$ABCD$ is a cyclic quadrilateral (see Figure). Find the angles of the cyclic quadrilateral.

(N/A) We know that the sum of the measures of opposite angles in a cyclic quadrilateral is $180^{\circ}$.
Therefore,$\angle A + \angle C = 180^{\circ}$.
$(4y + 20) + (-4x) = 180$
$-4x + 4y = 160$
$-x + y = 40$ $...(i)$
Also,$\angle B + \angle D = 180^{\circ}$.
$(3y - 5) + (-7x + 5) = 180$
$-7x + 3y = 180$ $...(ii)$
Multiplying equation $(i)$ by $3$,we obtain:
$-3x + 3y = 120$ $...(iii)$
Subtracting equation $(iii)$ from equation $(ii)$:
$(-7x + 3y) - (-3x + 3y) = 180 - 120$
$-4x = 60$
$x = -15$
Substituting $x = -15$ in equation $(i)$:
$-(-15) + y = 40$
$15 + y = 40$
$y = 25$
Now,calculating the angles:
$\angle A = 4y + 20 = 4(25) + 20 = 120^{\circ}$
$\angle B = 3y - 5 = 3(25) - 5 = 70^{\circ}$
$\angle C = -4x = -4(-15) = 60^{\circ}$
$\angle D = -7x + 5 = -7(-15) + 5 = 110^{\circ}$