Let’s go step-by-step to solve this system of equations. Given the equations are:

1. $8x - 6y = -20$
2. $-16x + 7y = 30$
3. $16x - 12y = -40$
4. $-5y = -10$, which simplifies to $y = 2$.

Step 1: Find $y$

Since the fourth equation gives $y = 2$, we can substitute this value of $y$ into the other equations to find $x$.

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Step 2: Substitute $y = 2$ into the first equation

$8x - 6(2) = -20$ $8x - 12 = -20$ Add 12 to both sides: $8x = -8$ Divide both sides by 8: $x = -1$

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Step 3: Verify with other equations

Now, let’s verify that $x = -1$ and $y = 2$ satisfy the other equations.

• Second equation:
  $-16(-1) + 7(2) = 30$ $16 + 14 = 30$ This is true.

• Third equation:
  $16(-1) - 12(2) = -40$ $-16 - 24 = -40$ This is also true.

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Final Answer

The solution is: $x = -1, \quad y = 2$

Do you need further details or have any questions?

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Related Questions:

1. How do you solve a system of equations using substitution?
2. What is the elimination method for solving systems of equations?
3. How do you determine if a system of equations has no solution or infinite solutions?
4. Can matrices be used to solve systems of equations?
5. What is Cramer's rule, and how does it apply to 2-variable systems?

Tip: Always substitute your solution into all original equations to ensure correctness.