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2.4 Solving Linear Equations

2.4 Solving Linear Equations. Objective: *Solve a linear equation in one variable. Recognize linear equations. Solve equations of the form x + B = C. Solve equations of the form Ax = C Solve equations of the form Ax + B = C. Definition.

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2.4 Solving Linear Equations

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  1. 2.4 Solving Linear Equations Objective: *Solve a linear equation in one variable

  2. Recognize linear equations. Solve equations of the form x + B = C. Solve equations of the form Ax = C Solve equations of the form Ax + B = C

  3. Definition A linear equation in one variable is any equation that can be written in the form ax + b = 0, where a and b are real numbers and a is not zero.

  4. Examples of Linear Equations • x + 30 = 330 • 10x = 50 • 2x + 3 = 13 Any equation that can be written in the form Ax + B = C, where A, B, C are real numbers.

  5. NOT Linear equations x2 + 5x -3 = 0 |x - 3| = 7 1/x = 12 √x = 25.

  6. Linear or Nonlinear? 5 = 2x 3 – s = ¼ 3 – t2 = ¼ 50 = ¼ r2

  7. Write an Equation for the Following x + 300 = 3,300 10 x = 500 I added B300 to my bank account and my balance is B3,300. What did I start with? I paid B500 for 10 bags of cherries. How much did each bag cost?

  8. Write an Equation for the Following 40 + 20x = 100 Parking costs a flat rate of B40 plus B20 per hour. I spent B100. How long did I parked?

  9. Solving Equations x + 3 = 7 x + 3 - 3 = 7 - 3 x = 4 When you solve equations, you have both sides of the scale! You can change the weight, but the scale must balance!

  10. IFL Solving equations always requires them to be balanced. An unbalanced or messy sequence ends in the wrong answer. *A satisfying life always requires a balance. Otherwise, life will become miserable. Dr. E. M. Role

  11. Inverse Operations To solve x + 3 = 7, you subtract 3 from both sides. To solve x - 3 = 7, you add 3 from both sides. Addition and subtraction areinverse operations. We always use inverse operations to solve equations.

  12. Examples x – 17 + 17 = 25 + 17 x = 42 12 – r –12 = 7 – 12 -r = -5 (-1)-r = -5(-1) r = 5 t – ½ + ½ = ¾ + ½ t = ¾ + 2/4 t = 5/4 x – 17 = 25 12 – r = 7 t – ½ = 3/4

  13. Examples 5t – 4t = 9 - 6 t = 3 4x + 2x -5x = 9 – 4 – 6+ 3 x = 2 Use the distributive property! 6 + 15x – 1 – 14x = 6 15x – 14x = 6 - 6 + 1 x = 1 5t – 4t + 6 = 9 4x + 6 + 2x – 3 = 9 + 5x – 4 3(2+5x) – (1+14x) = 6

  14. Examples -25 p = 50 -25 -25 p = -2 (3) 2 x= 6 (3) (2) 3 2 x = 9 (2) z = 19(2) (2) x = 38 Solve -25p = 50 Solve 2 x = 6 3 Solve z = 19 2

  15. Problem Set 2.4 (pp. 110-112) 8a + 2 = -6 8a = -6 -2 8a = -8 8 8 a = -1 1 = 4x + 2 2 2 1 = 2x + 1 0 = 2x 1 – 1 = 2x 0 = x 5.) 2(4a + 1) = -6 13.) 1 = 1 (4x + 2) 2

  16. Problem Set 2.4 (pp. 110-112) 1 x - 3 = 1 x + 1 2 2 4 4 1 x – 1 x = 1 + 3 2 4 4 2 2 x – 1 x = 1 + 6 44 4 4 1 x = 7 4 4 (4)1 x = 7(4) 4 4 x = 7 19.) 1 (x – 3) = 1 (x + 1) 2 4

  17. Problem Set 2.4 (pp. 110-112) 0.06x + 8 – 0.08x = 6.5 0.06x – 0.08x = 6.5 - 8 -0.02x = -1.5 -0.02x = -1.5 -0.02 -0.02 x = 75 27.) 0.06x + 0.08(100 – x) = 6.5

  18. Problem Set 2.4 (pp. 110 - 111) Individual Practice: Homework Multiples of 4 Nos. 4, 8, 12,…76, (88 only)

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