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The system of equations provided is:

\[
\begin{aligned}
\frac{3}{4}x + \frac{1}{2}y &= -2 \quad \text{(1)}\\
\frac{1}{5}x - \frac{1}{4}y &= -6 \quad \text{(2)}
\end{aligned}
\]

### To solve this system of equations:

#### Step 1: Clear the fractions.
Multiply each equation by the least common denominator (LCD) of the fractions in that equation to eliminate the fractions.

- For Equation (1): Multiply by 4 (LCD of 4 and 2)
- For Equation (2): Multiply by 20 (LCD of 5 and 4)

**Equation (1):**
\[
4 \cdot \left(\frac{3}{4}x + \frac{1}{2}y\right) = 4 \cdot (-2)
\]
\[
3x + 2y = -8 \quad \text{(3)}
\]

**Equation (2):**
\[
20 \cdot \left(\frac{1}{5}x - \frac{1}{4}y\right) = 20 \cdot (-6)
\]
\[
4x - 5y = -120 \quad \text{(4)}
\]

#### Step 2: Solve the system using the method of substitution or elimination.

I'll use the elimination method.

**Multiply Equation (3) by 5 to match the coefficient of \(y\) in Equation (4):**
\[
5(3x + 2y) = 5(-8)
\]
\[
15x + 10y = -40 \quad \text{(5)}
\]

**Now, multiply Equation (4) by 2 to match the coefficient of \(y\) in Equation (5):**
\[
2(4x - 5y) = 2(-120)
\]
\[
8x - 10y = -240 \quad \text{(6)}
\]

**Add Equation (5) and Equation (6) to eliminate \(y\):**
\[
(15x + 10y) + (8x - 10y) = -40 + (-240)
\]
\[
23x = -280
\]
\[
x = \frac{-280}{23} \approx -12.17
\]

#### Step 3: Substitute \(x\) back into one of the original equations to solve for \(y\).

Substituting \(x = \frac{-280}{23}\) into Equation (3):
\[
3\left(\frac{-280}{23}\right) + 2y = -8
\]
\[
\frac{-840}{23} + 2y = -8
\]
\[
2y = -8 + \frac{840}{23}
\]
\[
2y = \frac{-184 + 840}{23} = \frac{656}{23}
\]
\[
y = \frac{656}{46} \approx 14.26
\]

### Final Solution:
The solution to the system is approximately:
\[
x \approx -12.17, \quad y \approx 14.26
\]

Would you like more details or have any questions?

### Follow-up Questions:

1. How do you solve a system of equations using substitution?
2. What are the advantages of using the elimination method over substitution?
3. Can you explain how to find the least common denominator of multiple fractions?
4. How can systems of equations be applied in real-world problems?
5. What are the conditions for a system of linear equations to have no solution or infinitely many solutions?

### Tip:
When solving systems of equations, always check your solution by substituting the values back into the original equations to ensure they satisfy both equations.

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

Learn how to solve the system of linear equations (3/4)x + (1/2)y = -2 and (1/5)x - (1/4)y = -6. This step-by-step solution involves clearing fractions and applying the elimination method. Suitable for students in Grades 8-10.

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