Lord i offer my life

Name:David B. Victorioso
Year/section:IV-Acacia
Date:Saeptember 16,2013
Teacher:Mrs.Noceda
Age Problem
Problem 1 An eagle is 4 times as old as a falcon. Three years ago, the eagle was 7 times as old as the falcon. Find the present age of each bird.

Solution x = falcon's age now 4x = eagle's age now {the eagle is 4 times as old as falcon} x - 3 = falcon's age 3 years ago 4x - 3 = eagle's age 3 years ago 4x – 3 = 7(x – 3) {three years ago, eagle was 7 times the falcon} 4x – 3 = 7x – 21 {used distributive property} 4x = 7x -18 {added 3 to both sides} -3x = -18 {subtracted 7x from both sides} x = 6 {divided both sides by -3} 4x = 24 {substituted 6, in for x, into 4x}

falcon is 6 now eagle is 24 now
Problem 2 5 years from now Sharon will be twice as old as Tiffany. The current sum of the ages of Sharon and Tiffany is 86. How old is Tiffany right now?

Solution Sharon + 5 = 2 × (Tiffany + 5) Sharon + Tiffany = 86 Sharon = 86 - Tiffany 86 - Tiffany + 5 = 2 × (Tiffany + 5) 91 - Tiffany = 2 × Tiffany + 10 91 = 3 × Tiffany + 10 3 × Tiffany = 91 - 10 = 81 Tiffany = 27
Problem 3 If three times Kathy's age is decreased by 36, the result is twice Kathy's age. How old is Kathy?"
Solution
x = Kathy's age 3x - 36 = 2x {three times her age minus 36 equals twice her age} -36 = -x {subtracted 3x from both sides} x = 36 {divided both sides by -1}

Kathy is 36
Problem 4 Brenda is 4 years older than Walter, and Carol is twice as old as Brenda. Three years ago, the sum of their ages was 35.
How old is each now?

Solution x = Walter's age now x + 4 = Brenda's age now {Brenda is 4 yrs older than Walter} 2(x + 4) = 2x + 8 = Carol's age now {Carol is twice as old as Brenda, used distributive property} x - 3 = Walter's age 3 years ago