After substituting $y=2x-1$, rearrange each resulting equation so that $x$ is written as a fraction involving $k$. You should end up with two different expressions for $x$ in terms of $k$.

Because there is one point $(x,y)$ satisfying both equations and the line, the two expressions you found for $x$ must be equal. Set them equal and simplify to get an equation involving only $k$.

Your equation in $k$ should be a quadratic. You are asked for the sum of all possible $k$ values. Think about how the coefficients of a quadratic relate to the sum of its roots (Vieta’s formulas).

In Desmos, treat $k$ as the variable $x$. Type the equation you derived from setting the two expressions for $x$ equal:

x^2 - 15x + 40 = 0.

Desmos will display this parabola and can also show its intersection points with the $x$-axis (the roots).

Tap on the points where the graph crosses the $x$-axis to see the two $x$-values (these are the $k$ values). Then, in a new expression line, type their sum (for example, a + b if you stored them as a and b). The value Desmos returns is the required sum of all real $k$ values.

The solution $(x,y)$ lies on the line $y=2x-1$, so in each equation of the system replace $y$ with $2x-1$.

First equation:

Second equation:

Simplify the first substituted equation:

Now solve for $x$ (assuming $k\neq 0$ so $7k\neq 0$):

Simplify the second substituted equation:

Now solve for $x$ (assuming $5-2k\neq 0$):

The same point $(x,y)$ must satisfy both equations, so the two expressions for $x$ must be equal:

Cross-multiply to clear denominators:

Expand both sides:

Set them equal and gather all terms on one side:

This quadratic equation gives all $k$ for which the system's solution lies on $y=2x-1$ (with $k\neq 0$ and $k\neq \tfrac{5}{2}$ so the earlier divisions were valid).

For the quadratic

the sum of the roots is given by Vieta’s formula: sum of roots $= \dfrac{15}{1} = 15$.

There are two real roots because the discriminant $15^2 - 4\cdot 1\cdot 40 = 65$ is positive. These roots are not $0$ or $\tfrac{5}{2}$, so our earlier divisions are valid. The system has a unique solution exactly when the determinant of its coefficient matrix, $k^2 + 13k + 5$, is nonzero; this quadratic has different roots from $k^2 - 15k + 40$, so neither of our $k$ values makes the determinant zero.

Thus both $k$ values satisfy all the conditions, and the sum of all such real values of $k$ is 15.