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

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Presentation on theme: "10-5 Solving for a Variable Course 3 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation."— Presentation transcript:

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

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

3 Problem of the Day The formula A = 4r 2 gives the surface area of a geometric figure. Solve the formula for r. Can you identify what the geometric figure is? sphere Course Solving for a Variable

4 Learn to solve an equation for a variable. Course Solving 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. Course Solving for a Variable

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

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

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

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

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

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

13 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. A = h(b 1 + b 2 ) 1 2 Write the formula.2 A = 2 h(b 1 + b 2 ) 1 2 Multiply both sides by 2. 2A = h(b 1 + b 2 ) Course Solving for a Variable 2A (b 1 + b 2 ) = h Isolate h. 2A (b 1 + b 2 ) h(b 1 + b 2 ) (b 1 + b 2 ) = Divide both sides by (b 1 + b 2 ).

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

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

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

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

18 Additional Example 3: Solving for y and Graphing Solve for y and graph 3x + 2y = 8. 3x + 2y = 8 –3x 2y = –3x + 8 –3x y2y 2 =y = + 4 –3x 2 xy – Course Solving for a Variable

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

20 Try This: Example 3 Solve for y and graph 4x + 3y = 12. 4x + 3y = 12 –4x – 4x 3y = –4x + 12 –4x y3y 3 =y = + 4 –4x 3 xy – –4 Course Solving for a Variable

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

22 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 Insert Lesson Title Here C – Rt = S 1 3 C – S t = h 3V3V A = l P – 2w 2 Course Solving for a Variable

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


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