Solving Equations Objectives 1 To solve equations 2

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

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

Example #2: Using the Distributive Property Solve 3 x – 7(2 x – 13) = 3(-2 x + 9) 3 x – 14 x + 91= -6 x + 27 -11 x + 91 = -6 x + 27 +11 x 91 = 5 x + 27 - 27 64 = 5 x 5 5 12. 8 = x

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) 2 A = h (b 1+b 2) 2 A/ (b 1+b 2) = h

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 – bx ab = (a – b)x (a-b) ab/(a-b) = x

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 = 2 w + 2 l or P = 2(w + l) Since the length is 5 times the width, we’ll represent everything in terms of width. w = width 5 w = length

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

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 18 in. Find the lengths of the sides. Use the ratio and the perimeter to create an equation: 3 x + 4 x + 5 x = 18 12 12

Example #6: Using Ratios x = 1. 5 Now that you know what x is, it’s time to find 3 x, 4 x, 5 x 3 x = 3(1. 5) = 4. 5 4 x = 4(1. 5) = 6 5 x = 5(1. 5) = 7. 5

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

Example #7: Word Problems Find a relationship in the information. Distance for approaching plane + distance for the interceptor = 5000 mi 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

Example #7: Word Problems 800 t + 700 (t + 0. 25) = 5000 800 t +700 t + 175 = 5000 1500 t + 175 = 5000 -175 1500 t = 4825 1500 t = 3. 217 or 3 hr 13 min