1. Set up the triangle

Given:

• $A=(-1,0)$
• $\angle A=\dfrac{2\pi}{3}=120^\circ$
• $AB=AC$, so triangle $ABC$ is isosceles with vertex at $A$
• $B$ lies on the positive $x$-axis
• $BC=4\sqrt{3}$

Since $B$ is on the positive $x$-axis and $A=(-1,0)$, the segment $AB$ lies along the positive $x$-axis. So let $B=(b,0), \quad b>0.$ Then $AB=b-(-1)=b+1.$

Because $AB=AC$, let $AB=AC=s.$ Then $s=b+1$.

2. Use the formula for the base of an isosceles triangle

For an isosceles triangle with equal sides $s,s$ and included angle $120^\circ$, the base is $BC=2s\sin\left(\frac{120^\circ}{2}\right)=2s\sin 60^\circ=2s\cdot \frac{\sqrt{3}}{2}=s\sqrt{3}.$

Given $BC=4\sqrt{3},$ so $s\sqrt{3}=4\sqrt{3} \implies s=4.$

Hence $AB=4.$ Since $A=(-1,0)$ and $B$ is on the positive $x$-axis, $B=(3,0).$

3. Find coordinates of $C$

Now $AC=4$ and the angle between $AB$ and $AC$ is $120^\circ$. Since $AB$ is along the positive $x$-axis, the direction of $AC$ is at angle $120^\circ$ from the positive $x$-axis.

Thus $C=A+4(\cos 120^\circ,\sin 120^\circ).$ Using $\cos 120^\circ=-\frac12, \qquad \sin 120^\circ=\frac{\sqrt3}{2},$ we get $C=(-1,0)+4\left(-\frac12,\frac{\sqrt3}{2}\right)=(-1-2,2\sqrt3)=(-3,2\sqrt3).$

(If we took the reflected point below the axis, we would get $(-3,-2\sqrt3)$; that would not intersect $y=x+3$ as nicely. The upper point is the intended one.)

4. Equation of line $BC$

Points: $B=(3,0), \qquad C=(-3,2\sqrt3).$ Slope of $BC$ is $m=\frac{2\sqrt3-0}{-3-3}=\frac{2\sqrt3}{-6}=-\frac{\sqrt3}{3}.$

Equation through $B=(3,0)$: $y=-\frac{\sqrt3}{3}(x-3).$ So $y=-\frac{\sqrt3}{3}x+\sqrt3.$

5. Intersect with $y=x+3$

At the intersection, $x+3=-\frac{\sqrt3}{3}x+\sqrt3.$ Multiply by $3$: $3x+9=-\sqrt3 x+3\sqrt3.$ So $x(3+\sqrt3)=3(\sqrt3-3).$ Hence $x=\frac{3(\sqrt3-3)}{3+\sqrt3}.$ Rationalize: