You have a system of two linear equations:

3x + 5y = -4 \tag{1} x - 2y = -1 \tag{2}

Step 1: Solve Equation (2) for $x$

From equation (2):

$x = 2y - 1$

Step 2: Substitute $x$ into Equation (1)

Substitute $x = 2y - 1$ into equation (1):

$3(2y - 1) + 5y = -4$

Expand the equation:

$6y - 3 + 5y = -4$

Simplify:

$11y - 3 = -4$

Step 3: Solve for $y$

Add 3 to both sides:

$11y = -1$

Now, divide by 11:

$y = -\frac{1}{11}$

Step 4: Solve for $x$

Now, substitute $y = -\frac{1}{11}$ back into the expression for $x$:

$x = 2\left(-\frac{1}{11}\right) - 1$

Simplify:

$x = -\frac{2}{11} - 1 = -\frac{2}{11} - \frac{11}{11} = -\frac{13}{11}$

Final Answer:

The solution to the system is:

$x = -\frac{13}{11}, \quad y = -\frac{1}{11}$

Would you like more details on any of the steps, or do you have any other questions?

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Here are 5 related questions to consider:

1. How do you solve a system of equations using the substitution method?
2. Can you solve the same system using the elimination method?
3. What happens if both equations represent the same line?
4. How would the solution change if one equation had a different constant term?
5. Can you graph these equations to visualize the solution?

Tip: When solving systems of equations, always check your solution by substituting the values back into the original equations.