Because one equation already has $y$ by itself ($y=x^2$), try substituting $x^2$ for $y$ in the other equation $x+y=6$ to get an equation in just $x$.

After substitution, you will get a quadratic equation in $x$. Factor it (or use another method) to find the possible $x$-values, then plug back into $y=x^2$ to find $y$ for each.

Once you have the possible $(x,y)$ pairs from solving the system, compare them with the four answer choices and pick the one that matches.

In Desmos, enter the equations as two separate lines: y = x^2 and y = 6 - x. These represent the parabola and the line from the system.

Look for the points where the parabola and the line cross. Click on each intersection to see its coordinates; these are the solutions to the system.

Compare the intersection coordinates you see in Desmos with the four answer choices and select the option that exactly matches one of those intersection points.

You are told that $y=x^2$ and also that $x+y=6$.

Since $y=x^2$, substitute $x^2$ for $y$ in the second equation:

Rewrite this as a standard quadratic equation by moving all terms to one side:

Factor the quadratic $x^2 + x - 6$:

Set each factor equal to zero:

So the possible $x$-values are $x=-3$ and $x=2$. Each gives a possible solution point.

Use $y=x^2$ to find $y$ for each $x$:

• If $x=2$, then $y = 2^2 = 4$, so one possible solution is $(2,4)$.
• If $x=-3$, then $y = (-3)^2 = 9$, so another possible solution is $(-3,9)$.

Both points satisfy the system, but you must choose from the given answer options.

Look at the listed options: $(2,4)$, $(3,3)$, $(1,5)$, $(-1,7)$.

Only $(2,4)$ matches one of the solution points you found for the system (the other solution, $(-3,9)$, is not listed). Therefore, the correct answer is $(2,4)$.