Question

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

Explanation:

Step1: Graph \( y = \frac{1}{3}x + 2 \)

• Find two points:
• When \( x = 0 \), \( y = \frac{1}{3}(0) + 2 = 2 \), so point \((0, 2)\).
• When \( x = 3 \), \( y = \frac{1}{3}(3) + 2 = 1 + 2 = 3 \), so point \((3, 3)\).
• Plot these points and draw the line.

Step2: Graph \( y = 2x - 3 \)

• Find two points:
• When \( x = 0 \), \( y = 2(0) - 3 = -3 \), so point \((0, -3)\).
• When \( x = 2 \), \( y = 2(2) - 3 = 4 - 3 = 1 \), so point \((2, 1)\).
• Plot these points and draw the line.

Step3: Find Intersection

• The two lines intersect at a point. To find the exact coordinates, we can also solve the system algebraically (as a check):
• Set \( \frac{1}{3}x + 2 = 2x - 3 \)
• Subtract \( \frac{1}{3}x \) from both sides: \( 2 = \frac{5}{3}x - 3 \)
• Add 3 to both sides: \( 5 = \frac{5}{3}x \)
• Multiply both sides by \( \frac{3}{5} \): \( x = 3 \)
• Substitute \( x = 3 \) into \( y = 2x - 3 \): \( y = 2(3) - 3 = 3 \)
• So the intersection point is \((3, 3)\).

Answer:

The solution to the system is \( x = 3 \), \( y = 3 \) (or the point \((3, 3)\)).