2. Solving Equations with Variables on Both Sides Tutorial 6d

3. Introduction  In the previous lessons you learned how to solve one and two step equations, like: 4x = 12 or 7x – 5 = 16  In this lesson you will learn to solve equations with variables on both sides, for example: 3x – 5 = 5 + 2x or 4(2x + 1) = 6 – 5x  The steps in solving an equation that has variables on both sides is similar to that of two-step equations, except you must get the variable on just one side first.

4. Example #1  Solve: 2x + 3x – 3 = 2x + 12  Now it is a two-step equation. To undo the subtraction, add a 3 to each side of the equation. Always check your answers! 2x + 3x – 3 = 2x + 12; Does x =5? 25 + 35 – 5 = 25 + 12 is true! 3x – 3 = 12 x = 5 3 15 1 1 3x = 15  To undo the multiplication, divide by 3 on each side of the equation.  The goal is to get the variable on just one side of the equation.  That can be done by subtracting a 2x from both sides of the equation. + 3 5x – 3 = 2x + 12 -2x  Combine like terms.

5. Example #2  Solve: 3x – 40 = 2(x – 5)  Now it is a one-step equation. To undo the subtraction, add a 40 to each side of the equation. Always check your answers! 3x – 40 = 2(x – 5); Does x =30? 330 – 40 = 2(30 – 5) is true! x – 40 = -10 x = 30  The goal is to get the variable on just one side of the equation.  That can be done by subtracting a 2x from both sides of the equation. + 40 3x – 40 = 2x – 10 -2x  Distribute the 2 on the right side.

6. Problem Solving  Mary & Jocelyn are sisters. They left school at 3:00 P.M. and bicycled home along the same bike path. Mary rode at a speed of 12 mi/h. Jocelyn rode at 9 mi/h. Mary got home 15 minutes before Jocelyn. How long did it take Mary to get home? Define your variable: Let t = Mary’s time in hours Let t +.25 = Jocelyn’s time in hours. Since Mary got home 15 minutes before Jocelyn, Jocelyn must have been traveling for 15 minutes longer than Mary. 15 minutes equals 1 / 4 or.25 of an hour.

7. Problem Solving  Mary & Jocelyn are sisters. They left school at 3:00 P.M. and bicycled home along the same bike path. Mary rode at a speed of 12 mi/h. Jocelyn rode at 9 mi/h. Mary got home 15 minutes before Jocelyn. How long did it take Mary to get home? Define your variable: Let t = Mary’s time in hours Let t +.25 = Jocelyn’s time in hours. Mary’s distance Jocelyn’s distance We will need to use the formula: Distance = Rate Time d = r t You can see that they are both traveling the same distance.

8. Problem Solving  Mary & Jocelyn are sisters. They left school at 3:00 P.M. and bicycled home along the same bike path. Mary rode at a speed of 12 mi/h. Jocelyn rode at 9 mi/h. Mary got home 15 minutes before Jocelyn. How long did it take Mary to get home? Define your variable: Let t = Mary’s time in hours Let t +.25 = Jocelyn’s time in hours. Relate: Write: 12t = 9(t +.25) Solve: 12t = 9 (t +.25) Mary’s Distance (ratetime) equals Jocelyn’s distance (ratetime)

9. Problem Solving  Mary & Jocelyn are sisters. They left school at 3:00 P.M. and bicycled home along the same bike path. Mary rode at a speed of 12 mi/h. Jocelyn rode at 9 mi/h. Mary got home 15 minutes before Jocelyn. How long did it take Mary to get home?  Now it is a one-step equation. To undo the multiplication, divide each side of the equation by 3. 3t = 2.25 t =.75  The goal is to get the variable on just one side of the equation.  That can be done by subtracting a 9t from both sides of the equation. 12t = 9t + 2.25 -9t  Distribute the 9 on the right side. 12t = 9(t +.25) 12t = 9t + 9(.25) 3 1.75 1 1

10. Problem Solving  Mary & Jocelyn are sisters. They left school at 3:00 P.M. and bicycled home along the same bike path. Mary rode at a speed of 12 mi/h. Jocelyn rode at 9 mi/h. Mary got home 15 minutes before Jocelyn. How long did it take Mary to get home? 3t = 2.25 t =.75 12t = 9t + 2.25 -9t 12t = 9(t +.25) 12t = 9t + 9(.25) 3 1.75 1 1 It took Mary.75 hour or 45 minutes to get home.