1 / 35

Solve Equations With Variables on Both Sides

Solve Equations With Variables on Both Sides. Chapter 2: Lessons 4-5. Steps:. 1) Simplify each side: Distribute then Combine Like Terms 2) Cancel the variable on the RIGHT by doing the inverse (opposite) operation 3) Solve by using inverse operations add or subtract

rsherman
Télécharger la présentation

Solve Equations With Variables on Both Sides

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Solve Equations With Variables on Both Sides Chapter 2: Lessons 4-5

  2. Steps: • 1) Simplify each side: Distribute then Combine Like Terms • 2) Cancel the variable on the RIGHT by doing the inverse (opposite) operation • 3) Solve by using inverse operations add or subtract multiply or divide 4) Check

  3. Distribute (multiply) Combine like terms Cancel variable on right – use inverse operation Solve 2 step equation 2(x + 7) + 3 = 5x - 1 2x + 14+ 3 = 5x – 1 2x + 17 = 5x - 1 -5x -5x -3x +17 = - 1 -17 -17 -3x = -18 -3 -3 x = 6

  4. Replace X = 6 Checks Check - replace “x” with solution2(x + 7) + 3 = 5x - 1 2(6+ 7) + 3 =5(6) - 1 2(13)+ 3 =5(6) - 1 Follow order of operations on both sides of equation 26+ 3 =30- 1 29= 29  X = 6 is the solution to the equation

  5. Solve 2 step equation Use Steps to Solve Equation: -3x + 4 = 5x – 8 Cancel variable on right (-5x) x = 3/2

  6. distribute Solve 2 step equation Use Steps to Solve Equation:4(1 – 2x) = 4 – 6x Cancel variable on right (add 6x) x = 0

  7. Investigate this: 9 + 5x = 5x + 9

  8. 9 + 5x = 5x + 9 Cancel variable on right -5x -5x When solving, if you get a TRUE STATEMENT, then that means that ANY real number works. 9 = 9 This is called: Infinite Solutions

  9. 6x – 1 = 6x - 8 Try this:

  10. The variables zeroed out and remaining is afalsestatementwhere a number is equal to a different number, so there will be no number that will work in the equation. 6x – 1 = 6x – 8 -6x -6x -1 = - 8 x = no solution

  11. Review Steps: • 1) Simplify each side: • Distribute • Combine Like Terms • 2) Use inverse operation to cancel variable on RIGHT • 3) Solve 2-step equation using inverse operations

  12. Practice: 1. 2m – 6 + 4m = 12 2. 6a – 5(a – 1) = 11

  13. Solve the equation. 3. A charter bus company charges $11.25 per ticket plus a handling charge of $.50 per ticket, and a $15 fee for booking the bus. If a group pays $297 to charter a bus, how many tickets did they buy?

  14. ANSWER 11 ANSWER –4 You got this 1. 3b+2(b–4)=47 2. –6 + 4(2c + 1) = –34

  15. Classwork: Pg.

  16. ANSWER The solution is 2. Check by substituting 2 for x in the original equation. EXAMPLE 1 Solve an equation with variables on both sides Solve 7 – 8x= 4x– 17. 7 – 8x = 4x – 17 Write original equation. 7 – 8x + 8x = 4x – 17 + 8x Add 8x to each side. 7 = 12x – 17 Simplify each side. 24 = 12x Add 17 to each side. 2 = x Divide each side by 12.

  17. ? 7 – 8(2) = 4(2) – 17 ? –9 = 4(2) – 17 –9=–9 EXAMPLE 1 Solve an equation with variables on both sides CHECK 7 – 8x= 4x– 17 Write original equation. Substitute 2 for x. Simplify left side. Simplify right side. Solution checks.

  18. (16x + 60). Solve 9x–5= (16x + 60) 9x – 5= 1 1 4 4 EXAMPLE 2 Solve an equation with grouping symbols Write original equation. 9x – 5 = 4x + 15 Distributive property 5x – 5 = 15 Subtract 4x from each side. 5x =20 Add 5 to each side. x = 4 Divide each side by 5.

