After expanding, move all terms to one side of the equation so it looks like $(\text{expression in }k)x + (\text{expression in }k) = 0$.

For an equation like $Ax + B = 0$, consider what must be true about $A$ and $B$ for the equation to have no values of $x$ that make it true.

Set the coefficient of $x$ equal to $0$, but make sure the constant term is not $0$. Solve the resulting equation for $k$.

Type y1 = (k - 3)(2x + 5) and y2 = 4x + 10(k - 3). Desmos will create a slider for k so you can change its value.

Move the k slider and watch the two lines. For most values of k, the lines intersect at exactly one point (one solution for $x$).

Adjust k until the two graphs are parallel (same slope) but never cross. The value of k shown on the slider at that moment is the value that makes the equation have no solution (since the two sides are never equal for any $x$).

The equation

is linear in $x$, but the coefficients depend on the constant $k$. We want to find the value of $k$ that makes this linear equation have no solutions for $x$.

First expand the left side using the distributive property:

• $(k - 3)(2x + 5) = k(2x) + k(5) - 3(2x) - 3(5)$
• $= 2kx + 5k - 6x - 15$.

Expand the right side:

• $10(k - 3) = 10k - 30$,
• so the right side is $4x + 10k - 30$.

Now the equation is:

Subtract the entire right side from both sides to get $0$ on the right:

Combine like terms:

• $x$ terms: $2kx - 6x - 4x = 2kx - 10x = (2k - 10)x$,
• constant/$k$ terms: $5k - 15 - 10k + 30 = -5k + 15$.

So the equation becomes:

This has the form $Ax + B = 0$, where $A = 2k - 10$ and $B = -5k + 15$.

For a linear equation $Ax + B = 0$:

• If $A \neq 0$, there is one solution $x = -B/A$.
• If $A = 0$ and $B = 0$, there are infinitely many solutions (the equation is always true).
• If $A = 0$ and $B \neq 0$, there are no solutions (the equation is impossible).

We want no solutions, so we need:

• $2k - 10 = 0$ (so the $x$ term disappears), and
• $-5k + 15 \neq 0$ (so a nonzero constant remains).

First solve $2k - 10 = 0$:

Now check the constant term:

So when $k = 5$, the equation becomes an impossible statement (a nonzero constant equals $0$), meaning there is no solution for $x$.

Therefore, the value of $k$ that makes the equation have no solutions is $5$.