We are given the system of two linear equations:

$1) \ 2x - 7y = 20$ $2) \ -3x - 7y = 5$

To solve for $x$ and $y$, we can use the elimination method. The goal is to eliminate one of the variables by adding or subtracting the equations.

Step 1: Eliminate $y$

Notice that both equations have the term $-7y$ in them. If we add the two equations together, the $y$-terms will cancel out.

Add equation (1) and equation (2): $(2x - 7y) + (-3x - 7y) = 20 + 5$ This simplifies to: $2x - 3x - 7y - 7y = 25$ $-x - 14y = 25$ $-x = 25 + 14y$ Now we have an equation with just $x$, so we can solve for $x$.

Step 2: Solve for $x$

To isolate $x$, we multiply both sides by -1: $x = -25 - 14y$

Step 3: Substitute into one of the original equations

Substitute $x = -25 - 14y$ into equation (1): $2(-25 - 14y) - 7y = 20$ Simplifying: $-50 - 28y - 7y = 20$ $-50 - 35y = 20$ Add 50 to both sides: $-35y = 70$ Now, solve for $y$: $y = \frac{70}{-35} = -2$

Step 4: Solve for $x$

Now that we have $y = -2$, substitute this value into the equation $x = -25 - 14y$: $x = -25 - 14(-2)$ $x = -25 + 28$ $x = 3$

Final Solution:

The solution to the system is: $x = 3 \quad \text{and} \quad y = -2$

Would you like further explanation or details on the steps? Here are some related questions for practice:

1. How can the substitution method be used to solve this system instead of elimination?
2. What are the different methods for solving a system of linear equations?
3. How do you check if the solution is correct after solving a system?
4. What would happen if the system had no solution or infinitely many solutions?
5. How do you solve systems of equations that involve fractions or decimals?

Tip: When solving systems of equations, always check for opportunities to eliminate variables easily (like in this case with the matching $-7y$ terms) to simplify the process!