To solve the system of linear equations:

-8x + 2y = 8 \tag{1} 4x - 3y = -20 \tag{2}

We can use the substitution method or the elimination method. Let us use the elimination method for this example.

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Step 1: Multiply the equations to align the coefficients

To eliminate $y$, we make the coefficients of $y$ in both equations equal. Multiply Equation (1) by $3$ and Equation (2) by $2$:

3(-8x + 2y) = 3(8) \quad \implies \quad -24x + 6y = 24 \tag{3} 2(4x - 3y) = 2(-20) \quad \implies \quad 8x - 6y = -40 \tag{4}

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Step 2: Add the equations to eliminate $y$

Adding Equation (3) and Equation (4):

$(-24x + 6y) + (8x - 6y) = 24 + (-40)$ $-16x = -16$ $x = 1$

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Step 3: Substitute $x = 1$ into one of the original equations

Substitute $x = 1$ into Equation (1):

$-8(1) + 2y = 8$ $-8 + 2y = 8$ $2y = 16$ $y = 8$

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Final Solution:

The solution to the system of equations is:

$x = 1, \, y = 8$

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Would you like a step-by-step explanation of the method or any additional details? Here are some related questions for practice:

1. Solve $3x + 2y = 7$ and $4x - y = 5$.
2. Use substitution to solve $5x + y = 12$ and $2x - 3y = -6$.
3. Solve $7x + 6y = 0$ and $x - 2y = 3$ using elimination.
4. Verify the solution to $2x + 5y = 1$ and $3x - y = 7$ graphically.
5. Solve the system $-x + 4y = 3$ and $x + 3y = -7$ with fractions involved.

Tip: When using the elimination method, always check if you can simplify coefficients to minimize calculations.