Question

math 1
spiral review: february 2nd - 6th
solve each system by graphing.

1. $y = \frac{7}{3}x - 4$

$y = -\frac{1}{3}x + 4$

1. $y = \frac{1}{2}x + 4$

$y = -\frac{1}{4}x + 1$
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id: 1
solve each system by substitution.

Explanation:

Problem 1: Solve \(

$$\begin{cases} y = \frac{7}{3}x - 4 \\ y = -\frac{1}{3}x + 4 \end{cases}$$

\) by Graphing

Step 1: Analyze the first line \( y = \frac{7}{3}x - 4 \)

• Slope (\(m\)): \( \frac{7}{3} \) (positive, steep)
• Y-intercept (\(b\)): \( -4 \) (crosses y-axis at \( (0, -4) \))
• To graph, start at \( (0, -4) \), then rise 7, run 3 (or down 7, run -3) for another point.

Step 2: Analyze the second line \( y = -\frac{1}{3}x + 4 \)

• Slope (\(m\)): \( -\frac{1}{3} \) (negative, shallow)
• Y-intercept (\(b\)): \( 4 \) (crosses y-axis at \( (0, 4) \))
• To graph, start at \( (0, 4) \), then rise -1, run 3 (or up 1, run -3) for another point.

Step 3: Find the intersection

• The two lines intersect where \( \frac{7}{3}x - 4 = -\frac{1}{3}x + 4 \).
• Solve algebraically (to confirm):

\( \frac{7}{3}x + \frac{1}{3}x = 4 + 4 \)
\( \frac{8}{3}x = 8 \)
\( x = 8 \times \frac{3}{8} = 3 \)
Substitute \( x = 3 \) into \( y = -\frac{1}{3}x + 4 \):
\( y = -\frac{1}{3}(3) + 4 = -1 + 4 = 3 \)

• Thus, the intersection (solution) is \( (3, 3) \).

Problem 2: Solve \(

$$\begin{cases} y = \frac{1}{2}x + 4 \\ y = -\frac{1}{4}x + 1 \end{cases}$$

\) by Graphing

Step 1: Analyze the first line \( y = \frac{1}{2}x + 4 \)

• Slope (\(m\)): \( \frac{1}{2} \) (positive, moderate)
• Y-intercept (\(b\)): \( 4 \) (crosses y-axis at \( (0, 4) \))
• To graph, start at \( (0, 4) \), then rise 1, run 2 (or down 1, run -2) for another point.

Step 2: Analyze the second line \( y = -\frac{1}{4}x + 1 \)

• Slope (\(m\)): \( -\frac{1}{4} \) (negative, very shallow)
• Y-intercept (\(b\)): \( 1 \) (crosses y-axis at \( (0, 1) \))
• To graph, start at \( (0, 1) \), then rise -1, run 4 (or up 1, run -4) for another point.

Step 3: Find the intersection

• The two lines intersect where \( \frac{1}{2}x + 4 = -\frac{1}{4}x + 1 \).
• Solve algebraically (to confirm):

\( \frac{1}{2}x + \frac{1}{4}x = 1 - 4 \)
\( \frac{3}{4}x = -3 \)
\( x = -3 \times \frac{4}{3} = -4 \)
Substitute \( x = -4 \) into \( y = \frac{1}{2}x + 4 \):
\( y = \frac{1}{2}(-4) + 4 = -2 + 4 = 2 \)

• Thus, the intersection (solution) is \( (-4, 2) \).

Answer:

s:

1. The solution to \(

$$\begin{cases} y = \frac{7}{3}x - 4 \\ y = -\frac{1}{3}x + 4 \end{cases}$$

\) is \( \boldsymbol{(3, 3)} \).

1. The solution to \(

$$\begin{cases} y = \frac{1}{2}x + 4 \\ y = -\frac{1}{4}x + 1 \end{cases}$$

\) is \( \boldsymbol{(-4, 2)} \).