Question 42·Hard·Linear Equations in One Variable

After expanding, move all $x$ terms to one side and all constant (no $x$) terms to the other side so the equation looks like "(something)$x$ = (something)."

If you have an equation of the form $ax = b$ in the variable $x$, what must be true about $a$ and $b$ for there to be no possible value of $x$ that makes the equation true?

In your simplified equation, identify the coefficient of $x$ and the constant. Which value of $k$ makes the coefficient of $x$ zero while the constant term is not zero?

In one expression line, type y1 = (2k - 1)(x - 3) + 7. In another line, type y2 = (k + 5)x + 5k - 6. Desmos will create a slider for k.

Move the k slider and watch the graphs of y1 and y2. You are looking for the value of k where the two lines are parallel (same slope) but never intersect. Read that k value from the slider.

For that k value, create a table for y1 and y2 at several x-values. If the y-values are never equal for any x, that k value makes the original equation have no solution.

Start by distributing on the left side.

Left side:

Right side is already expanded:

So the equation becomes

Move all terms to one side so the equation is in the form (something)$x$ + (something) $= 0$.

Subtract $(k+5)x$ and $(5k - 6)$ from both sides:

Combine like terms:

• Coefficient of $x$:

• Constant term:

So the equation becomes

Rewriting, this is equivalent to

Think of the simplified equation as $ax = b$, where $a$ and $b$ depend on $k$.

Here,

• $a = k - 6$ (the coefficient of $x$)
• $b = 11k - 16$ (the constant term)

For an equation $ax = b$ in the variable $x$:

• If $a \neq 0$, there is exactly one solution.
• If $a = 0$ and $b = 0$, there are infinitely many solutions.
• If $a = 0$ and $b \neq 0$, there is no solution.

So for no solution, we need

• $k - 6 = 0$ (so $a = 0$), and
• $11k - 16 \neq 0$ (so $b \neq 0$).

Now find the value of $k$ that makes $k - 6 = 0$, and then check that $11k - 16$ is not zero for that value.

Solve $k - 6 = 0$:

Check $11k - 16$ at $k = 6$:

so the condition $a = 0$ and $b \neq 0$ is satisfied.

To confirm, substitute $k = 6$ back into the original equation:

Left side:

Right side:

The equation becomes

which simplifies to $-26 = 24$, an impossible statement. That means there is no solution in $x$ when $k = 6$.

So the value of $k$ that makes the equation have no solution is $6$.