To find the intersection of the two lines represented by the equations:

1. $y = -\frac{1}{2}x + \frac{3}{2}$
2. $y = -\frac{4}{3}x + \frac{2}{3}$

we need to solve the system algebraically by setting the equations equal to each other:

$-\frac{1}{2}x + \frac{3}{2} = -\frac{4}{3}x + \frac{2}{3}$

Step 1: Eliminate Fractions by Multiplying by a Common Denominator

The common denominator for 2 and 3 is 6. Multiply the entire equation by 6:

$6\left(-\frac{1}{2}x + \frac{3}{2}\right) = 6\left(-\frac{4}{3}x + \frac{2}{3}\right)$

This simplifies to:

$-3x + 9 = -8x + 4$

Step 2: Solve for $x$

Now, let's solve for $x$ by isolating it on one side:

$-3x + 8x = 4 - 9$

$5x = -5$

$x = -1$

Step 3: Substitute $x = -1$ into One of the Original Equations

Now substitute $x = -1$ into either equation to find $y$. Let's use the first equation:

$y = -\frac{1}{2}(-1) + \frac{3}{2}$

$y = \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2$

Final Result: Point of Intersection

The intersection point of the two lines is at:

$(x, y) = (-1, 2)$

This means the two lines cross at the point $(-1, 2)$ on the graph.

Would you like to see this process visualized on a graph, or do you have any other questions or tasks you'd like to complete?

Related Questions:

1. How do we verify the solution by substitution into both equations?
2. What is the significance of finding the intersection point graphically?
3. Can this method be used for systems with more than two variables?
4. How would the solution change if the coefficients were different?
5. What would happen if the system had no intersection point?
6. How does this algebraic method compare to graphing in terms of accuracy?
7. How can this method be extended to systems of inequalities?
8. How can we use matrices to solve systems of linear equations?

Tip: Always double-check your algebraic steps, as small errors can lead to incorrect intersection points.