Answer:
a. [tex]x=27,y=13,z=9[/tex]
b. Yes, p || r.
Step-by-step explanation:
a.
From the figure,
Since, p || q,
[tex]3(x-1)=4x-30[/tex] (∵ Alternate exterior angles are equal for parallel lines)
This gives,
[tex]3x-3=4x-30\\4x-3x=-3+30\\x=27[/tex]
Now, q || r
∴ [tex]6y=4x-30[/tex] (∵ Corresponding angles are equal for parallel lines)
This gives,
[tex]6y=4(27)-30\\6y=78\\y=13[/tex]
Now, since r is a straight line, [tex]6y[/tex] and [tex]6(z+8)[/tex] are supplementary angles.
[tex]6y+6(z+8)=180\\6(13)+6z+48=180\\78+48+6z=180\\6z=180-(78+48)\\6z=54\\z=\frac{54}{6}=9[/tex]
Therefore, [tex]x=27,y=13,z=9[/tex].
b.
[tex]3(x-1)=3(27-1)=3(26)=78\\6y=6\times 13=78[/tex]
Since, alternate exterior angles [tex]3(x-1)[/tex] and [tex]6y[/tex] are equal, the lines p and r are parallel because, alternate exterior angles are equal only if two lines are parallel.
