Question 885201
Typical rectangle dimensions using perimeter, problem.


w width
L length
2w+2L=750 and "the length exceeds by 5m", must really mean "The length is 5 meters more than THE WIDTH(or breadth)".  This would be L=5+w.


Solve the system:  2w+2L=750 AND L=w+5.


Simplify the perimeter equation:  w+L=375 
Substitute for L:  w+(w+5)=375
Solve that for w:   2w+5=375, 2w=370, w=185
Easily find L:    L=185+5, L=190.




SUMMARY:
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Breadth, w=185 meters
Length, L=190 meters
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