We can solve the given system of equations using the elimination method. The equations are:

1. $x + y = 27$
2. $\frac{x}{30} + \frac{y}{24} = 1$

Step 1: Simplify the second equation

Multiply through by the least common multiple (LCM) of 30 and 24, which is 120, to eliminate the fractions:

$120 \cdot \frac{x}{30} + 120 \cdot \frac{y}{24} = 120 \cdot 1$

This simplifies to:

4x + 5y = 120 \tag{2'}

Now we have the system:

1. $x + y = 27$
2. $4x + 5y = 120$

Step 2: Eliminate one variable

From the first equation, solve for $y$ in terms of $x$:

$y = 27 - x$

Substitute $y = 27 - x$ into the second equation $4x + 5y = 120$:

$4x + 5(27 - x) = 120$

Step 3: Simplify and solve for $x$

Expand and simplify:

$4x + 135 - 5x = 120$

$-x + 135 = 120$

$x = 15$

Step 4: Solve for $y$

Substitute $x = 15$ into $y = 27 - x$:

$y = 27 - 15 = 12$

Final Answer:

$x = 15, \; y = 12$

Verification:

• From the first equation: $15 + 12 = 27$ ✅
• From the second equation: $\frac{15}{30} + \frac{12}{24} = 0.5 + 0.5 = 1$ ✅

The solution is correct.

---

Would you like additional explanations or related examples?
Here are 5 related questions for practice:

1. Solve $x + y = 20$ and $\frac{x}{5} + \frac{y}{10} = 3$ using elimination.
2. Determine the solution to $x - y = 10$ and $2x + y = 25$.
3. Solve $3x + 4y = 40$ and $2x - y = 10$ using elimination.
4. Use substitution instead of elimination to solve $x + y = 18$ and $\frac{x}{6} + \frac{y}{9} = 2$.
5. Explain how to identify when the elimination method is more efficient than substitution.

Tip: Elimination is particularly useful when one variable can be easily canceled by scaling the equations.