Question

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

Explanation:

Step1: Analyze \( y = 2x - 6 \)

This is a linear equation in slope - intercept form (\( y=mx + b \), where \( m \) is the slope and \( b \) is the y - intercept). The slope \( m = 2 \) and the y - intercept \( b=-6 \). To graph this line, we can find two points. When \( x = 0 \), \( y=2(0)-6=-6 \), so one point is \( (0, - 6) \). When \( y = 0 \), \( 0=2x - 6\), solving for \( x \) gives \( 2x=6\), so \( x = 3 \). So another point is \( (3,0) \).

Step2: Analyze \( x + y=3 \)

We can rewrite this in slope - intercept form. Solve for \( y \): \( y=-x + 3 \). Here, the slope \( m=-1 \) and the y - intercept \( b = 3 \). To find points, when \( x = 0 \), \( y=3 \) (point \( (0,3) \)). When \( y = 0 \), \( x=3 \) (point \( (3,0) \)). Wait, no, when \( y = 0 \), \( x+0 = 3\), so \( x = 3 \), point \( (3,0) \); when \( x = 0 \), \( y = 3 \), point \( (0,3) \).

Step3: Find the intersection

The two lines \( y = 2x-6 \) and \( y=-x + 3 \) intersect where their \( x \) and \( y \) values are equal. We can also see from the points we found that for \( y = 2x-6 \) and \( y=-x + 3 \), set \( 2x-6=-x + 3 \). Add \( x \) to both sides: \( 3x-6 = 3 \). Add 6 to both sides: \( 3x=9 \), so \( x = 3 \). Substitute \( x = 3 \) into \( y=-x + 3 \), we get \( y=-3 + 3=0 \). So the intersection point is \( (3,0) \). But wait, let's check with the other equation: \( y=2(3)-6=6 - 6=0 \). So the solution to the system is the point where the two lines intersect, which is \( (3,0) \). But wait, maybe we made a mistake in the point for \( x + y=3 \). Wait, no, when we graph \( y = 2x-6 \), when \( x = 0 \), \( y=-6 \); when \( x = 3 \), \( y = 0 \). For \( y=-x + 3 \), when \( x = 0 \), \( y = 3 \); when \( x = 3 \), \( y = 0 \). So the two lines intersect at \( (3,0) \)? Wait, no, let's solve the system algebraically to confirm.
We have the system:
\(

$$\begin{cases}y = 2x-6\\x + y=3\end{cases}$$

\)
Substitute \( y = 2x-6 \) into \( x + y=3 \):
\( x+(2x - 6)=3 \)
\( 3x-6 = 3 \)
\( 3x=9 \)
\( x = 3 \)
Then \( y=2(3)-6=0 \)

Wait, but when we graph \( y=-x + 3 \), the y - intercept is \( (0,3) \) and slope - 1, so it goes down from left to right. The line \( y = 2x-6 \) has y - intercept \( (0,-6) \) and slope 2, goes up from left to right. They should intersect at \( (3,0) \)? Wait, no, let's re - check the equations. Wait, \( x + y=3 \), when \( x = 3 \), \( y = 0 \); \( y = 2x-6 \), when \( x = 3 \), \( y=0 \). So the solution is \( (3,0) \)? Wait, but let's take another approach. Let's find two points for each line correctly.

For \( y = 2x-6 \):

• \( x = 0 \), \( y=-6 \) (point \( (0,-6) \))
• \( x = 1 \), \( y=2(1)-6=-4 \) (point \( (1,-4) \))

For \( y=-x + 3 \):

• \( x = 0 \), \( y = 3 \) (point \( (0,3) \))
• \( x = 1 \), \( y=-1 + 3=2 \) (point \( (1,2) \))

Now, if we plot these lines, the line \( y = 2x-6 \) goes from \( (0,-6) \) through \( (1,-4) \) and \( (3,0) \). The line \( y=-x + 3 \) goes from \( (0,3) \) through \( (1,2) \) and \( (3,0) \). So they intersect at \( (3,0) \).

Answer:

The solution to the system of equations is \( x = 3 \), \( y = 0 \) (the point of intersection is \( (3,0) \)).