Solving Brothers' Ages Word Problem

This problem involves finding the ages of three brothers based on the sums of ages of pairs and then calculating the age difference between the eldest and the youngest. We can solve this by setting up a system of linear equations.

Understanding the Ages

Let the ages of the three brothers be denoted by variables. To make it easier, let's assume they are ordered from youngest to eldest.

• Let the youngest brother's age be \$x\$ years.
• Let the middle brother's age be \$y\$ years.
• Let the eldest brother's age be \$z\$ years.

We assume that \$x < y < z\$.

Setting Up the Equations for Brother's Ages

The problem gives us the sums of ages of two brothers at a time. Since we assumed the ages are ordered, the sums must correspond to specific pairs:

• The smallest sum (4 years) must be the sum of the ages of the two youngest brothers:

  \$x + y = 4 \quad \text{(Equation 1)}\$

• The largest sum (8 years) must be the sum of the ages of the two eldest brothers:

  \$y + z = 8 \quad \text{(Equation 2)}\$

• The remaining sum (6 years) must be the sum of the ages of the youngest and the eldest brothers:

  \$x + z = 6 \quad \text{(Equation 3)}\$

So we have a system of three linear equations with three variables:

Solving the System of Equations for Ages

There are several ways to solve this system. One efficient method is to add all three equations together:

\$(x + y) + (y + z) + (x + z) = 4 + 8 + 6\$

Combine like terms:

\$2x + 2y + 2z = 18\$

Factor out 2 from the left side:

\$2(x + y + z) = 18\$

Divide both sides by 2 to find the sum of all three ages:

\$x + y + z = \frac{18}{2}\$

\$x + y + z = 9 \quad \text{(Equation 4)}\$

Now we can use Equation 4 along with the original equations to find the individual ages:

• Substitute Equation 1 (\$x + y = 4\$) into Equation 4:

  \$(x + y) + z = 9\$

  \$4 + z = 9\$

  Subtract 4 from both sides:

  \$z = 9 - 4\$

  \$z = 5\$

  The eldest brother's age is 5 years.

• Substitute Equation 2 (\$y + z = 8\$) into Equation 4:

  \$x + (y + z) = 9\$

  \$x + 8 = 9\$

  Subtract 8 from both sides:

  \$x = 9 - 8\$

  \$x = 1\$

  The youngest brother's age is 1 year.

• Now that we have \$x\$ and \$z\$, we can use Equation 3 (\$x + z = 6\$) as a check, or use either Equation 1 or 2 to find \$y\$. Let's use Equation 1 (\$x + y = 4\$):

  \$1 + y = 4\$

  Subtract 1 from both sides:

  \$y = 4 - 1\$

  \$y = 3\$

  The middle brother's age is 3 years.

The ages of the three brothers are 1 year, 3 years, and 5 years. This satisfies the initial assumption \$x < y < z\$ (1 < 3 < 5).

Calculating the Age Difference

The question asks for the age difference between the eldest and the youngest brother.

• Eldest brother's age = \$z = 5\$ years
• Youngest brother's age = \$x = 1\$ year

Age difference = Eldest age - Youngest age

Age difference = \$5 - 1 = 4\$ years

The age difference between the eldest and the youngest brother is 4 years.

Summary of Ages and Differences

These sums match the information given in the problem.

Revision Table: Key Concepts for Age Problems

Additional Information: Alternative Solving Method

Another way to solve the system \$x + y = 4\$, \$y + z = 8\$, \$x + z = 6\$ is using substitution or elimination on pairs of equations:

1. From Equation 1, \$y = 4 - x\$.
2. From Equation 3, \$z = 6 - x\$.
3. Substitute these expressions for \$y\$ and \$z\$ into Equation 2 (\$y + z = 8\$):

   \$(4 - x) + (6 - x) = 8\$

   \$10 - 2x = 8\$

4. Subtract 10 from both sides:

   \$-2x = 8 - 10\$

   \$-2x = -2\$

5. Divide by -2:

   \$x = \frac{-2}{-2}\$

   \$x = 1\$

6. Now substitute \$x = 1\$ back into the expressions for \$y\$ and \$z\$:

   \$y = 4 - x = 4 - 1 = 3\$

   \$z = 6 - x = 6 - 1 = 5\$

This method also gives the ages as 1, 3, and 5 years, leading to the same age difference of 4 years between the eldest and youngest.