After you substitute $y = x^2$ into $2y = x + 1$, rearrange the equation so that everything is on one side and the other side is $0$.

Once you have a quadratic equation in $x$, try factoring it. If it factors into two binomials, set each factor equal to zero to find possible $x$-values, then see which one appears in the choices.

In Desmos, enter the two equations as functions: y = x^2 and y = (x + 1)/2. These are equivalent to the system $y = x^2$ and $2y = x + 1$.

Look for the points where the graphs of $y = x^2$ and $y = (x + 1)/2$ intersect. Click on each intersection point that Desmos shows and note the $x$-coordinates.

Compare the $x$-coordinates of the intersection points to the answer options $-2, 0, 1, 3$ and identify which option matches one of those $x$-values.

You are given the system

Because $y = x^2$, substitute $x^2$ for $y$ in the first equation.

So $2y = x + 1$ becomes:

To solve for $x$, move all terms to one side so the equation equals $0$.

Starting with

subtract $x$ and $1$ from both sides:

This is a quadratic equation in standard form $ax^2 + bx + c = 0$.

Now factor $2x^2 - x - 1$.

We look for two binomials of the form $(2x + m)(x + n)$ that multiply to give $2x^2 - x - 1$.

Trying $(2x + 1)(x - 1)$:

So the factored form is

Use the zero product property: if $(2x + 1)(x - 1) = 0$, then either $2x + 1 = 0$ or $x - 1 = 0$.

Solve each:

• From $2x + 1 = 0$: $2x = -1$, so $x = -\tfrac{1}{2}$.
• From $x - 1 = 0$: $x = 1$.

The possible $x$-values are $-\tfrac{1}{2}$ and $1$. Among the answer choices $-2, 0, 1, 3$, the only one that matches a solution is $1$, so the correct answer is $1$.