To solve the system of equations, we have:

1. $-5x + 8y = -21$
2. $3x - 4y = 15$

Step 1: Eliminate $y$ by finding a common multiple

To eliminate $y$, we can multiply the equations by values that make the coefficients of $y$ the same.

Multiply the first equation by $4$ and the second equation by $8$:

$-5x \cdot 4 + 8y \cdot 4 = -21 \cdot 4$ $3x \cdot 8 - 4y \cdot 8 = 15 \cdot 8$

This yields:

1. $-20x + 32y = -84$
2. $24x - 32y = 120$

Step 2: Add the equations

Now, add the two equations to eliminate $y$:

$(-20x + 32y) + (24x - 32y) = -84 + 120$

This simplifies to:

$4x = 36$

Step 3: Solve for $x$

Divide both sides by 4:

$x = 9$

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

Now that we have $x = 9$, substitute it into the second equation to solve for $y$:

$3(9) - 4y = 15$

$27 - 4y = 15$

Subtract 27 from both sides:

$-4y = -12$

Divide by $-4$:

$y = 3$

Solution

The solution to the system is:

$x = 9, \quad y = 3$

Would you like more details or have any questions?

Here are some related questions:

1. How do we solve a system of equations using substitution?
2. What is the elimination method in solving systems of equations?
3. How do we verify a solution to a system of equations?
4. Can a system of linear equations have no solution?
5. What is a consistent vs. inconsistent system of equations?

Tip: Always double-check your solution by plugging values back into the original equations to ensure accuracy.