We are given the following information:

1. The sum of two numbers is 132.
2. One-third of the smaller number exceeds one-sixth of the larger number by 8.

Let:

• $x$ be the smaller number.
• $y$ be the larger number.

From the given information, we can write two equations:

Equation 1:

$x + y = 132$

Equation 2:

$\frac{x}{3} = \frac{y}{6} + 8$

We can solve this system of equations step-by-step.

Step 1: Solve Equation 1 for $y$

$y = 132 - x$

Step 2: Substitute $y = 132 - x$ into Equation 2

$\frac{x}{3} = \frac{132 - x}{6} + 8$

Step 3: Multiply through by 6 to eliminate the denominators

$2x = 132 - x + 48$

Step 4: Simplify and solve for $x$

$2x + x = 132 + 48$ $3x = 180$ $x = \frac{180}{3} = 60$

Step 5: Substitute $x = 60$ back into Equation 1 to find $y$

$60 + y = 132$ $y = 132 - 60 = 72$

Thus, the two numbers are:

• The smaller number $x = 60$
• The larger number $y = 72$

Would you like more details, or do you have any questions?

Related Questions:

1. Can you derive other possible systems of equations from a similar problem?
2. How would the solution change if one-third of the smaller number exceeded one-sixth of the larger by 10 instead of 8?
3. What if the sum of the two numbers was a different value, say 150? How would you approach the problem?
4. Can you verify the solution by substituting the values back into both equations?
5. How does solving a system of equations graphically differ from solving it algebraically?

Tip:

When solving systems of equations, always check your solutions by substituting the values back into the original equations to verify accuracy.