Question

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

Explanation:

Step1: Analyze the first equation \( y = \frac{1}{2}x + 1 \)

This is a linear equation in slope - intercept form \( y=mx + b \), where the slope \( m=\frac{1}{2} \) and the y - intercept \( b = 1 \). To graph this line, we can start by plotting the y - intercept. The y - intercept is the point where \( x = 0 \), so when \( x=0 \), \( y=1 \). So we have the point \( (0,1) \). Then, using the slope (rise over run), since the slope is \( \frac{1}{2} \), from the point \( (0,1) \), we can move up 1 unit and then 2 units to the right to get the next point. For example, when we move up 1 (rise) and right 2 (run) from \( (0,1) \), we get the point \( (2,1 + 1)=(2,2) \).

Step2: Analyze the second equation \( y=-3x - 6 \)

This is also a linear equation in slope - intercept form \( y = mx + b \), where the slope \( m=-3 \) and the y - intercept \( b=-6 \). The y - intercept is the point where \( x = 0 \), so when \( x = 0 \), \( y=-6 \). So we have the point \( (0,-6) \). Using the slope, since the slope is - 3 (which can be written as \( \frac{-3}{1} \)), from the point \( (0,-6) \), we can move down 3 units (because the slope is negative) and 1 unit to the right, or up 3 units and 1 unit to the left. For example, moving down 3 units and 1 unit to the right from \( (0,-6) \), we get the point \( (1,-6-3)=(1,-9) \), or moving up 3 units and 1 unit to the left, we get the point \( (-1,-6 + 3)=(-1,-3) \).

Step3: Find the intersection point (graphically)

When we graph both lines, the solution to the system of equations is the point where the two lines intersect. To find the intersection algebraically (which will help us confirm the graphical solution), we can set the two equations equal to each other:
\[
\frac{1}{2}x+1=-3x - 6
\]
Multiply through by 2 to clear the fraction:
\[
x + 2=-6x-12
\]
Add \( 6x \) to both sides:
\[
x+6x + 2=-12
\]
\[
7x+2=-12
\]
Subtract 2 from both sides:
\[
7x=-12 - 2=-14
\]
Divide both sides by 7:
\[
x=-2
\]
Now substitute \( x = - 2 \) into one of the original equations, say \( y=\frac{1}{2}x + 1 \). Then \( y=\frac{1}{2}(-2)+1=-1 + 1=0 \). So the point of intersection is \( (-2,0) \).

Answer:

The solution to the system of equations (the point of intersection of the two lines) is \( x=-2,y = 0 \) or the ordered pair \( (-2,0) \).