  19. ANSWER 3 for Examples 1 and 2 GUIDED PRACTICE Solve the equation. Check your solution. 1. 24 – 3m= 5m

  20. ANSWER 9 for Examples 1 and 2 GUIDED PRACTICE Solve the equation. Check your solution. 2. 20 +c= 4c – 7

  21. ANSWER –8 for Examples 1 and 2 GUIDED PRACTICE Solve the equation. Check your solution. 3. 9 – 3k= 17k – 2k

  22. ANSWER 6 for Examples 1 and 2 GUIDED PRACTICE Solve the equation. Check your solution. 4. 5z– 2= 2(3z – 4)

  23. ANSWER 2 for Examples 1 and 2 GUIDED PRACTICE Solve the equation. Check your solution. 5. 3 – 4a= 5(a – 3)

  24. 6. (6y + 15) 8y–6= ANSWER 4 2 3 for Examples 1 and 2 GUIDED PRACTICE Solve the equation. Check your solution.

  25. CAR SALES A car dealership sold 78 new cars and 67 used cars this year. The number of new cars sold by the dealership has been increasing by 6 cars each year. The number of used cars sold by the dealership has been decreasing by 4 cars each year. If these trends continue, in how many years will the number of new cars sold be twice the number of used cars sold? EXAMPLE 3 Solve a real-world problem

  26. SOLUTION Let x represent the number of years from now. So, 6x represents the increase in the number of new cars sold over x years and –4x represents the decrease in the number of used cars sold over x years. Write a verbal model. 78 6x 67 (– 4 x) ) + + = 2 ( EXAMPLE 3 Solve a real-world problem

  27. ANSWER The number of new cars sold will be twice the number of used cars sold in 4 years. EXAMPLE 3 Solve a real-world problem 78 + 6x =2(67 – 4x) Write equation. 78 + 6x = 134 – 8x Distributive property 78 + 14x =134 Add 8x to each side. 14x = 56 Subtract 78 from each side. x = 4 Divide each side by 14.

  28. You can use a table to check your answer. CHECK EXAMPLE 3 Solve a real-world problem

  29. 7. WHAT IF? In Example 3, suppose the car dealership sold 50 new cars this year instead of 78. In how many years will the number of new cars sold be twice the number of used cars sold? ANSWER 6 yr for Example 3 GUIDED PRACTICE

  30. a. 3x = 3(x + 4) b. 2x + 10 = 2(x + 5) EXAMPLE 4 Identify the number of solutions of an equation Solve the equation, if possible. SOLUTION a. 3x = 3(x + 4) Original equation 3x = 3x + 12 Distributive property The equation 3x = 3x + 12 is not true because the number 3x cannot be equal to 12 more than itself. So, the equation has no solution. This can be demonstrated by continuing to solve the equation.

  31. 0 = 12 ANSWER The statement 0 = 12 is not true, so the equation has no solution. EXAMPLE 4 Identify the number of solutions of an equation 3x – 3x= 3x + 12 – 3x Subtract 3x from each side. Simplify.

  32. ANSWER Notice that the statement 2x + 10 = 2x + 10 is true for all values of x. So, the equation is an identity, and the solution is all real numbers. EXAMPLE 1 EXAMPLE 4 Identify the number of solutions of an equation b.2x + 10 = 2(x + 5) Original equation 2x + 10 = 2x + 10 Distributive property

  33. ANSWER no solution for Example 4 GUIDED PRACTICE Solve the equation, if possible. 8. 9z + 12= 9(z + 3)

  34. ANSWER 0 for Example 4 GUIDED PRACTICE Solve the equation, if possible. 9. 7w + 1= 8w + 1

  35. ANSWER identity for Example 4 GUIDED PRACTICE Solve the equation, if possible. 10. 3(2a + 2)= 2(3a + 3)

More Related