Pre-Algebra 10-5 Solving for a Variable 10-5 Solving for a Variable Pre-Algebra Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation

Pre-Algebra 10-5 Solving for a Variable Warm Up Solve. 1. 8x – 9 = x + 12 = 4x x – 8 = 7x x + 3 = x = 4 x = 5 x = –11 Pre-Algebra 10-5 Solving for a Variable x = –, or –5 3 4

Pre-Algebra 10-5 Solving for a Variable 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

Pre-Algebra 10-5 Solving for a Variable Learn to solve an equation for a variable.

Pre-Algebra 10-5 Solving for a Variable 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.

Pre-Algebra 10-5 Solving for a Variable 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.

Pre-Algebra 10-5 Solving for a Variable Solve for the indicated variable. Additional Example 1B: Solving for a Variable by Addition or Subtraction 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.

Pre-Algebra 10-5 Solving for a Variable 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.

Pre-Algebra 10-5 Solving for a Variable Solve for the indicated variable. Try This: Example 1B 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.

Pre-Algebra 10-5 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.

Pre-Algebra 10-5 Solving for a Variable 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 √A = sIsolate s. √A = √s 2 Take the square root of both sides.

Pre-Algebra 10-5 Solving for a Variable 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. Solve for the indicated variable. Assume all values are positive.

Pre-Algebra 10-5 Solving for a Variable 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 ) 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 ).

Pre-Algebra 10-5 Solving for a Variable Solve for the indicated variable. Assume all values are positive. Try This: Example 2A A. Solve w = g for g. w = g √ 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.

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

Pre-Algebra 10-5 Solving for a Variable C. Solve s = 180(n – 2) for n. s = 180(n – 2) 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

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

Pre-Algebra 10-5 Solving for a Variable 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 –

Pre-Algebra 10-5 Solving for a Variable Additional Example 3 Continued 3x + 2y = 8

Pre-Algebra 10-5 Solving for a Variable 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

Pre-Algebra 10-5 Solving for a Variable Try This: Example 3 Continued x y –4 – –2 –4 4 4x + 3y = 12 –6

Pre-Algebra 10-5 Solving for a Variable 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 1 3 C – S t = h 3V3V A = l P – 2w 2

Pre-Algebra 10-5 Solving for a Variable Lesson Quiz: Part 2 5. Solve for y and graph 2x + 7y = 14. y = – + 2 2x 2x 7