multiple/divide Solve the equation. Check your answer. 4b + 2 = 3b 4b + 2 = 3b –3b –3b b+2= 0 –2 –2 b = –2 1: Move variable to one side 2: add/subtract Check 4b + 2 = 3b 4(–2) + 2 –8 + 2 –6 3(–2) –6 –6 Solve the equation. Check your answer. 0.5 + 0.3y = 0.7y – 0.3 0.5 + 0.3y = 0.7y – 0.3 –0.3y 0.5 +0.3 0.8 –0.3y = 0.4y – 0.3 + 0.3 1: Move variable to one side 2: add/subtract 3: Multiple/divide = 0.4y Check 0.5 + 0.3y = 0.7y – 0.3 0.5 + 0.3(2) 2=y 0.5 + 0.6 1.1 0.7(2) – 0.3 1.4 – 0.3 1.1 1.4a – 3 = 2a + 7 a=5 2.-3r + 9 = -4r + 5 r = -4 3.3d + 8 = 2d – 17 d = -25 4.2x – 7 = 5x – 10 x=1 To solve more complicated equations, you may need to first simplify by using the Distributive Property or combining like terms. Don’t call me after midnight 1. D= Distributive property 2. C= combine like term 3. M = move variable to one side 4. A = addition/subtraction 5. M = Multiplication/division Solve the equation. Check your answer. 3x + 15 – 9 = 2(x + 2) D: distributive Property 3x + 15 – 9 = 2(x + 2) 3x + 15 – 9 = 2(x) + 2(2) 3x + 15 – 9 = 2x + 4 3x + 6 = 2x + 4 –2x –2x x+6= 4 –6 –6 x = –2 C: Combine like term M: move variable to one side A: add/subtract Simplifying Each Side Before Solving Equations 4 – 6a + 4a = –1 – 5(7 – 2a) D: Distributive P. 4 – 6a + 4a = –1 –5(7 – 2a) 4 – 6a + 4a = –1 –5(7) –5(–2a) 4 – 6a + 4a = –1 – 35 + 10a 4 – 2a = –36 + 10a +2a +2a 4 = -36 + 12a C: Combine like terms M: move variable to One side 4 + 36 40 = -36 + 12a A: add/subtract +36 = 12a M: multiply/divide Check 3x + 15 – 9 = 2(x + 2) 3(–2) + 15 – 9 2(–2 + 2) –6 + 15 – 9 2(0) 0 0 An identity is an equation that is always true, no matter what value is substituted for the variable. The solution set of an identity is all real numbers. Some equations are always false. Their solution sets are empty. In other words, their solution sets contain no elements. Infinitely Many Solutions or No Solutions Solve the equation. 10 – 5x + 1 = 7x + 11 – 12x 10 – 5x + 1 = 7x + 11 – 12x C: combine like term 11 – 5x = 11 – 5x The statement 11 – 5x = 11 – 5x is true for all values of x. The equation 10 – 5x + 1 = 7x + 11 – 12x is an identity. All values of x will make the equation true. In other words, all real numbers are solutions. Infinitely Many Solutions or No Solutions Solve the equation. 12x – 3 + x = 5x – 4 + 8x 12x – 3 + x = 5x – 4 + 8x 13x – 3 = 13x – 4 –13x –13x –3 = –4 C: combine like term M: move variable to one side The equation 12x – 3 + x = 5x – 4 + 8x is always false. There is no value of x that will make the equation true. There are no solutions. Writing Math The empty set can be written as or {}. Check It Out! Example 3a Solve the equation. 4y + 7 – y = 10 + 3y 4y + 7 – y = 10 + 3y Identify like terms. 3y + 7 = 3y + 10 –3y –3y 7 = 10 Subtract 3y from both sides. False statement; the solution set is . The equation 4y + 7 – y = 10 + 3y is always false. There is no value of y that will make the equation true. There are no solutions. Check It Out! Example 3b Solve the equation. 2c + 7 + c = –14 + 3c + 21 2c + 7 + c = –14 + 3c + 21 3c + 7 = 3c + 7 Identify like terms. Combine like terms on the left and the right. The statement 3c + 7 = 3c + 7 is true for all values of c. The equation 2c + 7 + c = –14 + 3c + 21 is an identity. All values of c will make the equation true. In other words, all real numbers are solutions. Additional Example 4: Application Jon 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? Person Bulbs Jon 60 bulbs plus 44 bulbs per hour Sara 96 bulbs plus 32 bulbs per hour Additional Example 4 Continued Let h represent hours, and write expressions for the number of bulbs planted. When is 60 60 plus bulbs + 44h 44 bulbs each hour = the same as 96 96 bulbs + plus 32 bulbs ? each hour 32h 60 + 44h = 96 + 32h To collect the variable terms on one side, subtract 32h –32h –32h from both sides. 60 + 12h = 96 Additional Example 4 Continued 60 + 12h = 96 –60 – 60 12h = 36 h=3 Since 60 is added to 12h, subtract 60 from both sides. Since h is multiplied by 12, divide both sides by 12 to undo the multiplication. Additional Example 4 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 h = 3: 60 + 44h = 60 + 44(3) = 60 + 132 = 192 96 + 32h = 96 + 32(3) = 96 + 96 = 192 After 3 hours, Jon and Sara will each have planted 192 bulbs. Check It Out! Example 4 Four 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. Four times Greg's age 4g decreased by 3 – 3 is equal to = three times Greg's age 3g increased by + 7 7 . Check It Out! Example 4 Continued 4g – 3 = 3g + 7 –3g –3g g–3= +3 g= 7 +3 10 Greg is 10 years old. To collect the variable terms on one side, subtract 3g from both sides. Since 3 is subtracted from g, add 3 to both sides. Lesson Quiz Solve each equation. 1. 7x + 2 = 5x + 8 3 2. 4(2x – 5) = 5x + 4 8 3. 6 – 7(a + 1) = –3(2 – a) 4. 4(3x + 1) – 7x = 6 + 5x – 2 all real numbers 5. 1 20 hours 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?

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