Question 366881
Let x=set fee and y=hourly charge
x+3y=125----eq1
and
x+6y=200----eq2
subtract eq1 from eq2:
3y=75
y=$25 per hour----hourly charge
substitute y=$25 into eq1:
x+75=125 or
x=$50 set fee

CK
50+75=125
and
50+6*25=200
Hope this helps---ptaylor