Solving Equations with Variables on Both Sides Course 3 Solving Equations with Variables on Both Sides To solve multi-step equations with variables on both sides, first combine like terms and clear fractions. Then add or subtract variable terms to both sides so that the variable occurs on only one side of the equation. Then use properties of equality to isolate the variable.

Example 1: Solving Equations with Variables on Both Sides Course 3 Solving Equations with Varibles on Both Sides Example 1: Solving Equations with Variables on Both Sides Solve. 4x + 6 = x 4x + 6 = x – 4x – 4x Subtract 4x from both sides. 6 = –3x 6 –3 –3x = Divide both sides by –3. –2 = x

Example 2: Solving Equations with Variables on Both Sides Course 3 Solving Equations with Variables on Both Sides Example 2: Solving Equations with Variables on Both Sides Solve. 9b – 6 = 5b + 18 9b – 6 = 5b + 18 – 5b – 5b Subtract 5b from both sides. 4b – 6 = 18 + 6 + 6 Add 6 to both sides. 4b = 24 4b 4 24 = Divide both sides by 4. b = 6

Solving Equations with Variables on Both Sides Course 3 Solving Equations with Variables on Both Sides if the variables in an equation are eliminated and the resulting statement is false, the equation has no solution. Helpful Hint

Example 3: Solving Equations with Variables on Both Sides Course 3 Solving Equations with Variables on Both Sides Example 3: Solving Equations with Variables on Both Sides Solve. 9w + 3 = 9w + 7 9w + 3 = 9w + 7 – 9w – 9w Subtract 9w from both sides. 3 ≠ 7 No solution. There is no number that can be substituted for the variable w to make the equation true.

Solving Equations with Variables on Both Sides Course 3 Solving Equations with Variables on Both Sides if the variables in an equation are eliminated and the resulting statement is true, the equation has many solutions. Helpful Hint

Solving Equations with Variables on Both Sides Course 3 Solving Equations with Variables on Both Sides Example 4 Solve. 5x + 8 = 5x + 8 5x + 8 = 5x + 8 – 5x – 5x Subtract 5x from both sides. 8 = 8 True statement. Many solutions. There is an infinite amount of numbers that can be substituted for the variable x to make the equation true.

Solving Equations with Variables on Both Sides Course 3 Solving Equations with Variables on Both Sides Example 5 Solve. 3b – 2 = 2b + 12 3b – 2 = 2b + 12 – 2b – 2b Subtract 2b from both sides. b – 2 = 12 + 2 + 2 Add 2 to both sides. b = 14 One solution. There is only one number that can be substituted for the variable b to make the equation true.

Solving Equations with Variables on Both Sides Course 3 Solving Equations with Variables on Both Sides Example 6 Solve. 3w + 1 = 3w + 8 3w + 1 = 3w + 8 – 3w – 3w Subtract 3w from both sides. 1 ≠ 8 No solution. There is no number that can be substituted for the variable w to make the equation true.

Example 7: Solving Multi-Step Equations with Variables on Both Sides Course 3 Solving Equations with Variables on Both Sides Example 7: Solving Multi-Step Equations with Variables on Both Sides Solve. 10z – 15 – 4z = 8 – 2z - 15 10z – 15 – 4z = 8 – 2z – 15 6z – 15 = –2z – 7 Combine like terms. + 2z + 2z Add 2z to both sides. 8z – 15 = – 7 + 15 +15 Add 15 to both sides. 8z = 8 8z 8 8 = Divide both sides by 8. z = 1

Business Application Example Course 3 Solving Equations with Variables on Both Sides Business Application Example Daisy’s Flowers sell a rose bouquet for $39.95 plus $2.95 for every rose. A competing florist sells a similar bouquet for $26.00 plus $4.50 for every rose. Find the number of roses that would make both florist’s bouquets cost the same price.

Business Application Example Continued Course 3 Solving Equations with Variables on Both Sides Business Application Example Continued Let r represent the price of one rose. 39.95 + 2.95r = 26.00 + 4.50r Subtract 2.95r from both sides. – 2.95r – 2.95r 39.95 = 26.00 + 1.55r Subtract 26.00 from both sides. – 26.00 – 26.00 13.95 = 1.55r 13.95 1.55 1.55r 1.55 = Divide both sides by 1.55. 9 = r The two services would cost the same when using 9 roses.

Insert Lesson Title Here Course 3 Solving Equations with Variables on Both Sides Insert Lesson Title Here Practice Solve. 1. 4x + 16 = 2x 2. 8x – 4 = 4(2x – 1) 3. 2(3x + 11) = 6x + 4 4. x = x – 9 5. An apple has about 30 calories more than an orange. Five oranges have about as many calories as 3 apples. How many calories are in each? x = –8 Many Solutions no solution 1 4 1 2 x = 36 An orange has 45 calories. An apple has 75 calories.