Question 171·Medium·Algebra

After distributing, move all the x terms to one side of the equation and the constant terms (without x) to the other side.

If you end up with something like $Ax = B$, what has to be true about $A$ and $B$ so that there is no value of $x$ that makes the equation true?

Identify the expression that plays the role of $A$ (the coefficient of $x$) and the expression that plays the role of $B$ (the constant), and use the no-solution condition on them to find $k$.

In Desmos, type y = k(3x - 2) on one line and y = 3x + 5 on another line. Desmos will automatically create a slider for k.

Move the k slider and watch how the graph of y = k(3x - 2) changes compared with y = 3x + 5. You are looking for the k value where the two lines have the same slope (are parallel) but never intersect.

Stop when the two graphs are parallel and do not touch at any point. The current value shown for k on the slider is the value that makes the equation have no solution.

Start by expanding the left-hand side of the equation.

Distribute $k$:

Get all the x terms on one side and the constants on the other.

Subtract $3x$ from both sides:

Factor out $x$ from the first two terms:

Add $2k$ to both sides:

Now the equation looks like $(3k - 3)x = 2k + 5$.

For an equation of the form $Ax = B$:

• If $A \neq 0$, there is exactly one solution for $x$.
• If $A = 0$ and $B = 0$, there are infinitely many solutions.
• If $A = 0$ and $B \neq 0$, there is no solution.

Here, $A = 3k - 3$ and $B = 2k + 5$. For no solution, we need $3k - 3 = 0$ and $2k + 5 \neq 0$.

Solve $3k - 3 = 0$:

Check the constant term when $k = 1$: $2k + 5 = 2(1) + 5 = 7$, which is not $0$.

So, when $k = 1$, the equation becomes $0x = 7$, which is impossible, meaning the equation has no solution.

The required value of $k$ is $1$.