- Identify the two lines by their slopes and \(y\)-intercepts, then determine points on each to facilitate graphing.
- Plot both lines on a Cartesian coordinate system using the points identified.
- Find the intersection point of the two lines, which represents the solution to the system.
- Verify the intersection point algebraically to ensure accuracy: \(\boxed{-1, 2}\).
Explanation
- Analyzing the Equations
To solve the system of linear equations \(y = 3x + 5\) and \(y = -3x - 1\), we recognize these as two linear equations that describe straight lines on a Cartesian coordinate system. Each equation is in the slope-intercept form \(y = mx + b\), where \(m\) is the slope and \(b\) is the \(y\)-intercept. For the first line, the slope is \(3\) and the \(y\)-intercept is \((0, 5)\). For the second line, the slope is \(-3\) and the \(y\)-intercept is \((0, -1)\). Our objective is to determine the point \((x, y)\) where these two lines intersect.
- Graphing the Lines
To graph these lines, we select two distinct points for each.
For the line \(y = 3x + 5\):
- If \(x = 0\), then \(y = 3(0) + 5 = 5\), giving the point \((0, 5)\).
- If \(x = -1\), then \(y = 3(-1) + 5 = 2\), giving the point \((-1, 2)\).
For the line \(y = -3x - 1\):
- If \(x = 0\), then \(y = -3(0) - 1 = -1\), giving the point \((0, -1)\).
- If \(x = 1\), then \(y = -3(1) - 1 = -4\), giving the point \((1, -4)\).
By plotting these points and drawing lines through them, we visually inspect where they meet.
- Solving the System Algebraically
To confirm the intersection point analytically, we set the two expressions for \(y\) equal to each other, since at the point of intersection, the \(x\) and \(y\) values must be the same for both equations:
\[3x + 5 = -3x - 1\]
Adding \(3x\) to both sides gives:
\[6x + 5 = -1\]
Subtracting \(5\) from both sides:
\[6x = -6\]
Dividing by \(6\) results in:
\[x = -1\]
Now, substitute \(x = -1\) into the first equation to find \(y\):
\[y = 3(-1) + 5 = -3 + 5 = 2\]
Thus, the point of intersection is \((-1, 2)\). This satisfies both equations, as checking the second equation also yields \(y = -3(-1) - 1 = 3 - 1 = 2\).
- Conclusion
The intersection of the two lines on the coordinate plane is at the point \((-1, 2)\). This means the solution to the system of equations is \(x = -1\) and \(y = 2\). Therefore, the solution point is \(\boxed{-1, 2}\).
Examples
This method of solving systems of linear equations is useful in many real-life situations, such as comparing the costs of two different service plans. For instance, if one plan has a flat fee plus a per-minute rate and another plan has a different flat fee and rate, graphing the cost functions allows you to find the exact duration where the plans cost the same, helping you make an informed decision on which plan is more economical for your usage.