A boy is 10 years older than his brother. In 4 years he will be twice 
as old as his brother. Find the present age of each.

Let the present age of the boy be \( x \) and the present age of his brother be \( y \).

### Step 1: Translate the problem into equations
We know:
1. The boy is 10 years older than his brother: 
   \[
   x = y + 10 \tag{1}
   \]
2. In 4 years, the boy will be twice as old as his brother: 
   \[
   x + 4 = 2(y + 4) \tag{2}
   \]

### Step 2: Solve the system of equations
Substitute equation (1) into equation (2):
\[
(y + 10) + 4 = 2(y + 4)
\]
Simplify:
\[
y + 14 = 2y + 8
\]
Solve for \( y \):
\[
14 - 8 = 2y - y
\]
\[
y = 6
\]

### Step 3: Find \( x \)
Now, substitute \( y = 6 \) into equation (1):
\[
x = 6 + 10 = 16
\]

### Conclusion:
- The boy is 16 years old.
- The brother is 6 years old.

Would you like more details on the solving process, or have any questions?

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Here are 5 related questions for practice:
1. A man is 5 years older than his wife. In 7 years, he will be 1.5 times as old as her. What are their present ages?
2. A father is 3 times as old as his son. In 10 years, the father will be twice as old as his son. What are their present ages?
3. A girl is 15 years older than her sister. In 5 years, she will be twice her sister's age. How old are they now?
4. The sum of the ages of a mother and her son is 45. In 3 years, the mother will be 3 times as old as her son. What are their present ages?
5. A woman is 20 years older than her daughter. In 8 years, the mother will be 2.5 times as old as her daughter. How old are they now?

**Tip:** When solving age-related problems, always set up variables and translate the given conditions into equations step-by-step.

A boy is 10 years older than his brother. In 4 years he will be twice as old as his brother. Find the present age of each.

Solve the age problem where a boy is 10 years older than his brother and in 4 years will be twice as old. This solution involves setting up linear equations and solving step-by-step. Ideal for students in Grades 6-8.

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