Problem on Solving Age Problem Using System of Linear Equation (Again:'<)?

Question

1. Irene is one-third the age of Bea. Three years ago, Irene was twice as old as Irene. How old is each?

System Obtain:

Method Used:

Solving:

X:

Y:

2. Buddy is thrice the age of Nimfa. Six years later, Buddy will be twice as old as Nimfa. How old is each?

System Obtain:

Method Used:

Solving:

X:

Y:

3. Susan is half the age of Amy. Five years ago, Susan was three-fourths as old as Amy. How old is each?

System Obtain:

Method Used:

Solving:

X:

Y:

Each correct answer = points:'>

Answer 1

1. Let the present age of Irene be X and of Bea be Y.

Equation 1 (by reading the first sentence) : X = Y/3

Equation 2 (second sentence) : (X - 3) = 2 * (Y - 3)  (correct your question here, irene is written twice, not possible, i have assumed second irene to be Bea in second sentence)

Solving :

Subtract Eq. 2 from Eq. 1, we get

3 = Y/3 - 2 * (Y - 3)

=&gt; age of Bea = Y = 9/5 yrs.,  Bea is 9/5 years old.  Now put value replace Y by its value in any of the equation, say Eq 1, we get

X = 3/5, Irene is 3/5 yrs old.

2. Approx. the same question,

Let present age of buddy be X and of Nimfa be Y

Eq. 1 =&gt; X = 3 * Y

Eq. 2 =&gt; X + 6 = 2 * (Y + 6)

Solving :

Subtract Eq 1 from Eq 2 we get

6 = 2 * Y + 12 - 3 * Y

=&gt; age of Nimfa = Y = 6 yrs., Nimfa is 6 years old.

Replace Y in any Eq. 1 by its value. We get age of Buddy is X = 18 yrs old.

3. Let Susan&#039;s present age be X and that of Amy be Y

Eq. 1 =&gt; X = Y/2

Eq. 2 =&gt; X - 5 = 3 * (Y - 5)/4

Subtract Eq. 2 from Eq. 1 we will get

5 = Y/2 - 3 * (Y - 5)/4

=&gt;Y = -5 !! Impossible! age cannot be negative, 3rd question has a bug. Please correct it.

I hope this illustrates the basics of age-related word problems in mathematics.