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To solve the given system of equations, let's first rewrite the equations for clarity:

1. \(\frac{1}{6}x - \frac{1}{3}y = -6\)
2. \(\frac{1}{4}x + \frac{3}{5}y = 2\)

### Step 1: Eliminate Fractions by Multiplying Both Sides by a Common Denominator

For the first equation:

\[
\frac{1}{6}x - \frac{1}{3}y = -6
\]

The least common multiple (LCM) of 6 and 3 is 6. Multiply everything by 6:

\[
x - 2y = -36 \quad \text{(Equation 1)}
\]

For the second equation:

\[
\frac{1}{4}x + \frac{3}{5}y = 2
\]

The LCM of 4 and 5 is 20. Multiply everything by 20:

\[
5x + 12y = 40 \quad \text{(Equation 2)}
\]

### Step 2: Solve the System of Equations Using the Elimination Method

Now, we have the system:

1. \(x - 2y = -36\)
2. \(5x + 12y = 40\)

Let's eliminate \(x\) by making the coefficients of \(x\) the same in both equations. We can multiply the first equation by 5:

\[
5(x - 2y) = 5(-36)
\]

\[
5x - 10y = -180 \quad \text{(Equation 3)}
\]

Now, we have:

1. \(5x - 10y = -180\)
2. \(5x + 12y = 40\)

Subtract Equation 1 from Equation 2 to eliminate \(x\):

\[
(5x + 12y) - (5x - 10y) = 40 - (-180)
\]

\[
5x + 12y - 5x + 10y = 220
\]

\[
22y = 220
\]

\[
y = 10
\]

### Step 3: Substitute \(y\) Back to Find \(x\)

Substitute \(y = 10\) back into Equation 1:

\[
x - 2(10) = -36
\]

\[
x - 20 = -36
\]

\[
x = -16
\]

### Solution

The solution to the system of equations is:

\[
x = -16, \quad y = 10
\]

Would you like more details or have any questions?

### Related Questions

1. How do you solve a system of equations using the substitution method?
2. What are the common applications of solving systems of equations?
3. How can you determine if a system of equations has no solution, one solution, or infinitely many solutions?
4. What is the graphical interpretation of solving a system of equations?
5. How does the determinant of a coefficient matrix relate to solving systems of linear equations?

### Tip

When dealing with fractions in a system of equations, it's often helpful to clear the fractions by multiplying through by the least common denominator. This simplifies the equations and makes further algebraic manipulation easier.

Solve the following system of equations: 1/6x - 1/3y = -6 and 1/4x + 3/5y = 2

Learn how to solve the system of equations 1/6x - 1/3y = -6 and 1/4x + 3/5y = 2 using the elimination method. This problem covers key algebraic concepts and is suitable for students in Grades 9-11.

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