Question

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Explanation:

Step1: Solve System a (Substitution)

We have the system:
\( y = 2x - 1 \)
\( y = -5x + 6 \)

Set the two expressions for \( y \) equal:
\( 2x - 1 = -5x + 6 \)

Add \( 5x \) to both sides:
\( 2x + 5x - 1 = 6 \)
\( 7x - 1 = 6 \)

Add 1 to both sides:
\( 7x = 6 + 1 \)
\( 7x = 7 \)

Wait, there was a mistake in the handwritten work. Correctly, \( 2x - 1 = -5x + 6 \) → \( 2x + 5x = 6 + 1 \) → \( 7x = 7 \) → \( x = 1 \).

Substitute \( x = 1 \) into \( y = 2x - 1 \):
\( y = 2(1) - 1 = 1 \).

Step2: Solve System b (Substitution)

We have the system:
\( y = x + 1 \)
\( y = -3x - 15 \)

Set the two expressions for \( y \) equal:
\( x + 1 = -3x - 15 \)

Add \( 3x \) to both sides:
\( x + 3x + 1 = -15 \)
\( 4x + 1 = -15 \)

Subtract 1 from both sides:
\( 4x = -15 - 1 \)
\( 4x = -16 \)

Divide by 4:
\( x = -4 \).

Substitute \( x = -4 \) into \( y = x + 1 \):
\( y = -4 + 1 = -3 \).

Answer:

• System a: Solution is \( (x, y) = (1, 1) \)
• System b: Solution is \( (x, y) = (-4, -3) \)