Question

Solve the following system of equations graphically on the set of axes below.
y, equals, minus, x, minus, 2
y=−x−2
y, equals, one sixth, x, plus, 5
y=
6
1

x+5
Plot two lines by clicking the graph.
Click a line to delete it.
x
y
x

Answers

GPT-4o mini
To solve the system of equations graphically, we will plot the two equations:

1. \( y = -x - 2 \)
2. \( y = \frac{1}{6}x + 5 \)

### Step 1: Plot the first equation

For the first equation \( y = -x - 2 \):

- Find the y-intercept: When \( x = 0 \),
\[
y = -0 - 2 = -2
\]
So the point is (0, -2).

- Find another point: When \( x = 2 \),
\[
y = -2 - 2 = -4
\]
So the point is (2, -4).

### Step 2: Plot the second equation

For the second equation \( y = \frac{1}{6}x + 5 \):

- Find the y-intercept: When \( x = 0 \),
\[
y = \frac{1}{6}(0) + 5 = 5
\]
So the point is (0, 5).

- Find another point: When \( x = 6 \),
\[
y = \frac{1}{6}(6) + 5 = 1 + 5 = 6
\]
So the point is (6, 6).

### Step 3: Graphing the lines

Now you can plot the points you've found:

1. For \( y = -x - 2 \):
- Plot (0, -2) and (2, -4) and draw the line through these points.

2. For \( y = \frac{1}{6}x + 5 \):
- Plot (0, 5) and (6, 6) and draw the line through these points.

### Step 4: Finding the Intersection

The solution to the system of equations is the point where the two lines intersect. From the graph, identify the point where the two lines cross.

### Step 5: Analyzing the Graph

You will likely find that the lines intersect at a certain point (let's say \( (x, y) \)). You can also compute the intersection analytically by setting the two equations equal to each other:

\[
-x - 2 = \frac{1}{6}x + 5
\]

Combine like terms and solve for \( x \):

1. Multiply through by 6 to eliminate the fraction:
\[
-6x - 12 = x + 30
\]
\[
-7x = 42
\]
\[
x = -6
\]

2. Substitute \( x = -6 \) back into one of the original equations to find \( y \):
\[
y = -(-6) - 2 = 6 - 2 = 4
\]

So the solution to the system of equations is \( (-6, 4) \).

You can confirm this graphically by ensuring the lines intersect at this point on your plotted graph.

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