Answer:
[tex]\large\boxed{18)\ x=13\ and\ y=9}\\\boxed{20)\ AO=8}[/tex]
Step-by-step explanation:
18)
If a quadrilateral is inscribed in a circle, the opposite angles add up to 180°.
Therefore we have the system of equations:
[tex]\left\{\begin{array}{ccc}(6x+23)+(8y+7)=180\\(6x-4)+(12y-2)=180\end{array}\right\\\\\left\{\begin{array}{ccc}6x+8y+(23+7)=180\\6x+12y+(-4-2)=180\end{array}\right\\\left\{\begin{array}{ccc}6x+8y+30=180&|\text{subtract 30 from both sides}\\6x+12y-6=180&|\text{add 6 to both sides}\end{array}\right\\\left\{\begin{array}{ccc}6x+8y=150\\6x+12y=186&|\text{multiply both sides by (-1)}\end{array}\right[/tex]
[tex]\underline{+\\\left\{\begin{array}{ccc}6x+8y=150\\-6x-12y=-186\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad\qquad-4y=-36\qquad|\text{divide both sides by (-4)}\\.\qquad\qquad\boxed{y=9}\\\\\text{Put the value of y to the first equation and solve them for x:}\\\\6x+8(9)=150\\6x+72=150\qquad\text{subtract 72 from both sides}\\6x=78\qquad\text{divide both sides by 6}\\\boxed{x=13}[/tex]
20)
If line AB is tangent to the circle, then the angle A is a right angle
[tex]m\angle A=90^o[/tex]
Therefore the triangle is a right triangle. We can use the Pythagorean theorem:
Let O - center of the circle. Therefoe OA is the radius of a circle.
From a Pythagorean theorem:
[tex]AO^2+15^2=(9+8)^2\\\\AO^2+225=17^2\\\\AO^2+225=289\qquad\text{subtract 225 from both sides}\\\\AO^2=64\to AO=\sqrt{64}\\\\AO=8[/tex]
