Question

solve the following system of equations graphically on the set of axes below.
$y = x + 3$
$y = -dfrac{1}{2}x - 6$
plot two lines by clicking the graph.
click a line to delete it.

Explanation:

Step1: Analyze \( y = x + 3 \)

This is a linear equation in slope - intercept form \( y=mx + b \), where the slope \( m = 1 \) and the y - intercept \( b = 3 \). To graph this line, we start by plotting the y - intercept at the point \( (0,3) \). Then, using the slope (rise over run, since \( m = 1=\frac{1}{1} \)), we can find another point. From \( (0,3) \), we move 1 unit up (rise) and 1 unit to the right (run) to get the point \( (1,4) \), or 1 unit down and 1 unit to the left to get \( (- 1,2) \).

Step2: Analyze \( y=-\frac{1}{2}x - 6 \)

This is also in slope - intercept form with slope \( m=-\frac{1}{2} \) and y - intercept \( b=-6 \). We first plot the y - intercept at \( (0, - 6) \). Using the slope \( m =-\frac{1}{2}=\frac{-1}{2} \), from the point \( (0,-6) \), we move 1 unit down (rise) and 2 units to the right (run) to get the point \( (2,-7) \), or 1 unit up and 2 units to the left to get \( (-2,-5) \).

Step3: Find the intersection point

The solution to the system of linear equations is the point where the two lines intersect. To find this point algebraically (to confirm the graphical solution), we set the two equations equal to each other:
\[
x + 3=-\frac{1}{2}x-6
\]
Add \( \frac{1}{2}x \) to both sides:
\[
x+\frac{1}{2}x + 3=-\frac{1}{2}x+\frac{1}{2}x-6
\]
\[
\frac{2x + x}{2}+3=-6
\]
\[
\frac{3x}{2}+3=-6
\]
Subtract 3 from both sides:
\[
\frac{3x}{2}+3 - 3=-6 - 3
\]
\[
\frac{3x}{2}=-9
\]
Multiply both sides by \( \frac{2}{3} \):
\[
x=-9\times\frac{2}{3}=-6
\]
Now substitute \( x = - 6 \) into the first equation \( y=x + 3 \):
\[
y=-6 + 3=-3
\]

Answer:

The solution to the system of equations is the point of intersection of the two lines, which is \( (-6,-3) \)