We are tasked with solving the system of equations using the elimination method:

1. $2x - 5y = -13$ $\text{(Equation 1)}$
2. $-3y + 4 = -5x$ $\text{(Equation 2)}$

Step 1: Rearrange the second equation

To align both equations better, we need to rearrange Equation 2 into a form similar to Equation 1:

$-3y + 4 = -5x$ Add $5x$ to both sides: $5x - 3y = 4 \quad \text{(Equation 3)}$

Now, we have the system:

$2x - 5y = -13 \quad \text{(Equation 1)}$ $5x - 3y = 4 \quad \text{(Equation 3)}$

Step 2: Multiply to eliminate one variable

We will eliminate $x$. To do that, let's multiply the first equation by 5 and the second equation by 2, so that the coefficients of $x$ will be the same:

Multiply Equation 1 by 5: $(5)(2x - 5y) = 5(-13) \quad \Rightarrow \quad 10x - 25y = -65 \quad \text{(Equation 4)}$

Multiply Equation 3 by 2: $(2)(5x - 3y) = (2)(4) \quad \Rightarrow \quad 10x - 6y = 8 \quad \text{(Equation 5)}$

Step 3: Subtract the equations

Now subtract Equation 5 from Equation 4 to eliminate $x$:

$(10x - 25y) - (10x - 6y) = (-65) - (8)$ This simplifies to: $10x - 25y - 10x + 6y = -65 - 8$ $-19y = -73$ Solving for $y$: $y = \frac{-73}{-19} = \frac{73}{19} = 3.84 \quad \text{(rounded to 2 decimal places)}.$

Step 4: Substitute $y$ back into one of the original equations

Let's substitute $y = 3.84$ into Equation 1 to solve for $x$:

$2x - 5(3.84) = -13$ $2x - 19.2 = -13$ Add 19.2 to both sides: $2x = 6.2$ Solving for $x$: $x = \frac{6.2}{2} = 3.1.$

Final Solution Set:

$x = 3.1, \quad y = 3.84.$

This is the solution set for the system of equations.

Would you like further details or clarifications on any steps?

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Related Questions:

1. How does the substitution method compare to elimination?
2. Can you solve the same system graphically?
3. What happens when a system of equations has no solution?
4. How do you solve systems of three variables using elimination?
5. What are other methods for solving systems of linear equations?

Tip: Always check your solution by substituting it back into the original equations.