Setting Up Algebraic Equations

Let's represent the unknown numbers using variables. Let '$L$' be the larger number and '$S$' be the smaller number.

From Condition 1, we can write the first equation:

$L - S = 16$

From Condition 2, we can write the second equation. "One-third of the smaller number" is $\frac{1}{3} S$. "One-seventh of the larger number" is $\frac{1}{7} L$. The condition states that $\frac{1}{3} S$ is greater than $\frac{1}{7} L$ by 4.

$\frac{1}{3} S = \frac{1}{7} L + 4$

Solving the System of Equations

Now we have a system of two linear equations with two variables:

1. $L - S = 16$
2. $\frac{1}{3} S = \frac{1}{7} L + 4$

Let's use substitution to solve for '$L$'. First, rearrange Equation 1 to express '$S$' in terms of '$L$':

$S = L - 16$

Now, substitute this expression for '$S$' into Equation 2:

$\frac{1}{3} (L - 16) = \frac{1}{7} L + 4$

To eliminate the fractions, we can multiply the entire equation by the least common multiple (LCM) of 3 and 7, which is 21:

$21 \times \left( \frac{1}{3} (L - 16) \right) = 21 \times \left( \frac{1}{7} L + 4 \right)

Distribute the multiplication:

$7(L - 16) = 3L + 84

Expand the left side:

$7L - 112 = 3L + 84

Now, gather the '$L$' terms on one side and the constants on the other. Subtract '$3L$' from both sides:

$7L - 3L - 112 = 84

$4L - 112 = 84

Add 112 to both sides:

$4L = 84 + 112

$4L = 196

Finally, divide by 4 to find the value of '$L$':

$L = \frac{196}{4}

$L = 49

Verifying the Solution

We found the larger number '$L$' to be 49. Let's find the smaller number '$S$' using $S = L - 16$:

$S = 49 - 16 = 33

Now, let's check if these numbers satisfy the second condition:

• One-third of the smaller number: $\frac{1}{3} \times 33 = 11$
• One-seventh of the larger number: $\frac{1}{7} \times 49 = 7$

Is 11 greater than 7 by 4? Yes, $11 = 7 + 4$. Both conditions are satisfied.

Conclusion: The Larger Number

The calculations confirm that the larger number is 49.