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10-5

10-5. Solving for a Variable. Warm Up. Problem of the Day. Lesson Presentation. Pre-Algebra. PA HOMEWORK Answers. Page 517 #14-26 EVENS. Pre-Algebra HOMEWORK. Page 521 #1-11 ALL NO WORK= ZERO CREDIT! NO WORK= ZERO CREDIT !. Our Learning Goal.

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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. PA HOMEWORK Answers Page 517 #14-26 EVENS

  3. Pre-Algebra HOMEWORK Page 521 #1-11 ALL NO WORK= ZERO CREDIT! NO WORK= ZERO CREDIT!

  4. Our Learning Goal Students will be able to solve multi-step equations with multiple variables, solve inequalities and graph the solutions on a number line.

  5. Our Learning Goal Assignments • Learn to solve two-step equations. • Learn to solve multistep equations. • Learn to solve equations with variables on both sides of the equal sign. • Learn to solve two-step inequalities and graph the solutions of an inequality on a number line. • Learn to solve an equation for a variable. • Learn to solve systems of equations.

  6. Today’s Learning Goal Assignment Learn to solve an equation for a variable.

  7. Remember!!! 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.

  8. 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.

  9. 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.

  10. 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.

  11. 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.

  12. As the equations get more complicated…. REMEMBER… 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.

  13. √A = √s2 Take the square root of both sides. √A = sIsolate s. 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

  14. √w – 4 = √g2 √ w – 4 = gIsolate g. 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. = Divide both sides by I. IR V = RIsolate R. I I V I 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

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

  17. 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 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)

  18. s 180(n – 2) = 180 180 s s + 2 = n = (n – 2) 180 180 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.

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

  20. 2y = 2 –3x + 8 2 –3x y = + 4 2 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

  21. Additional Example 3 Continued 3x + 2y = 8

  22. 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

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

  24. 3V = h 1 C – S = l A 3 t P – 2w 2 Don’t forget your proper heading! Trade & Grade! 10-5 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

  25. y 2x 4 7 2 –4 –2 2 4 –2 y = – + 2 –4 Lesson Quiz: Part 2 5. Solve for y and graph 2x + 7y = 14.

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