Question

directions - solve the system by graphing. \

$$\begin{cases} line 1: \\, y = \\frac{1}{3}x - 3 \\\\ line 2: \\, y = -x + 1 \\end{cases}$$

Explanation:

Step1: Graph Line 1 (\(y = \frac{1}{3}x - 3\))

• Y-intercept: When \(x = 0\), \(y = -3\). So one point is \((0, -3)\).
• Slope: The slope is \(\frac{1}{3}\), which means for every 3 units we move to the right (increase \(x\) by 3), we move up 1 unit (increase \(y\) by 1). From \((0, -3)\), moving right 3 and up 1 gives \((3, -2)\).

Step2: Graph Line 2 (\(y = -x + 1\))

• Y-intercept: When \(x = 0\), \(y = 1\). So one point is \((0, 1)\).
• Slope: The slope is \(-1\), which means for every 1 unit we move to the right (increase \(x\) by 1), we move down 1 unit (decrease \(y\) by 1). From \((0, 1)\), moving right 1 and down 1 gives \((1, 0)\).

Step3: Find the Intersection

The two lines intersect where their \(x\) and \(y\) values are equal. By looking at the graph (or solving the equations algebraically: \(\frac{1}{3}x - 3 = -x + 1\)), we find that \(x = 3\) and \(y = -2\) (wait, no, let's solve it properly:

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

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

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

Add 3 to both sides: \(\frac{4}{3}x = 4\)

Multiply both sides by \(\frac{3}{4}\): \(x = 3\)

Then \(y = -3 + 1 = -2\)? Wait, no, \(y = -x + 1\), so \(y = -3 + 1 = -2\)? Wait, but when \(x = 3\), Line 1: \(y = \frac{1}{3}(3) - 3 = 1 - 3 = -2\). Line 2: \(y = -3 + 1 = -2\). So the intersection is at \((3, -2)\). Wait, but let's check the graph again. Wait, maybe I made a mistake. Wait, Line 2: \(y = -x + 1\). When \(x = 3\), \(y = -2\). Line 1: \(y = \frac{1}{3}(3) - 3 = -2\). So yes, the intersection is at \((3, -2)\). Wait, but let's re-examine the graphing steps.

Wait, Line 1: \(y = \frac{1}{3}x - 3\). When \(x = 3\), \(y = -2\). When \(x = 6\), \(y = \frac{1}{3}(6) - 3 = 2 - 3 = -1\). When \(x = -3\), \(y = \frac{1}{3}(-3) - 3 = -1 - 3 = -4\).

Line 2: \(y = -x + 1\). When \(x = 3\), \(y = -2\). When \(x = 1\), \(y = 0\). When \(x = 2\), \(y = -1\). When \(x = -1\), \(y = 2\).

So the two lines intersect at \((3, -2)\).

Answer:

The solution to the system is \((3, -2)\) (the point where the two lines intersect when graphed).