Question

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

Explanation:

Step1: Analyze the first equation \( y = -\frac{1}{2}x + 2 \)

To plot this line, we can find two points. When \( x = 0 \), \( y = -\frac{1}{2}(0)+2 = 2 \). So one point is \( (0, 2) \). When \( x = 4 \), \( y = -\frac{1}{2}(4)+2 = -2 + 2 = 0 \). So another point is \( (4, 0) \).

Step2: Analyze the second equation \( x - y = 1 \)

Rewrite it in slope - intercept form (\( y=mx + b \)): \( y=x - 1 \). When \( x = 0 \), \( y=0 - 1=-1 \). So one point is \( (0,-1) \). When \( x = 1 \), \( y=1 - 1 = 0 \). So another point is \( (1, 0) \).

Step3: Find the intersection point

To find the solution of the system, we can also solve the two equations algebraically and then verify graphically.
From \( x - y=1 \), we have \( y=x - 1 \).
Substitute \( y=x - 1 \) into \( y = -\frac{1}{2}x+2 \):
\( x - 1=-\frac{1}{2}x + 2 \)
Add \( \frac{1}{2}x \) to both sides: \( x+\frac{1}{2}x-1=2 \)
\( \frac{2x + x}{2}-1 = 2 \)
\( \frac{3x}{2}=3 \)
Multiply both sides by \( \frac{2}{3} \): \( x = 2 \)
Substitute \( x = 2 \) into \( y=x - 1 \), we get \( y=2 - 1=1 \)

So the solution of the system of equations is the point of intersection of the two lines, which is \( (2,1) \). When we plot the two lines \( y = -\frac{1}{2}x + 2 \) (passing through \( (0,2) \) and \( (4,0) \)) and \( y=x - 1 \) (passing through \( (0, - 1) \) and \( (1,0) \)), they intersect at \( (2,1) \).

Answer:

The solution of the system of equations is \( x = 2,y = 1 \) (or the point \( (2,1) \)).