To solve the system of linear equations:

1. $8x + 14y = 4$
2. $-6x - 7y = -10$

We can use either substitution or elimination. Here, elimination seems convenient because we can eliminate $y$ by multiplying the second equation by 2. Let's proceed:

Step 1: Multiply the second equation by 2 to match the coefficient of $y$ in both equations.

$2(-6x - 7y) = 2(-10)$ This gives: $-12x - 14y = -20$

Step 2: Add the two equations to eliminate $y$:

$(8x + 14y) + (-12x - 14y) = 4 + (-20)$ $(8x - 12x) + (14y - 14y) = -16$ $-4x = -16$

Step 3: Solve for $x$:

$x = \frac{-16}{-4} = 4$

Step 4: Substitute $x = 4$ into one of the original equations (let's use the first equation):

$8(4) + 14y = 4$ $32 + 14y = 4$

Step 5: Solve for $y$:

$14y = 4 - 32$ $14y = -28$ $y = \frac{-28}{14} = -2$

Solution:

$x = 4, \quad y = -2$

Would you like further details or have any questions?

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Here are 5 related questions to explore further:

1. What is the graphical representation of the two equations?
2. How do you solve a system of equations using substitution?
3. Can this system be solved using matrices or determinants?
4. What are some real-life applications of solving systems of equations?
5. What happens if both equations represent the same line?

Tip: Always check your solution by substituting the values back into both original equations to verify accuracy.