Greetings!
Using the Parallel Line Theorem (c-formation), we can conclude that:
[tex](5z+21)+(3z-11)=180[/tex]
Combine like terms:
[tex]5z+21+3z-11=180[/tex]
[tex]8z+10=180[/tex]
Add -10 to both sides:
[tex](8z+10)+(-10)=(180)+(-10)[/tex]
[tex]8z=170[/tex]
Divide both sides by 8:
[tex] \frac{8z}{8}= \frac{170}{8} [/tex]
[tex]\boxed{z=\frac{85}{4}}[/tex]
Using the Parallel Line Theorem (z-formation), we can also conclude that:
[tex](3z-11)=y[/tex]
Input the value for z:
[tex](3(\frac{85}{4})-11)=y[/tex]
Combine like terms:
[tex](\frac{255}{4})-11=y[/tex]
[tex]\frac{255}{4}-\frac{44}{4}=y[/tex]
[tex]\frac{211}{4}=y[/tex]
[tex]\boxed{y=\frac{211}{4}}[/tex]
Lastly, using the Parallel Line Theorem (f-formation), we can conclude that:
[tex]y=(7x+7)[/tex]
Input the value for y:
[tex]\frac{211}{4}=(7x+7)[/tex]
Add -7 to both sides:
[tex](\frac{211}{4})+(-7)=(7x+7)+(-7)[/tex]
[tex]\frac{211}{4}-\frac{28}{4}=7x[/tex]
[tex]\frac{183}{4}=7x[/tex]
Divide both sides by 7:
[tex] \frac{\frac{183}{4}}{7}= \frac{7x}{7} [/tex]
[tex](\frac{183}{4})(\frac{1}{7})=x[/tex]
[tex](\frac{183}{28})=x[/tex]
[tex]\boxed{x=\frac{183}{28}}[/tex]
The Solution Is:
[tex]\boxed{\frac{183}{28},\frac{211}{4},\frac{85}{4}}[/tex]
I hope this helped!
-Benjamin
