Vincent propses an idea to the student government to install a basket hoop along with a court marked with all the shooting lines and boundary lines at his school for students to use at recess. He presents a plan to install a half-court design as shown below. After checking with the school administration he is told it will be approved if it fits on the empty lot that measures 25 feet by 75 feet on the school property

I think your question is missed of key information, allow me to add in and hope it will fit the original one.  Please have a look at the attached photo.

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Scale Drawing Lengths:   .     .     / .  

Actual Court Lengths       .     .     .  

Scale Factor: inch corresponds to ( ∙ ) inches, or inches, so the scale factor is .  

Let = , represent the scale drawing lengths in inches, and represent the actual court lengths in inches. The -values must be converted from feet to inches.  

To find actual length:

= =

() = inches, or feet  

To find actual width:  

= = ( )  

= / ∙ /  

= inches, or feet  

The actual court measures feet by feet. Yes, the lot is big enough for the court Vincent planned. The court will take up the entire width of the lot.

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.

step-by-step explanation:

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