# Lesson Solving Age Problems

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 This Lesson (Solving Age Problems) was created by by algebrahouse.com(1079)  : View Source, ShowAbout algebrahouse.com: Visit: www.algebrahouse.com to ask questions and for notes, examples, and more When solving age problems, you need to represent the following in terms of a variable: - the present ages of the people or things involved - the age, at the other specified time, of the people or things involved Then, form an equation based on these representations. Here are some examples: -------------------------------------------------------------------------------- Cary is 9 years older than Dan. In 7 years, the sum of their ages will equal 93. Find both of their ages now. x = Dan's age now x + 9 = Cary's age now {Cary is 9 yrs older than Dan} x + 7 = Dan's age in 7 years x + 16 = Cary's age in 7 years x + 7 + x + 16 = 93 {in seven years the sum of their ages will be 93} 2x + 23 = 93 {combined like terms} 2x = 70 {subtracted 23 from both sides} x = 35 {divided both sides by 35} x + 9 = 44 {substituted 35, in for x, into x + 9} Dan is 35 Cary is 44 www.algebrahouse.com -------------------------------------------------------------------------------- Fred is 4 times as old as his niece, Selma. Ten years from now, he will be twice as old as she will be. How old is each now? x = Selma's age now 4x = Fred's age now {Fred is 4 times as old as Selma} x + 10 = Selma's age in 10 years 4x + 10 = Fred's age in 10 years 4x + 10 = 2(x + 10) {his age in 10 yrs is twice her age in 10 years} 4x + 10 = 2x + 20 {used distributive property} 2x + 10 = 20 {subtracted 2x from both sides} 2x = 10 {subtracted 10 from both sides} x = 5 {divided both sides by 2} 4x = 20 {substituted 5, in for x, into 4x} Selma is 5 now Fred is 20 now www.algebrahouse.com -------------------------------------------------------------------------------- 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. 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 www.algebrahouse.com -------------------------------------------------------------------------------- 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? 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 {subtracted 3 from x} x + 1 = Brenda's age 3 years ago {subtracted 3 from x + 4} 2x + 5 = Carol's age 3 years ago {subtracted 3 from 2x + 8} (x - 3) + (x + 1) + (2x + 5) = 35 {sum of ages, 3 years ago, was 35} x - 3 + x + 1 + 2x + 5 = 35 {took out parentheses} 4x + 3 = 35 {combined like terms} 4x = 32 {subtracted 3 from both sides} x = 8 = Walter now {divided both sides by 4} x + 4 = 12 = Brenda now {substituted 8, in for x, into x + 4} 2(x + 4) = 24 = Carol now {substituted 8, in for x, into 2(x + 4)} Walter is 8 now Brenda is 12 now Carol is 24 now www.algebrahouse.com -------------------------------------------------------------------------------- This lesson has been accessed 34404 times.