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solve the following system of equations graphically on the set of axes …

Question

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

$y = -x - 4$

$x - 2y = -4$

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

Explanation:

Step1: Analyze the first equation \( y = -x - 4 \)

This is a linear equation in slope - intercept form (\( y=mx + b \)), where the slope \( m=- 1 \) and the y - intercept \( b=-4 \). To graph this line, we can start by plotting the y - intercept. The y - intercept is the point where \( x = 0 \), so when \( x = 0 \), \( y=-4 \). Then, using the slope (which is \( \frac{\text{rise}}{\text{run}}=-1=\frac{- 1}{1} \)), from the point \( (0,-4) \), we can move 1 unit to the right (run) and 1 unit down (rise) to get the next point \( (1,-5) \), or 1 unit to the left and 1 unit up to get \( (-1,-3) \).

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

We need to rewrite this equation in slope - intercept form (\( y=mx + b \)).
First, solve for \( y \):

$$ LATEXBLOCK0 $$

Here, the slope \( m = \frac{1}{2} \) and the y - intercept \( b = 2 \). To graph this line, we start by plotting the y - intercept. When \( x = 0 \), \( y = 2 \). Then, using the slope \( \frac{1}{2}=\frac{\text{rise}}{\text{run}} \), from the point \( (0,2) \), we can move 2 units to the right (run) and 1 unit up (rise) to get the point \( (2,3) \), or 2 units to the left and 1 unit down to get \( (-2,1) \).

Step3: Find the intersection point

When we graph both lines, the point where they intersect is the solution to the system of equations. Let's also solve the system algebraically to confirm.
Substitute \( y=-x - 4 \) into \( x - 2y=-4 \):

$$ LATEXBLOCK1 $$

Now substitute \( x = - 4 \) into \( y=-x - 4 \):
\( y=-(-4)-4=4 - 4 = 0 \)

So the two lines intersect at the point \( (-4,0) \). When we plot the two lines on the given coordinate plane, the first line \( y=-x - 4 \) passes through \( (0,-4) \) and \( (-4,0) \) (since when \( y = 0 \), \( 0=-x - 4\Rightarrow x=-4 \)), and the second line \( y=\frac{1}{2}x + 2 \) passes through \( (0,2) \) and \( (-4,0) \) (since when \( x=-4 \), \( y=\frac{1}{2}(-4)+2=-2 + 2=0 \)).

Answer:

The solution to the system of equations is \( x=-4,y = 0 \) or the point \( (-4,0) \)