Question 240·Hard·Algebra

Ignore $y$ for a moment and compare $3x - 4$ and $-\tfrac12 x + 6$. For there to be any $y$ that satisfies both inequalities at once, should $3x - 4$ be less than or greater than $-\tfrac12 x + 6$?

Set up and solve an inequality that compares $3x - 4$ and $-\tfrac12 x + 6$. Then remember you still must also satisfy $x > -1$ from the third inequality.

Once you have an inequality for $x$ from the first two conditions and the inequality $x > -1$ from the third, express the combined condition on $x$ as a single compound inequality.

In Desmos, type y < -1/2 x + 6 and y >= 3x - 4. You should see two shaded regions: one below the line with slope $-\tfrac12$ and one on or above the line with slope $3$.

Type x > -1. Desmos will shade to the right of the vertical line $x = -1$. Now look for the region where all three shadings overlap; this is the solution set of the system.

Find the leftmost and rightmost $x$-values where all three shaded regions overlap. The left boundary comes from the vertical line $x = -1$. The right boundary is the $x$-coordinate of the intersection point of the two lines $y = -1/2 x + 6$ and $y = 3x - 4$ (click that intersection point to see its coordinates).

Use the two boundary $x$-values you observed (from the vertical line and from the intersection point) to write a compound inequality describing all $x$-coordinates in the overlapping region, and then match that inequality to one of the choices.

The system

• $y < -\tfrac12 x + 6$
• $y \ge 3x - 4$
• $x > -1$

describes all $(x,y)$ points that satisfy all three inequalities. The question asks for only the $x$-coordinates of such points—that is, all $x$ for which there exists at least one $y$ satisfying the first two inequalities and also $x > -1$.

For a fixed value of $x$:

• $y \ge 3x - 4$ means $y$ is at or above the line $y = 3x - 4$ (this is a lower bound for $y$).
• $y < -\tfrac12 x + 6$ means $y$ is below the line $y = -\tfrac12 x + 6$ (this is an upper bound for $y$).

So for a given $x$ to work, there must be at least one $y$ that lies between these two lines:

This is only possible if the lower bound is strictly less than the upper bound.

Require the lower bound to be less than the upper bound:

Solve this inequality:

So any $x$ that can possibly work with the first two inequalities must satisfy $x < \tfrac{20}{7}$.

We now have two conditions on $x$:

• From the third inequality: $x > -1$.
• From comparing the $y$-bounds: $x < \tfrac{20}{7}$.

Together, they describe all possible $x$-coordinates of points that satisfy the system:

This matches answer choice D.