Question

solve the following system of equations graphically on the set of axes below.
$y = -dfrac{1}{3}x + 1$
$2x - y = 6$
plot two lines by clicking the graph.
click a line to delete it.

Explanation:

Step1: Analyze the first equation \( y = -\frac{1}{3}x + 1 \)

This is in slope - intercept form \( y=mx + b \), where the slope \( m=-\frac{1}{3} \) and the y - intercept \( b = 1 \). To find two points on this line:

• When \( x = 0 \), \( y=-\frac{1}{3}(0)+1=1 \). So one point is \( (0,1) \).
• When \( x = 3 \), \( y=-\frac{1}{3}(3)+1=- 1 + 1=0 \). So another point is \( (3,0) \).

Step2: Analyze the second equation \( 2x - y=6 \)

Rewrite it in slope - intercept form (\( y=mx + b \)):
\( y = 2x-6 \)
Here, the slope \( m = 2 \) and the y - intercept \( b=-6 \). To find two points on this line:

• When \( x = 0 \), \( y=2(0)-6=-6 \). So one point is \( (0,-6) \).
• When \( x = 3 \), \( y=2(3)-6=6 - 6 = 0 \). So another point is \( (3,0) \).

Step3: Find the intersection point

The two lines intersect where they have the same \( x \) and \( y \) values. From the points we found, both lines pass through \( (3,0) \). We can also verify by substituting \( x = 3 \) and \( y = 0 \) into both equations:

• For \( y=-\frac{1}{3}x + 1 \): \( 0=-\frac{1}{3}(3)+1=-1 + 1 \), which is true.
• For \( 2x - y=6 \): \( 2(3)-0=6 \), which is true.

Answer:

The solution to the system of equations is \( x = 3 \), \( y = 0 \) (or the ordered pair \( (3,0) \))