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10-5

10-5. Solving for a Variable. Warm Up. Problem of the Day. Lesson Presentation. Pre-Algebra. Solving for a Variable. 10-5. 1. 23. 3. 1. 2. 4. 4. 8. x = – , or –5. Pre-Algebra. 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 Warm Up Problem of the Day Lesson Presentation Pre-Algebra

  2. Solving for a Variable 10-5 1 23 3 1 2 4 4 8 x = – , or –5 Pre-Algebra 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. 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. Learn to solve an equation for a variable.

  5. 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. 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. 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. 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. 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. 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.

  11. = Divide both sides by I. IR V = RIsolate R. I I V I Additional Example 2A: Solving for a Variable by Division or Square Roots Solve for the indicated variable. Assume all values are positive. A. Solve V = IR for R. V = IR

  12. A = h(b1 + b2) Write the formula. 2  A = 2 h(b1 + b2) 1 1 2 2 Isolate h. Divide both sides by (b1 + b2). = h = h(b1 + b2) 2A 2A (b1 +b2) (b1 +b2) (b1 +b2) Additional Example 2B: Solving for a Variable by Division or Square Roots B. 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)

  13. = Divide both sides by l. lw A = wIsolate w. l l A l Try This: Example 2A Solve for the indicated variable. Assume all values are positive. A. Solve A = lw for w. A = lw

  14. s 180(n – 2) = 180 180 s s + 2 = n = (n – 2) 180 180 Try This: Example 2B Solve for the indicated variable. Assume all values are positive. B. Solve s = 180(n – 2) for n. s = 180(n – 2) Divide both sides by 180. + 2 + 2 Add 2 to both sides.

  15. Remember! To find solutions (x, y), choose values for x substitute to find y.

  16. 2y = 2 –3x + 8 2 –3x y = + 4 2 Additional Example 3: Solving for y Solve for y: 3x + 2y = 8. 3x + 2y = 8 –3x–3x 2y = –3x + 8

  17. 3y = 3 –4x + 12 3 –4x y = + 4 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

  18. Try This: Example 3 Continued y 10 8 6 4x + 3y = 12 4 2 x –4 –2 2 4 6 –2 –4 –6

  19. 3V = h 1 C – S = l A 3 t P – 2w 2 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

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