Once you substitute your expression for $y$ into $2x + 5y = 19$, combine like terms so that you have an equation of the form "(something with k) times $x$ equals a constant" and then solve for $x$.

Your expression for $x$ will be a fraction with denominator $5k + 2$. For which integer values of $k$ from 1 to 10 does that denominator divide the numerator evenly so that $x$ is an integer?

Once you find a possible $k$, plug it back in to find $x$ and $y$ and make sure both turn out to be integers.

In Desmos, type the expression x(k) = 14/(5k + 2) to define $x$ as a function of $k$ based on the algebra you did by hand.

In the same line, add {1 <= k <= 10} after the expression so Desmos knows the range for $k$, or simply create a slider for k and set its range from 1 to 10 (step 1).

Move the slider for $k$ through the integers 1 to 10 and watch the value of $x(k)$. You are looking for a value of $k$ where $x(k)$ is an integer. For that same $k$, compute y = 1 + kx(k) in a new line and confirm that $y$ is also an integer; this $k$ is the solution.

Start with the second equation:

Solve for $y$ in terms of $x$ and $k$:

Substitute $y = 1 + kx$ into the first equation $2x + 5y = 19$:

Distribute and combine like terms:

So,

We are told $x$ is an integer. That means $5k + 2$ must be a (positive) divisor of 14, because $x = 14/(5k+2)$.

The positive divisors of 14 are $1, 2, 7$, and $14$. Also, since $1 \le k \le 10$, we have $5k + 2 \ge 5\cdot 1 + 2 = 7$ and $5k + 2 \le 5\cdot 10 + 2 = 52$. So the only possible values for $5k + 2$ that divide 14 and lie between 7 and 52 are $7$ and $14$.

So we must have either $5k + 2 = 7$ or $5k + 2 = 14$. Solve each for $k$.

Case 1: $5k + 2 = 7$:

Case 2: $5k + 2 = 14$:

which is not an integer, so it is not allowed.

Thus the only valid value in the range $1 \le k \le 10$ is $k = 1$.

Check quickly: if $k = 1$, then $x = 14/(5\cdot 1 + 2) = 14/7 = 2$, and $y = 1 + kx = 1 + 1\cdot 2 = 3$, both integers.

Therefore, the value of $k$ is $1$.