After combining like terms, you should get something of the form $ax + b \ge 0$. Identify $a$ and $b$ in terms of $k$.

If $a \ne 0$ in $ax + b \ge 0$, think about what happens when $x$ becomes very large positive or very large negative. Can the inequality still hold for every real $x$?

Once you know what must be true about $a$ for $ax + b \ge 0$ to hold for all $x$, write that as an equation involving $k$ and solve.

For each answer choice, substitute that value of $k$ into the inequality to get a specific inequality in $x$. For example, if $k = 0$, you get $-3x + 7 \ge x + 4$, which simplifies to $-3x + 3 \ge 0$.

For a chosen $k$-value, rewrite the inequality so everything is on one side, like $f(x) \ge 0$. In Desmos, enter the expression $f(x)$ (for example, $f(x) = -3x + 3$) as a function.

Look at the graph of $f(x)$ in Desmos. If any part of the graph goes below the x-axis (where $f(x) < 0$), then that $k$ does not make the original inequality true for all $x$.

Repeat this process for each answer choice. The correct $k$-value is the one for which the graph of the corresponding $f(x)$ never goes below the x-axis, meaning $f(x) \ge 0$ for every $x$ shown.

Start with

Subtract $(3k+1)x$ and $4$ from both sides so that the right side becomes $0$:

Combine like terms:

• For $x$ terms: $(k-2) - (3k+1) = -2k - 3$.
• For constants: $7 - 4 = 3$.

So the inequality becomes

We now have an inequality of the form

for all real $x$, where here $a = -2k - 3$ and $b = 3$.

Ask: When can $ax + b \ge 0$ be true for every real $x$?

• If $a > 0$, then for very large negative $x$, $ax + b$ becomes a large negative number, so the inequality fails.
• If $a < 0$, then for very large positive $x$, $ax + b$ becomes a large negative number, so the inequality fails.

The only way $ax + b$ can stay $\ge 0$ for all $x$ is if $a = 0$, so the expression is constant, and then that constant $b$ must be $\ge 0$.

From the reasoning above, for

to hold for every real $x$:

1. The coefficient of $x$ must be $0$:

2. The constant term must satisfy $3 \ge 0$, which is already true, so it gives no extra restriction on $k$.

So all we need now is to solve the equation $-2k - 3 = 0$.

Solve

Add $3$ to both sides:

Divide both sides by $-2$:

This value makes the inequality equivalent to $3 \ge 0$ (a true statement) for all real $x$, so the correct answer is $k = -\frac{3}{2}$.