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10-5

10-5. Solving for a Variable. Course 3. Warm Up. Problem of the Day. Lesson Presentation. Solving for a Variable. 10-5. 1. 23. 3. 1. 2. 4. 4. 8. x = – , or –5. Course 3. Warm Up Solve. 1. 8 x – 9 = 23 2. 9 x + 12 = 4 x + 37 3. 6 x – 8 = 7 x + 3 4. x + 3 = .

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10-5

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  1. 10-5 Solving for a Variable Course 3 Warm Up Problem of the Day Lesson Presentation

  2. Solving for a Variable 10-5 1 23 3 1 2 4 4 8 x = – , or –5 Course 3 Warm Up Solve. 1.8x – 9 = 23 2. 9x + 12 = 4x + 37 3. 6x – 8 = 7x + 3 4.x + 3 = x = 4 x = 5 x = –11

  3. Solving for a Variable 10-5 Course 3 Problem of the Day The formula A = 4r2 gives the surface area of a geometric figure. Solve the formula for r. Can you identify what the geometric figure is? sphere

  4. Solving for a Variable 10-5 Course 3 Learn to solve an equation for a variable.

  5. Solving for a Variable 10-5 Course 3 If an equation contains more than one variable, you can sometimes isolate one of the variables by using inverse operations. You can add and subtract any variable quantity on both sides of an equation.

  6. Solving for a Variable 10-5 Course 3 Additional Example 1A: Solving for a Variable by Addition or Subtraction Solve for the indicated variable. A. Solve a – b + 1 = c for a. a – b + 1 = c Add b and subtract 1 from both sides. + b – 1+ b – 1 a = c + b – 1 Isolate a.

  7. Solving for a Variable 10-5 Course 3 Additional Example 1B: Solving for a Variable by Addition or Subtraction Solve for the indicated variable. B. Solve a – b + 1 = c for b. a – b + 1 = c Subtract a and 1 from both sides. – a – 1– a – 1 –b = c – a – 1 Isolate b. –1  (–b) = –1 (c – a – 1) Multiply both sides by –1. b = –c + a + 1 Isolate b.

  8. Solving for a Variable 10-5 Course 3 Try This: Example 1A Solve for the indicated variable. A. Solve y – b + 3 = c for y. y – b + 3 = c Add b and subtract 3 from both sides. + b – 3+ b – 3 y = c + b – 3 Isolate y.

  9. Solving for a Variable 10-5 Course 3 Try This: Example 1B Solve for the indicated variable. B. Solve p – w + 4 = f for w. p – w + 4 = f Subtract p and 4 from both sides. – p – 4– p – 4 –w = f – p – 4 Isolate w. –1  (–w) = –1 (f – p – 4) Multiply both sides by –1. w = –f + p + 4 Isolate w.

  10. Solving for a Variable 10-5 Course 3 To isolate a variable, you can multiply or divide both sides of an equation by a variable if it can never be equal to 0. You can also take the square root of both sides of an equation that cannot have negative values.

  11. Solving for a Variable 10-5 √A = √s2 Take the square root of both sides. √A = sIsolate s. Course 3 Additional Example 2A: Solving for a Variable by Division or Square Roots Solve for the indicated variable. Assume all values are positive. A. Solve A = s2 for s. A = s2

  12. Solving for a Variable 10-5 = Divide both sides by I. IR V = RIsolate R. I I V I Course 3 Additional Example 2B: Solving for a Variable by Division or Square Roots Solve for the indicated variable. Assume all values are positive. B. Solve V = IR for R. V = IR

  13. Solving for a Variable 10-5 A = h(b1 + b2) Write the formula. 2 •A = 2 • h(b1 + b2) 1 1 2 2 Divide both sides by (b1 + b2). Isolate h. = = h 2A h(b1 + b2) 2A (b1 +b2) (b1 +b2) (b1 +b2) Course 3 Additional Example 2C: Solving for a Variable by Division or Square Roots C. Solve the formula for the area of a trapezoid for h. Assume all values are positive. Multiply both sides by 2. 2A = h(b1 + b2)

  14. Solving for a Variable 10-5 √w – 4 = √g2 √ w – 4 = gIsolate g. Course 3 Try This: Example 2A Solve for the indicated variable. Assume all values are positive. A. Solve w = g2 + 4 for g. w = g2 + 4 – 4– 4Subtract 4 from both sides. w – 4 = g2 Take the square root of both sides.

  15. Solving for a Variable 10-5 = Divide both sides by l. lw A = wIsolate w. l l A l Course 3 Try This: Example 2B Solve for the indicated variable. Assume all values are positive. B. Solve A = lw for w. A = lw

  16. Solving for a Variable 10-5 s 180(n – 2) = 180 180 s s + 2 = n = (n – 2) 180 180 Course 3 Try This: Example 2C Solve for the indicated variable. Assume all values are positive. C. Solve s = 180(n – 2) for n. s = 180(n – 2) Divide both sides by 180. + 2 + 2 Add 2 to both sides.

  17. Solving for a Variable 10-5 Remember! To find solutions (x, y), choose values for x substitute to find y. Course 3

  18. Solving for a Variable 10-5 2y = 2 –3x + 8 2 –3x y = + 4 2 Course 3 Additional Example 3: Solving for y and Graphing Solve for y and graph 3x + 2y = 8. 3x + 2y = 8 –3x–3x x y 2y = –3x + 8 –2 7 0 4 2 1 4 –2

  19. Solving for a Variable 10-5 Course 3 Additional Example 3 Continued 3x + 2y = 8

  20. Solving for a Variable 10-5 3y = 3 –4x + 12 3 –4x y = + 4 3 Course 3 Try This: Example 3 Solve for y and graph 4x + 3y = 12. 4x + 3y = 12 x y –4x– 4x –3 8 3y = –4x + 12 0 4 3 0 6 –4

  21. Solving for a Variable 10-5 Course 3 Try This: Example 3 Continued y 10 8 6 4x + 3y = 12 4 2 x –4 –2 2 4 6 –2 –4 –6

  22. Solving for a Variable 10-5 3V = h 1 C – S = l A 3 t P – 2w 2 Course 3 Insert Lesson Title Here Lesson Quiz: Part 1 Solve for the indicated variable. 1.P = R – C for C. 2.P = 2l+ 2w for l. 3.V = Ah for h. 4.R = for S. C = R - P C – Rt = S

  23. Solving for a Variable 10-5 2x 7 y = – + 2 Course 3 Insert Lesson Title Here Lesson Quiz: Part 2 5. Solve for y and graph 2x + 7y = 14.

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