Question

solve the following system of equations graphically on the set of axes below.

$y = -x - 6$

$x - 2y = -6$

plot two lines by clicking the graph.
click a line to delete it.

Explanation:

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

For the line \( y = -x - 6 \), it's in slope - intercept form (\( y=mx + b \)) where the slope \( m=-1 \) and the y - intercept \( b = - 6 \). To plot this line, we can find two points. When \( x = 0 \), \( y=-0 - 6=-6 \), so the point is \( (0,-6) \). When \( y = 0 \), \( 0=-x - 6\), which gives \( x=-6 \), so the point is \( (-6,0) \).

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

First, we rewrite the equation \( x - 2y=-6 \) in slope - intercept form. Solve for \( y \):
\[ \begin{align*} x-2y&=-6\\ -2y&=-x - 6\\ y&=\frac{1}{2}x + 3 \end{align*} \]
The slope \( m = \frac{1}{2} \) and the y - intercept \( b = 3 \). To plot this line, when \( x = 0 \), \( y = 3 \), so the point is \( (0,3) \). When \( y = 0 \), \( 0=\frac{1}{2}x+3\), then \( \frac{1}{2}x=-3 \), and \( x=-6 \), so the point is \( (-6,0) \).

Step3: Find the intersection

The two lines intersect at the point where their \( x \) and \( y \) values are equal. From the points we found, both lines pass through \( (-6,0) \)? Wait, no, let's check again. Wait, for \( y=-x - 6 \), when \( x=-6 \), \( y = 0 \). For \( y=\frac{1}{2}x + 3 \), when \( x=-6 \), \( y=\frac{1}{2}\times(-6)+3=-3 + 3=0 \). Wait, but also, let's find another point for \( y=-x - 6 \), say \( x = - 5 \), then \( y=-(-5)-6=5 - 6=-1 \). For \( y=\frac{1}{2}x + 3 \), when \( x=-4 \), \( y=\frac{1}{2}\times(-4)+3=-2 + 3 = 1 \). Wait, actually, let's solve the system algebraically to check the intersection.
We have \( y=-x - 6 \) and \( x-2y=-6 \). Substitute \( y=-x - 6 \) into \( x - 2y=-6 \):
\[ \begin{align*} x-2(-x - 6)&=-6\\ x + 2x+12&=-6\\ 3x&=-18\\ x&=-6 \end{align*} \]
Then substitute \( x = - 6 \) into \( y=-x - 6 \), we get \( y=-(-6)-6=6 - 6 = 0 \)? Wait, no, wait \( y=-x - 6 \), if \( x=-6 \), \( y=-(-6)-6=6 - 6=0 \). But when we plugged \( x = - 6 \) into \( y=\frac{1}{2}x + 3 \), we got \( y = 0 \). Wait, but let's check with \( x = - 2 \) in \( y=-x - 6 \), \( y=-(-2)-6=2 - 6=-4 \). In \( y=\frac{1}{2}x + 3 \), \( y=\frac{1}{2}\times(-2)+3=-1 + 3 = 2 \). Wait, I made a mistake earlier. Let's solve the system correctly.
Substitute \( y=-x - 6 \) into \( x-2y=-6 \):
\[ \begin{align*} x-2(-x - 6)&=-6\\ x + 2x+12&=-6\\ 3x&=-18\\ x&=-6 \end{align*} \]
Then \( y=-(-6)-6=6 - 6 = 0 \)? Wait, but when we rewrite \( x - 2y=-6 \) as \( y=\frac{1}{2}x + 3 \), if \( x=-6 \), \( y=\frac{1}{2}\times(-6)+3=-3 + 3 = 0 \). But let's check with \( x = - 4 \) in \( y=-x - 6 \), \( y = -(-4)-6=4 - 6=-2 \). In \( y=\frac{1}{2}x + 3 \), \( y=\frac{1}{2}\times(-4)+3=-2 + 3 = 1 \). Wait, no, the two lines intersect at \( (-6,0) \)? Wait, no, let's plot the lines properly.
For \( y=-x - 6 \): when \( x = 0 \), \( y=-6 \); when \( y = 0 \), \( x=-6 \).
For \( y=\frac{1}{2}x + 3 \): when \( x = 0 \), \( y = 3 \); when \( y = 0 \), \( x=-6 \). Wait, so they intersect at \( (-6,0) \)? Wait, but let's check with \( x=-2 \) in \( y=-x - 6 \), \( y=-(-2)-6=-4 \); in \( y=\frac{1}{2}x + 3 \), \( y=\frac{1}{2}\times(-2)+3 = 2 \). No, that's not the same. Wait, I think I made a mistake in solving \( x - 2y=-6 \) for \( y \). Let's do it again:
\( x-2y=-6 \)
Subtract \( x \) from both sides: \( -2y=-x - 6 \)
Divide both sides by - 2: \( y=\frac{1}{2}x + 3 \). That's correct.
Now, let's find a common point by plugging in \( x=-2 \) into \( y=-x - 6 \), we get \( y=-(-2)-6=-4 \). Plug \( x=-2 \) into \( y=\frac{1}{2}x + 3 \), we get \( y=\frac{1}{2}\times(-2)+3=-1 + 3 = 2 \). Not the same. Plug \( x=-4 \) into \( y=-x - 6 \), \( y=-(-4)-6=-2 \). Plug \( x=-4 \) into \( y=\frac{1}{2}x + 3 \), \( y=\frac{1}{2}\times(-4)+3=1 \). Not the same. Plug \( x = - 6 \) into \( y=-x - 6 \), \( y = 0 \). Plug \( x=-6 \) into \( y=\frac{1}{2}x + 3 \), \( y = 0 \). Ah, so they intersect at \( (-6,0) \)? Wait, but when \( x = 0 \), \( y=-6 \) for the first line and \( y = 3 \) for the second line. So the two lines intersect at \( (-6,0) \)? Wait, no, let's use the two - point form to plot the lines.
Line 1 (\( y=-x - 6 \)): passes through \( (0,-6) \) and \( (-6,0) \).
Line 2 (\( y=\frac{1}{2}x + 3 \)): passes through \( (0,3) \) and \( (-6,0) \).
So the intersection point is \( (-6,0) \). Wait, but let's check with the system of equations.
Substitute \( x=-6 \) and \( y = 0 \) into \( y=-x - 6 \): \( 0=-(-6)-6=6 - 6=0 \), which is true.
Substitute \( x=-6 \) and \( y = 0 \) into \( x - 2y=-6 \): \( -6-2\times0=-6 \), which is true. So the solution is \( x=-6 \), \( y = 0 \).

Answer:

The solution to the system of equations is \( x=-6 \), \( y = 0 \) (the intersection point of the two lines is \( (-6,0) \)).