To solve the given system of equations using the elimination method, follow these steps:

Step 1: Write down the system of equations.

1. $4x + 5y = 2$
2. $x + 2y = -2$

Step 2: Multiply the second equation by a constant to align the $x$-terms for elimination.

Since the coefficient of $x$ in the second equation is 1, and in the first equation it's 4, multiply the entire second equation by 4:

$4(x + 2y) = 4(-2)$ This gives: $4x + 8y = -8$

Now the system becomes:

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

Step 3: Subtract the first equation from the second to eliminate $x$.

$(4x + 8y) - (4x + 5y) = -8 - 2$ Simplifying: $3y = -10$

Step 4: Solve for $y$.

$y = \frac{-10}{3}$ Thus, $y = -\frac{10}{3}$.

Step 5: Substitute the value of $y$ back into one of the original equations to solve for $x$.

Substitute $y = -\frac{10}{3}$ into the second equation $x + 2y = -2$:

$x + 2\left(-\frac{10}{3}\right) = -2$

This simplifies to: $x - \frac{20}{3} = -2$

Multiply both sides by 3 to clear the fraction: $3x - 20 = -6$

Add 20 to both sides: $3x = 14$

Now, divide by 3: $x = \frac{14}{3}$

Final Solution: The solution to the system is: $x = \frac{14}{3}, \quad y = -\frac{10}{3}$

Would you like further details or have any questions?

Related Questions:

1. How would the elimination method change if both equations had the same coefficients for $x$?
2. Can you solve the same system using the substitution method?
3. What is the graphical interpretation of the solution to this system?
4. How does the elimination method compare to the matrix method for solving systems of equations?
5. What happens if a system of equations has no solution or infinite solutions?

Tip: Always check your solution by substituting the values of $x$ and $y$ back into the original equations.