Since both expressions equal $y$, try setting $x^3 - 3x$ equal to $2x + 1$ and simplify the resulting equation. What kind of equation in $x$ do you get?

After simplifying, you get a cubic equation. Instead of solving it exactly, think about how many times its graph crosses the $x$-axis by checking the sign of the function at several $x$-values.

If a continuous function changes sign between two $x$-values, it must have a root between them. For a cubic, what is the maximum possible number of real roots it can have?

In Desmos, enter y = x^3 - 3x on one line and y = 2x + 1 on another. Make sure both graphs are visible in the same window.

Click on the points where the line and the cubic curve intersect. Desmos will mark each intersection point and display its coordinates.

Count how many distinct intersection points appear; that count is the number of solutions to the system, since each intersection corresponds to a solution $(x,y)$.

Each equation represents a graph in the $xy$-plane:

• $y = x^3 - 3x$ is a cubic curve (S-shaped).
• $y = 2x + 1$ is a straight line.

Solutions to the system are the intersection points of these two graphs. Our job is to figure out how many intersection points there are.

Since both expressions equal $y$, set them equal to each other:

Move everything to one side:

Any solution $x$ of this equation gives an intersection point (and therefore a solution to the system). So we need to know how many real solutions the equation $x^3 - 5x - 1 = 0$ has.

Let $f(x) = x^3 - 5x - 1$. This is a polynomial, so its graph is continuous (no jumps or gaps). If $f(x)$ changes sign between two $x$-values, there is at least one root between them.

Compute $f(x)$ at some convenient points:

• $f(-3) = (-3)^3 - 5(-3) - 1 = -27 + 15 - 1 = -13$ (negative)
• $f(-1) = (-1)^3 - 5(-1) - 1 = -1 + 5 - 1 = 3$ (positive)
• $f(0) = 0^3 - 5(0) - 1 = -1$ (negative)
• $f(1) = 1^3 - 5(1) - 1 = 1 - 5 - 1 = -5$ (negative)
• $f(3) = 3^3 - 5(3) - 1 = 27 - 15 - 1 = 11$ (positive)

Now look for sign changes:

• Between $x=-3$ (negative) and $x=-1$ (positive), there is at least one root.
• Between $x=-1$ (positive) and $x=0$ (negative), there is at least one root.
• Between $x=1$ (negative) and $x=3$ (positive), there is at least one root.

These are three separate intervals, so there are at least 3 distinct real roots of $f(x)$.

The equation $x^3 - 5x - 1 = 0$ is a cubic, so it can have at most 3 real roots. From the sign changes, we found there are at least 3 distinct real roots.

Putting this together, the cubic has exactly 3 real solutions for $x$. Each of these corresponds to one intersection point of the two original graphs and thus one solution $(x,y)$ to the system.

Therefore, there are exactly 3 solutions.