Since both expressions equal $y$, set $x^4 - 4x^2 + 3$ equal to $x + 1$ and rearrange to form one equation equal to 0.

Once you have the quartic equation in $x$, try to factor it by finding simple integer roots like $-2,-1,0,1,2$, then use division to reduce the degree.

If part of the expression factors to a quadratic that you cannot factor easily, apply the quadratic formula to find its roots.

In one expression line, type y = x^4 - 4x^2 + 3. In another line, type y = x + 1 to graph both equations on the same axes.

Use zoom out (−) and drag the graph so you can see where the quartic curve and the line cross each other over a reasonable range of $x$ (for example, from about $x = -3$ to $x = 3$).

Click on each point where the line and the quartic curve intersect; Desmos will display their coordinates. Count how many distinct intersection points you see—that count is the number of solutions to the system.

The solutions are the intersection points of the two graphs, so their $x$- and $y$-coordinates must satisfy both equations.

Set the right sides equal:

Move everything to one side:

Now we just need to solve this quartic (degree 4) equation for $x$.

Try to factor $x^4 - 4x^2 - x + 2$ by finding simple roots (values of $x$ that make it zero).

Test small integers:

• For $x = -1$:

so $x = -1$ is a root, and $(x + 1)$ is a factor.

Divide $x^4 - 4x^2 - x + 2$ by $(x+1)$ (using synthetic or long division) to get:

Now factor the cubic $x^3 - x^2 - 3x + 2$.

Test $x = 2$:

so $x = 2$ is a root, and $(x - 2)$ is a factor. Divide to get:

So the full factorization is

Set each factor equal to zero:

• $x + 1 = 0 \Rightarrow x = -1$
• $x - 2 = 0 \Rightarrow x = 2$
• $x^2 + x - 1 = 0$.

Solve $x^2 + x - 1 = 0$ using the quadratic formula:

So the four distinct real $x$-values where the graphs intersect are

Each $x$ gives one point $(x,y)$ on both graphs, with $y = x + 1$. Therefore, there are exactly 4 solutions to the system.