Ask yourself: if $x=4$, what happens to $(x-4)(y+3)$? Can that ever equal 3?

For each possible $x$ from the answer choices (other than 4), use $(x-4)(y+3)=3$ to find $y$, then plug both $x$ and $y$ into $x^2+y^2=25$ to see if it works.

In one expression line, type (x-4)(y+3)=3. In another line, type x^2 + y^2 = 25. Desmos will plot both the curve defined by the product equation and the circle of radius 5 centered at the origin.

Use the cursor or tap/click on the points where the two curves intersect. Desmos will display the coordinates $(x,y)$ of each intersection point.

Look at the x-coordinate(s) of the intersection point(s) that Desmos shows. Compare those x-values with the options 3, 4, 5, and 6, and choose the option that matches one of the intersection x-values.

Start with the equation $(x-4)(y+3)=3$.

If $x=4$, then $x-4=0$ and the left side becomes $0\cdot(y+3)=0$, which can never equal $3$. So $x=4$ cannot be part of any solution.

This immediately eliminates answer choice B) 4.

Try $x=3$.

From $(x-4)(y+3)=3$:

• Substitute $x=3$: $(3-4)(y+3)=3$.
• This gives $-1\cdot (y+3)=3$, so $y+3=-3$ and $y=-6$.

Now check the second equation $x^2+y^2=25$:

• $x^2+y^2 = 3^2 + (-6)^2 = 9 + 36 = 45$, which is not $25$.

So $x=3$ does not give a valid solution pair.

Try $x=6$.

From $(x-4)(y+3)=3$:

• Substitute $x=6$: $(6-4)(y+3)=3$.
• This gives $2(y+3)=3$, so $y+3=\dfrac{3}{2}$ and $y=-\dfrac{3}{2}$.

Now check $x^2+y^2=25$:

• $x^2+y^2 = 6^2 + \left(-\dfrac{3}{2}\right)^2 = 36 + \dfrac{9}{4} = \dfrac{144}{4} + \dfrac{9}{4} = \dfrac{153}{4}$.
• $\dfrac{153}{4}$ is not equal to $25$.

So $x=6$ also does not give a valid solution pair.

The only remaining option is $x=5$.

From $(x-4)(y+3)=3$:

• Substitute $x=5$: $(5-4)(y+3)=3$.
• This gives $1\cdot (y+3)=3$, so $y+3=3$ and $y=0$.

Now check $x^2+y^2=25$:

• $x^2+y^2 = 5^2 + 0^2 = 25 + 0 = 25$, which does satisfy the second equation.

Therefore, the possible value of $x$ from the choices given is 5.