We are given the system of linear equations:

1. $3c - 8d = 7$
2. $c + 2d = -7$

To solve this system, we'll use the substitution or elimination method. Let's use the substitution method for this case.

Step 1: Solve for one variable from one of the equations

We can solve for $c$ from the second equation:

$c + 2d = -7$ $c = -7 - 2d$

Step 2: Substitute this expression for $c$ into the first equation

Now substitute $c = -7 - 2d$ into the first equation $3c - 8d = 7$:

$3(-7 - 2d) - 8d = 7$ $-21 - 6d - 8d = 7$ $-21 - 14d = 7$

Step 3: Solve for $d$

Now isolate $d$:

$-14d = 7 + 21$ $-14d = 28$ $d = \frac{28}{-14} = -2$

Step 4: Substitute $d = -2$ back into the expression for $c$

Substitute $d = -2$ into the equation $c = -7 - 2d$:

$c = -7 - 2(-2)$ $c = -7 + 4 = -3$

Final Solution:

Thus, the solution to the system of equations is:

$c = -3, \quad d = -2$

Let me know if you'd like more details or if you have any questions!

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Here are 5 related questions to explore further:

1. How would the solution change if the signs in the equations were reversed?
2. What is the graphical interpretation of this system of equations?
3. Can this system be solved using the elimination method instead of substitution?
4. What would happen if the coefficients of $c$ and $d$ in both equations were equal?
5. How can you check the solution for $c$ and $d$ by substituting back into both equations?

Tip: You can always check your solution by substituting the values of $c$ and $d$ back into both original equations to confirm that both are satisfied!