Question

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Explanation:

Problem 1: Solve the system \( y = -\frac{1}{2}x - 2 \) and \( y = -\frac{3}{2}x + 2 \)

Step 1: Set the two equations equal

Since both equal \( y \), set \( -\frac{1}{2}x - 2 = -\frac{3}{2}x + 2 \)

Step 2: Solve for \( x \)

Add \( \frac{3}{2}x \) to both sides: \( \frac{3}{2}x - \frac{1}{2}x - 2 = 2 \)
Simplify: \( x - 2 = 2 \)
Add 2 to both sides: \( x = 4 \)

Step 3: Solve for \( y \)

Substitute \( x = 4 \) into \( y = -\frac{1}{2}x - 2 \):
\( y = -\frac{1}{2}(4) - 2 = -2 - 2 = -4 \)

Step 4: Sum of \( x \) and \( y \)

\( x + y = 4 + (-4) = 0 \)

Step 1: Set the two equations equal

\( 3x + 18 = -\frac{1}{2}x - 17 \)

Step 2: Solve for \( x \)

Add \( \frac{1}{2}x \) to both sides: \( \frac{7}{2}x + 18 = -17 \)
Subtract 18: \( \frac{7}{2}x = -35 \)
Multiply by \( \frac{2}{7} \): \( x = -10 \)

Step 3: Solve for \( y \)

Substitute \( x = -10 \) into \( y = 3x + 18 \):
\( y = 3(-10) + 18 = -30 + 18 = -12 \)

Step 4: Sum of \( x \) and \( y \)

\( x + y = -10 + (-12) = -22 \)

Step 1: Solve \( 2x + y = 20 \) for \( y \)

\( y = 20 - 2x \)

Step 2: Substitute into \( 6x - 5y = 12 \)

\( 6x - 5(20 - 2x) = 12 \)
Simplify: \( 6x - 100 + 10x = 12 \)
Combine like terms: \( 16x - 100 = 12 \)
Add 100: \( 16x = 112 \)
Divide by 16: \( x = 7 \)

Step 3: Solve for \( y \)

Substitute \( x = 7 \) into \( y = 20 - 2x \):
\( y = 20 - 14 = 6 \)

Step 4: Sum of \( x \) and \( y \)

\( x + y = 7 + 6 = 13 \)

Answer:

\( (4, -4) \), Sum of \( x \& y \): \( 0 \)

Problem 2: Solve the system \( y = 3x + 18 \) and \( y = -\frac{1}{2}x - 17 \)