- Identify the given system of linear equations: \(y = \frac{2}{3}x + 1\) and \(y = -\frac{2}{3}x - 1\).
- Equate the expressions for \(y\) to find the x-coordinate: \(\frac{2}{3}x + 1 = -\frac{2}{3}x - 1\), which simplifies to \(\frac{4}{3}x = -2\), so \(x = -1.5\).
- Substitute the value of \(x\) into one of the equations to find the y-coordinate: \(y = \frac{2}{3}(-1.5) + 1 = 0\).
- Conclude that the intersection point for the system is \(\boxed{(-1.5, 0)}\).
Explanation
- Analysis of the System of Equations
The problem requires finding the intersection point of two linear equations:
- \(y = \frac{2}{3}x + 1\)
- \(y = -\frac{2}{3}x - 1\)
The first equation, \(y = \frac{2}{3}x + 1\), represents a line with a positive slope of \(\frac{2}{3}\) and a y-intercept of \((0, 1)\). The second equation, \(y = -\frac{2}{3}x - 1\), represents a line with a negative slope of \(-\frac{2}{3}\) and a y-intercept of \((0, -1)\). The solution to this system of equations is the point \((x, y)\) where these two lines intersect.
- Solving for the Intersection Point
To find the point of intersection, we set the expressions for \(y\) from both equations equal to each other:
\[\frac{2}{3}x + 1 = -\frac{2}{3}x - 1\]
Next, we solve for \(x\). Adding \(\frac{2}{3}x\) to both sides gives:
\[\frac{2}{3}x + \frac{2}{3}x + 1 = -1\]
\[\frac{4}{3}x + 1 = -1\]
Subtracting 1 from both sides:
\[\frac{4}{3}x = -2\]
Multiplying both sides by \(\frac{3}{4}\):
\[x = -2 \cdot \frac{3}{4} = -\frac{6}{4} = -1.5\]
With the value of \(x = -1.5\), we substitute it back into the first equation to find the corresponding \(y\)-coordinate:
\[y = \frac{2}{3}(-1.5) + 1 = \frac{2}{3}\left(-\frac{3}{2}\right) + 1 = -1 + 1 = 0\]
Thus, the intersection point is \((-1.5, 0)\).
- Conclusion
The intersection point of the two lines is \((-1.5, 0)\). This means the graph should show the line \(y = \frac{2}{3}x + 1\) crossing the x-axis at \(-1.5\) and having a y-intercept at \((0, 1)\), and the line \(y = -\frac{2}{3}x - 1\) crossing the x-axis at \(-1.5\) and having a y-intercept at \((0, -1)\). Any correct graph representing this system must show the intersection occurring exactly at this point on the Cartesian plane.
Examples
Systems of linear equations are used extensively in economics and finance to find the "equilibrium point" between two different forces. For example, if one equation represents the supply of a product based on its price and another represents the demand for that same product, the intersection point tells producers and economists the price at which the supply exactly meets the demand. This helps businesses determine optimal pricing strategies for their goods.