Question

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Explanation:

1. Intersection of Equation 1 (\(y = x + 2\)) and Equation 2 (\(y = -2x - 4\))

Step1: Set equations equal

Since \(y\) is expressed in both, set \(x + 2=-2x - 4\).

Step2: Solve for \(x\)

Add \(2x\) to both sides: \(3x + 2=-4\). Subtract 2: \(3x=-6\), so \(x = -2\).

Step3: Find \(y\)

Substitute \(x=-2\) into \(y = x + 2\): \(y=-2 + 2 = 0\).

Step1: Set equations equal

\(-\frac{1}{4}x + 3=-2x - 4\).

Step2: Solve for \(x\)

Multiply all terms by 4: \(-x + 12=-8x - 16\). Add \(8x\): \(7x + 12=-16\). Subtract 12: \(7x=-28\), so \(x=-4\).

Step3: Find \(y\)

Substitute \(x = -4\) into \(y=-2x - 4\): \(y=-2(-4)-4 = 8 - 4 = 4\).

Step1: Set equations equal

\(\frac{2}{3}x=x + 2\).

Step2: Solve for \(x\)

Subtract \(\frac{2}{3}x\): \(0=\frac{1}{3}x + 2\). Subtract 2: \(\frac{1}{3}x=-2\), so \(x=-6\).

Step3: Find \(y\)

Substitute \(x=-6\) into \(y=\frac{2}{3}x\): \(y=\frac{2}{3}(-6)=-4\).

Answer:

\((-2, 0)\)

2. Intersection of Equation 4 (\(y = -\frac{1}{4}x + 3\)) and Equation 2 (\(y = -2x - 4\))