The system of equations provided in the image is:

4x - 6y = 10y + 2 \tag{1} ty = \frac{1}{2} + 2x \tag{2}

We are asked to find the value of $t$ such that the system has no solution. Let's proceed step by step:

Step 1: Simplify the first equation

Starting with equation (1):

$4x - 6y = 10y + 2$

Rearrange to get all terms involving $y$ on one side:

$4x = 16y + 2$

Now, solve for $x$:

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

Step 2: Substitute into the second equation

From equation (2):

$ty = \frac{1}{2} + 2x$

Substitute $x = 4y + \frac{1}{2}$ into this equation:

$ty = \frac{1}{2} + 2 \left( 4y + \frac{1}{2} \right)$

Simplify:

$ty = \frac{1}{2} + 8y + 1$ $ty = 8y + \frac{3}{2}$

Step 3: Analyze conditions for no solution

For the system to have no solution, the lines represented by these equations must be parallel, which means that the coefficients of $y$ in both equations must be equal. From the first equation, we already know that:

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

So, the coefficient of $y$ in equation (2) must also match the coefficient of $y$ in the equation we derived. Therefore, we equate the coefficients of $y$:

$t = 8$

Thus, the value of $t$ that causes the system to have no solution is $\boxed{8}$.

Would you like further details on any of these steps or have additional questions?

Here are five related questions you might explore:

1. How do we verify that two lines are parallel in general?
2. What happens when $t \neq 8$? Does the system have one or infinitely many solutions?
3. How would the problem change if $t$ was a function instead of a constant?
4. Can this method be extended to systems with three variables?
5. What conditions are needed for two lines to be perpendicular instead of parallel?

Tip: When dealing with systems of equations, always look at the relationship between the slopes of the lines to determine whether they are parallel, intersecting, or identical (coincident).