1. Solving Equations Objectives: 1)To solve equations 2)To solve problems by writing equations

2. Solution to An Equation A number that makes the equation true is a solution to the equation. Ex: 5x + 23 = 88 -23 -23 5x = 65 5 5 x = 13 5(13) + 23 = 88 65 + 23 = 88 88 = 88 Since x = 13 satisfies as an answer to 5x + 23 = 88, we call 13 the solution Now, check your answer

3. Example #2: Using the Distributive Property Solve 3x – 7(2x – 13) = 3(-2x + 9) 3x – 14x + 91= -6x + 27 -11x + 91 = -6x + 27 +11x +11x 91 = 5x + 27 - 27 - 27 64 = 5x 5 5 12.8 = x

4. Example #3: Solving a Formula for One of its Variables The formula for the area of a trapezoid is A = ½ h(b 1 +b 2 ). Solve the formula for h 2(A) = 2( ½) h(b 1 +b 2 ) 2A = h (b 1 +b 2 ) (b 1 +b 2 ) 2A/ (b 1 +b 2 ) = h

5. Example #4: Solving an Equation for One of its Variables Solve x/a + 1 = x/b for any restrictions on a & b ab(x/a) + (ab)(1) = (ab)(x/b) bx + ab = ax ab = ax – bx ab = (a – b)x (a-b) ab/(a-b) = x

6. Example #5 Word Problems A dog kennel owner has 100 ft of fencing to enclose a rectangular dog run. She wants it to be 5 times as long as it is wide. Find the dimensions of the dog run. 1 st, remember the formula for the perimeter of a rectangle. P = 2w + 2l or P = 2(w + l) Since the length is 5 times the width, we’ll represent everything in terms of width. w = width 5w = length

7. Example #5 Word Problems Write the equation: 2w + 2(5w) = 100 2w + 10w = 100 12w = 100 12 12 w = 8 1/3 Remember that w represents the width, so 5w = 41 2/3 is the length

8. Example #6: Using Ratios The lengths of the sides of a triangle are in the ratio 3:4:5. The perimeter of the triangle is 18in. Find the lengths of the sides. Use the ratio and the perimeter to create an equation: 3x + 4x + 5x = 18 12x = 18 12 12

9. Example #6: Using Ratios x = 1.5 Now that you know what x is, it’s time to find 3x, 4x, 5x 3x = 3(1.5) = 4.5 4x = 4(1.5) = 6 5x = 5(1.5) = 7.5

10. Example #7: Word Problems Radar detected an unidentified plane 5000mi away, approaching at 700mi/h. Fifteen minutes later an interceptor plane was dispatched, traveling at 800mi/h. How long did the interceptor take to reach the approaching plane?

11. Example #7: Word Problems Find a relationship in the information. Distance for approaching plane + distance for the interceptor = 5000mi Let t = the time in hours for the interceptor t + 0.25 = the time in hours for the approaching plane. 800 t + 700(t + 0.25) = 5000

12. Example #7: Word Problems 800 t + 700 (t + 0.25) = 5000 800t +700t + 175 = 5000 1500t + 175 = 5000 -175 -175 1500t = 4825 1500 1500 t = 3.217 or 3hr 13min