8y=-x-20               2y=-5x+14                               

You have a system of two linear equations:

1. \( 8y = -x - 20 \)
2. \( 2y = -5x + 14 \)

We can solve this system using either substitution or elimination. Let's use substitution by solving one equation for one variable and then substituting it into the other equation.

### Step 1: Solve for \( y \) in the first equation
From the first equation:

\[
8y = -x - 20
\]
Divide both sides by 8:

\[
y = \frac{-x - 20}{8} \quad \text{or} \quad y = \frac{-x}{8} - \frac{20}{8}
\]
Simplifying:

\[
y = \frac{-x}{8} - 2.5
\]

### Step 2: Substitute \( y = \frac{-x}{8} - 2.5 \) into the second equation
Now substitute this expression for \( y \) into the second equation \( 2y = -5x + 14 \).

\[
2\left( \frac{-x}{8} - 2.5 \right) = -5x + 14
\]
Distribute the 2:

\[
\frac{-2x}{8} - 5 = -5x + 14
\]
Simplify the fraction:

\[
\frac{-x}{4} - 5 = -5x + 14
\]

### Step 3: Solve for \( x \)
Now, let's eliminate the fraction by multiplying everything by 4:

\[
-x - 20 = -20x + 56
\]
Add \( 20x \) to both sides:

\[
19x - 20 = 56
\]
Add 20 to both sides:

\[
19x = 76
\]
Finally, divide by 19:

\[
x = 4
\]

### Step 4: Solve for \( y \)
Now, substitute \( x = 4 \) back into the expression for \( y \):

\[
y = \frac{-x}{8} - 2.5 = \frac{-4}{8} - 2.5 = -0.5 - 2.5 = -3
\]

### Solution
The solution to the system of equations is:

\[
x = 4, \quad y = -3
\]

Would you like further details or have any questions?

Here are 5 related questions:
1. How do you check if the solution is correct for both equations?
2. What is the graphical interpretation of solving a system of linear equations?
3. Can this system of equations be solved using the elimination method? How?
4. How would you represent this system in matrix form?
5. How can you determine if a system of equations has no solution or infinitely many solutions?

**Tip:** Always verify your solution by substituting the values back into the original equations to ensure they satisfy both equations.

Learn how to solve a system of linear equations: 8y = -x - 20 and 2y = -5x + 14 using the substitution method. This problem involves algebraic manipulation and step-by-step solution techniques. Suitable for Grades 9-12.

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