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Warm Up. Problem of the Day. Lesson Presentation. Lesson Quizzes. x. 4. Warm Up Solve. 1. x + 14 = 27 2. 50 = 2 x – 16 3. = 12 4. 6 x = 72. 13. 33. 48. 12. Problem of the Day
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Warm Up Problem of the Day Lesson Presentation Lesson Quizzes
x 4 Warm Up Solve. 1.x + 14= 27 2.50 = 2x – 16 3. = 12 4. 6x = 72 13 33 48 12
Problem of the Day Sandra has some pencils. Kylie has twice as many pencils as Sandra. Together they have 42 pencils. Write and solve an equation to find how many pencils each girl has. x + 2x = 42; Sandra: 14, Kylie: 28
Sunshine State Standards MA.8.A.4.1 Solve literal equations for a specified variable.
Vocabulary literal equation
A literal equation is an equation with two or more variables.
Additional Example 1A: Solving for Variables in Formulas Solve each equation for the given variable. A = lw for l A = lw Divide both sides by w to isolate l.
Additional Example 1B: Solving for Variables in Formulas Solve each equation for the given variable. I = prt for t I = prt Divide both sides by pr to isolate t.
Check It Out: Example 1A Solve each equation for the given variable. A = ℓw for w. ℓwℓ Aℓ = Aℓ = w
Check It Out: Example 1B Solve each equation for the given variable. I = Prt for P. Prtrt Irt = Irt = P
Additional Example 2: Application The area of a rectangular table is 30 ft2. What is the width of the table if the length is 7.5 feet? First solve the area formula for w because you want to find the width. Then use the given values to find w. A = lw Divide both sides by l to isolate 1.
Additional Example 2 Continued The area of a rectangular table is 30 ft2. What is the width of the table if the length is 7.5 feet? w = 4 The width of the table is 4 feet.
Check It Out: Example 2 The formula V = Bh gives the volume V of a rectangular prism with base area B and height h. The volume of a rectangular prism is 96 cm3. What is the height of the prism if base area is 12 cm? VB V = Bh h = VB BhB 9612 = = VB = h h = 8 The height of the prism is 8 cm.
Additional Example 3A: Solving Literal Equations with Addition or Subtraction Solve each equation for the given variable. c = d – e for d Locate d in the equation. c = d – e Add e to both sides of the equation. c + e = d +e + e
–1 – 1 +3x + 3x Additional Example 3B: Solving Literal Equations with Addition or Subtraction Solve each equation for the given variable. 4x = y + 1 – 3x for y Locate y in the equation. 4x = y + 1 – 3x Subtract 1 from both sides of the equation. 4x – 1 = y– 3x Add 3x to both sides of the equation. 7x – 1 = y
Check It Out: Example 3A Solve each equation for the given variable. 3x –2y = y + z for z 3x – 2y = y + z 3x –3y = z z = 3x –3y –y –y
Check It Out: Example 3B Solve each equation for the given variable. 4t + 2s = 2t + r – s for r 4t + 2s = 2t + r – s 4t + 3s = 2t + r –2t –2t +s +s 2t + 3s = r
Additional Example 4: Solving Literal Equations for a Variable The equation t = m + 10e gives the test score t for a student who answers m multiple-choice questions and e essay questions correctly. Solve this equation for e. t = m + 10e Locate e in the equation. t = m + 10e Since m is added to 10e, subtract m from both sides. t – m = 10e –m –m Since e is multiplied 10, divide both sides by 10. t – m = 10e 10 10 t – m = e 10
Check It Out! Example 4 Solve 2k - 4d = r for k 2k – 4d = r r + 4d2 + 4d +4d k = r + 4d2 2k2 =
Lesson Quizzes Standard Lesson Quiz Lesson Quiz for Student Response Systems
Lesson Quiz 1. Solve 3Y = F for Y. 2. Solve g = hk for h. 3. Solve 2b + 3c = d for b. 4. Solve 6 + 2m – n = 3n + 9 for m. 5. What would be the length of a rectangle if its perimeter was 72 meters and its width was 16 meters? 20 m
Lesson Quiz for Student Response Systems 1. Solve for D. A. B.D = V C.D = mV D.
Lesson Quiz for Student Response Systems 2. Solve for m. A. B.D = Vm C.DV = m D.