Look at the coefficients of $x$ and $y$ in both equations. Can you factor or simplify them to see if one equation is already a constant multiple of the other in those coefficients?

Once you notice the $x$- and $y$-coefficients in the second equation are a fixed multiple of those in the first, ask: what equation must the constant terms satisfy so that the entire second equation is that same multiple of the first?

In Desmos, type the two equations using a parameter (for example, use a instead of $t$):

• (5a - 7)x + 2y = 3
• (10a - 14)x + 4y = 6a - 1 Desmos will prompt you to create a slider for a. Add the slider.

Move the slider for a and watch how the two lines change. You are looking for the value of a where the two graphs lie exactly on top of each other, not just intersect at a point.

Adjust the slider carefully until the two lines are indistinguishable (only one line is visible because both equations describe it). The value shown for a at that moment is the value of $t$ for which the system has infinitely many solutions.

For a system of two linear equations in $x$ and $y$ to have infinitely many solutions, the two equations must represent the same line.

That means the ratios of the corresponding coefficients must all be equal:

• ratio of the $x$-coefficients,
• ratio of the $y$-coefficients,
• ratio of the constant terms.

So we want all three ratios (for $x$, $y$, and the constants) to be the same number.

Write the system:

Look at the $x$-coefficients:

• First equation: $5t - 7$
• Second equation: $10t - 14 = 2(5t - 7)$

Look at the $y$-coefficients:

• First equation: $2$
• Second equation: $4 = 2 \cdot 2$

So, for any $t$, the $x$- and $y$-coefficients in the second equation are already 2 times the corresponding coefficients in the first equation.

Therefore, for the lines to be the same, the constant term in the second equation must also be 2 times the constant term in the first equation.

The constant term in the first equation is $3$.

If the second equation is exactly $2$ times the first, then its constant term (right-hand side) should be:

• $2 \cdot 3 = 6$.

But the constant term in the second equation is $6t - 1$.

So we must have:

This equation will give the value of $t$ that makes the two equations identical.

Solve the equation from the previous step:

When $t = \frac{7}{6}$, every coefficient (including the constant term) in the second equation is exactly $2$ times the corresponding coefficient in the first, so the two equations represent the same line and the system has infinitely many solutions.

Therefore, the correct answer is $\boxed{\tfrac{7}{6}}$.