Question

7 solve the system of linear equations by graphing. (3 pts)
$y = \frac{1}{3}x + 2$
$y = 2x - 3$

Explanation:

Step1: Find intercepts for \( y = \frac{1}{3}x + 2 \)

• When \( x = 0 \), \( y = 2 \), so point \( (0, 2) \).
• When \( y = 0 \), \( 0 = \frac{1}{3}x + 2 \Rightarrow x = -6 \), so point \( (-6, 0) \). Plot these, draw line.

Step2: Find intercepts for \( y = 2x - 3 \)

• When \( x = 0 \), \( y = -3 \), so point \( (0, -3) \).
• When \( y = 0 \), \( 0 = 2x - 3 \Rightarrow x = \frac{3}{2} = 1.5 \), so point \( (1.5, 0) \). Plot these, draw line.

Step3: Find intersection

The lines intersect where \( \frac{1}{3}x + 2 = 2x - 3 \). Solve:
\( 2 + 3 = 2x - \frac{1}{3}x \Rightarrow 5 = \frac{5}{3}x \Rightarrow x = 3 \).
Substitute \( x = 3 \) into \( y = 2x - 3 \): \( y = 2(3) - 3 = 3 \). So intersection is \( (3, 3) \).

Answer:

The solution to the system of linear equations is \( (3, 3) \).