Five times Early’s age now exceeds 4 times Lucy’s age now by 38. In 5 years, twice Early’s age will exceed Lucy’s age by 8. How old are Early and Lucy now?

Let Early's current age be \( E \) and Lucy's current age be \( L \).

We are given two pieces of information:

1. Five times Early's current age exceeds four times Lucy's current age by 38:
   \[
   5E = 4L + 38
   \]
   
2. In 5 years, twice Early's age will exceed Lucy's age by 8:
   \[
   2(E + 5) = (L + 5) + 8
   \]

### Step 1: Simplify the second equation
\[
2(E + 5) = L + 5 + 8
\]
\[
2E + 10 = L + 13
\]
\[
2E - L = 3 \quad \text{(Equation 2)}
\]

### Step 2: Solve the system of equations

We now have the system:
\[
5E = 4L + 38 \quad \text{(Equation 1)}
\]
\[
2E - L = 3 \quad \text{(Equation 2)}
\]

#### Solve Equation 2 for \( L \):
\[
L = 2E - 3
\]

#### Substitute into Equation 1:
\[
5E = 4(2E - 3) + 38
\]
\[
5E = 8E - 12 + 38
\]
\[
5E = 8E + 26
\]
\[
-3E = 26
\]
\[
E = -\frac{26}{3}
\]

### Solution: \( E = 

Discover how to solve a system of linear equations to find Early and Lucy’s ages. The problem uses algebraic concepts such as linear equations and systems of equations. Suitable for students in Grades 8-10.

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