 # Holt Algebra Solving Equations with Variables on Both Sides Warm Up Simplify. 1. 4x – 10x 2. –7(x – 3) – (x – 2) Solve. 5. 3x + 2 = 8 6.

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Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides Warm Up Simplify. 1. 4x – 10x 2. –7(x – 3) 3. 4. 15 – (x – 2) Solve. 5. 3x + 2 = 8 6. –6x –7x + 21 17 – x 2x + 3 2 28

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides Solve equations in one variable that contain variable terms on both sides. Objective

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides Solve 7n – 2 = 5n + 6. Example 1: Solving Equations with Variables on Both Sides To collect the variable terms on one side, subtract 5n from both sides. 7n – 2 = 5n + 6 –5n 2n – 2 = 6 Since n is multiplied by 2, divide both sides by 2 to undo the multiplication. 2n = 8 + 2 n = 4

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides Solve 4b + 2 = 3b. Check It Out! Example 1a To collect the variable terms on one side, subtract 3b from both sides. 4b + 2 = 3b –3b b + 2 = 0 b = –2 – 2 – 2

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides Solve 0.5 + 0.3y = 0.7y – 0.3. Check It Out! Example 1b To collect the variable terms on one side, subtract 0.3y from both sides. 0.5 + 0.3y = 0.7y – 0.3 –0.3y 0.5 = 0.4y – 0.3 0.8 = 0.4y +0.3 + 0.3 2 = y Since 0.3 is subtracted from 0.4y, add 0.3 to both sides to undo the subtraction. Since y is multiplied by 0.4, divide both sides by 0.4 to undo the multiplication.

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides To solve more complicated equations, you may need to first simplify by using the Distributive Property or combining like terms.

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides Solve 4 – 6a + 4a = –1 – 5(7 – 2a). Example 2: Simplifying Each Side Before Solving Equations Combine like terms. Distribute –5 to the expression in parentheses. 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 +36 40 – 2a = 10a + 2a +2a 40 = 12a Since –36 is added to 10a, add 36 to both sides. To collect the variable terms on one side, add 2a to both sides.

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides Solve 4 – 6a + 4a = –1 – 5(7 – 2a). Example 2 Continued 40 = 12a Since a is multiplied by 12, divide both sides by 12.

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides Solve. Check It Out! Example 2A Since 1 is subtracted from b, add 1 to both sides. Distribute to the expression in parentheses. 1 2 + 1 3 = b – 1 To collect the variable terms on one side, subtract b from both sides. 1 2 4 = b

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides Solve 3x + 15 – 9 = 2(x + 2). Check It Out! Example 2B Combine like terms. Distribute 2 to the expression in parentheses. 3x + 15 – 9 = 2(x + 2) 3x + 15 – 9 = 2(x) + 2(2) 3x + 15 – 9 = 2x + 4 3x + 6 = 2x + 4 –2x x + 6 = 4 – 6 – 6 x = –2 To collect the variable terms on one side, subtract 2x from both sides. Since 6 is added to x, subtract 6 from both sides to undo the addition.

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides An identity is an equation that is true for all values of the variable. An equation that is an identity has infinitely many solutions. A contradiction is an equation that is not true for any value of the variable. It has no solutions.

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides WORDS Identity When solving an equation, if you get an equation that is always true, the original equation is an identity, and it has infinitely many solutions. NUMBERS 2 + 1 = 2 + 1 3 = 3 ALGEBRA 2 + x = 2 + x –x –x 2 = 2 Identities and Contradictions

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides Contradiction When solving an equation, if you get a false equation, the original equation is a contradiction, and it has no solutions. WORDS x = x + 3 –x –x 0 = 3  1 = 1 + 2 1 = 3  ALGEBRA NUMBERS Identities and Contradictions

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides Solve 10 – 5x + 1 = 7x + 11 – 12x. Example 3A: Infinitely Many Solutions or No Solutions Add 5x to both sides. Identify like terms. 10 – 5x + 1 = 7x + 11 – 12x 11 – 5x = 11 – 5x 11 = 11 + 5x + 5x True statement. Combine like terms on the left and the right. 10 – 5x + 1 = 7x + 11 – 12x 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.

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides Solve 12x – 3 + x = 5x – 4 + 8x. Example 3B: Infinitely Many Solutions or No Solutions Subtract 13x from both sides. Identify like terms. 12x – 3 + x = 5x – 4 + 8x 13x – 3 = 13x – 4 –3 = –4 –13x False statement. Combine like terms on the left and the right. 12x – 3 + x = 5x – 4 + 8x  The equation 12x – 3 + x = 5x – 4 + 8x is a contradiction. There is no value of x that will make the equation true. There are no solutions.

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides Solve 4y + 7 – y = 10 + 3y. Check It Out! Example 3a Subtract 3y from both sides. Identify like terms. 4y + 7 – y = 10 + 3y 3y + 7 = 3y + 10 7 = 10 –3y –3y False statement. Combine like terms on the left and the right. 4y + 7 – y = 10 + 3y  The equation 4y + 7 – y = 10 + 3y is a contradiction. There is no value of y that will make the equation true. There are no solutions.

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides Solve 2c + 7 + c = –14 + 3c + 21. Check It Out! Example 3b Subtract 3c both sides. Identify like terms. 2c + 7 + c = –14 + 3c + 21 3c + 7 = 3c + 7 7 = 7 –3c –3c True statement. Combine like terms on the left and the right. 2c + 7 + c = –14 + 3c + 21 The equation 2c + 7 + c = –14 + 3c + 21 is an identity. All values of c will make the equation true. All real numbers are solutions.

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides 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? Example 4: Application PersonBulbs Jon60 bulbs plus 44 bulbs per hour Sara96 bulbs plus 32 bulbs per hour

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides Example 4: Application Continued Let b represent bulbs, and write expressions for the number of bulbs planted. 60 bulbs plus 44 bulbs each hour the same as 96 bulbs plus 32 bulbs each hour When is ? 60 + 44b = 96 + 32b – 32b To collect the variable terms on one side, subtract 32b from both sides. 60 + 12b = 96

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides Example 4: Application Continued Since 60 is added to 12b, subtract 60 from both sides. 60 + 12b = 96 –60 – 60 12b = 36 Since b is multiplied by 12, divide both sides by 12 to undo the multiplication. b = 3

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides Example 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: 60 + 44b = 60 + 44(3) = 60 + 132 = 192 96 + 32b = 96 + 32(3) = 96 + 96 = 192 After 3 hours, Jon and Sara will each have planted 192 bulbs.

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides Four times Greg's age, decreased by 3 is equal to 3 times Greg's age increased by 7. How old is Greg? Check It Out! Example 4 Let g represent Greg's age, and write expressions for his age. four times Greg's age decreased by 3 is equal to three times Greg's age increased by 7. 4g – 3 = 3g + 7

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides Check It Out! Example 4 Continued 4g – 3 = 3g + 7 To collect the variable terms on one side, subtract 3g from both sides. g – 3 = 7 –3g Since 3 is subtracted from g, add 3 to both sides. + 3 + 3 g = 10 Greg is 10 years old.

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides Lesson Quiz Solve each equation. 1. 7x + 2 = 5x + 8 2. 4(2x – 5) = 5x + 4 3. 6 – 7(a + 1) = –3(2 – a) 4. 4(3x + 1) – 7x = 6 + 5x – 2 5. 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? 3 8 all real numbers 1 20 hours

Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides Home work pg. 103 16-46 evens 64,68,70

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