Question
What is the apparent solution to the system of equations?
y=12x+2y=2x−1
Graph the system of equations using the Line tool. Plot a point at the apparent solution to the system using the Point tool.
Keyboard Instructions
Initial graph state
The horizontal axis goes from -10.8 to 10.8 with ticks spaced every 1 unit(s).
The vertical axis goes from -10.8 to 10.8 with ticks spaced every 1 unit(s).
y=12x+2y=2x−1
Graph the system of equations using the Line tool. Plot a point at the apparent solution to the system using the Point tool.
Keyboard Instructions
Initial graph state
The horizontal axis goes from -10.8 to 10.8 with ticks spaced every 1 unit(s).
The vertical axis goes from -10.8 to 10.8 with ticks spaced every 1 unit(s).
Answers
GPT-4o mini
To solve the system of equations given by:
1. \(y = 12x + 2\)
2. \(y = 2x - 1\)
we will set the two equations equal to each other to find the intersection point, which is the solution to the system:
\[
12x + 2 = 2x - 1
\]
Now, we'll solve for \(x\):
1. Move \(2x\) to the left side:
\[
12x - 2x + 2 = -1
\]
which simplifies to:
\[
10x + 2 = -1
\]
2. Next, subtract \(2\) from both sides:
\[
10x = -3
\]
3. Now, divide by \(10\):
\[
x = -\frac{3}{10} = -0.3
\]
Now that we have the \(x\)-coordinate, we can substitute \(x = -0.3\) back into either of the original equations to find the \(y\)-coordinate. We'll use the second equation:
\[
y = 2(-0.3) - 1 = -0.6 - 1 = -1.6
\]
So, the apparent solution to the system of equations is:
\[
(x, y) = \left(-0.3, -1.6\right)
\]
### Graphing the System of Equations
1. **Graph the First Equation \(y = 12x + 2\)**:
- When \(x = 0\): \(y = 2\)
- When \(x = 1\): \(y = 14\)
- When \(x = -1\): \(y = -10\)
2. **Graph the Second Equation \(y = 2x - 1\)**:
- When \(x = 0\): \(y = -1\)
- When \(x = 1\): \(y = 1\)
- When \(x = -1\): \(y = -3\)
### Plotting Points
After plotting both lines based on the coordinates calculated:
- The line for \(y = 12x + 2\) is steep, and crosses the y-axis at (0,2).
- The line for \(y = 2x - 1\) rises more gradually and crosses the y-axis at (0,-1).
To locate the apparent solution on the graph:
- Plot the point \((-0.3, -1.6)\).
This point lies somewhere between the intersections of the two lines, where they intersect at the calculated coordinates.
### Final Step
Ensure to highlight the point \((-0.3, -1.6)\) clearly while graphing. If you're using digital drawing or graphing software, you can enable the line and point tools to accomplish this.
1. \(y = 12x + 2\)
2. \(y = 2x - 1\)
we will set the two equations equal to each other to find the intersection point, which is the solution to the system:
\[
12x + 2 = 2x - 1
\]
Now, we'll solve for \(x\):
1. Move \(2x\) to the left side:
\[
12x - 2x + 2 = -1
\]
which simplifies to:
\[
10x + 2 = -1
\]
2. Next, subtract \(2\) from both sides:
\[
10x = -3
\]
3. Now, divide by \(10\):
\[
x = -\frac{3}{10} = -0.3
\]
Now that we have the \(x\)-coordinate, we can substitute \(x = -0.3\) back into either of the original equations to find the \(y\)-coordinate. We'll use the second equation:
\[
y = 2(-0.3) - 1 = -0.6 - 1 = -1.6
\]
So, the apparent solution to the system of equations is:
\[
(x, y) = \left(-0.3, -1.6\right)
\]
### Graphing the System of Equations
1. **Graph the First Equation \(y = 12x + 2\)**:
- When \(x = 0\): \(y = 2\)
- When \(x = 1\): \(y = 14\)
- When \(x = -1\): \(y = -10\)
2. **Graph the Second Equation \(y = 2x - 1\)**:
- When \(x = 0\): \(y = -1\)
- When \(x = 1\): \(y = 1\)
- When \(x = -1\): \(y = -3\)
### Plotting Points
After plotting both lines based on the coordinates calculated:
- The line for \(y = 12x + 2\) is steep, and crosses the y-axis at (0,2).
- The line for \(y = 2x - 1\) rises more gradually and crosses the y-axis at (0,-1).
To locate the apparent solution on the graph:
- Plot the point \((-0.3, -1.6)\).
This point lies somewhere between the intersections of the two lines, where they intersect at the calculated coordinates.
### Final Step
Ensure to highlight the point \((-0.3, -1.6)\) clearly while graphing. If you're using digital drawing or graphing software, you can enable the line and point tools to accomplish this.