Question

solve the following system of equations graphically on the set of axes below.
$y = -x + 1$
$x - 2y = 4$
plot two lines by clicking the graph.
click a line to delete it.

Explanation:

Step1: Analyze the first equation \( y = -x + 1 \)

This is in slope - intercept form \( y=mx + b \), where the slope \( m=- 1 \) and the y - intercept \( b = 1 \). To graph this line, we can find two points. When \( x = 0 \), \( y=-0 + 1=1 \), so the point \( (0,1) \) is on the line. When \( y = 0 \), \( 0=-x + 1\), which gives \( x = 1 \), so the point \( (1,0) \) is on the line.

Step2: Analyze the second equation \( x-2y = 4 \)

We can rewrite it in slope - intercept form. Solve for \( y \):
\[

$$\begin{align*} x-2y&=4\\ -2y&=-x + 4\\ y&=\frac{1}{2}x-2 \end{align*}$$

\]
The slope \( m=\frac{1}{2} \) and the y - intercept \( b=-2 \). To find points on this line, when \( x = 0 \), \( y=\frac{1}{2}(0)-2=-2 \), so the point \( (0,-2) \) is on the line. When \( y = 0 \), \( 0=\frac{1}{2}x-2\), \( \frac{1}{2}x=2 \), \( x = 4 \), so the point \( (4,0) \) is on the line.

Step3: Find the intersection point

After graphing the two lines \( y=-x + 1 \) and \( y=\frac{1}{2}x-2 \), we look for the point where they intersect. We can also solve the system algebraically to find the intersection point (which should match the graphical solution). Set \( -x + 1=\frac{1}{2}x-2 \):
\[

$$\begin{align*} -x-\frac{1}{2}x&=-2 - 1\\ -\frac{3}{2}x&=-3\\ x&=(-3)\times(-\frac{2}{3})\\ x&=2 \end{align*}$$

\]
Substitute \( x = 2 \) into \( y=-x + 1 \), we get \( y=-2 + 1=-1 \). So the intersection point is \( (2,-1) \), which is the solution to the system of equations.

Answer:

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