Respuestas
Respuesta dada por:
1
a: number of tables with 6 seats
b: number of tables with 4 seats
so when cafe is full
ax6 + bx4 = 100 seats
then
3a + 2b = 50 seats.... (i)
also
a + b = 20 tables
a + b = 20
a = 20-b ...(ii)
(ii) in (i)
3(20-b)+2b=50
60-3b+2b=50
b=10
how a+b=20
then a = 10
there are 10 tables of 6 seats and 10 tables of 4 seats
b: number of tables with 4 seats
so when cafe is full
ax6 + bx4 = 100 seats
then
3a + 2b = 50 seats.... (i)
also
a + b = 20 tables
a + b = 20
a = 20-b ...(ii)
(ii) in (i)
3(20-b)+2b=50
60-3b+2b=50
b=10
how a+b=20
then a = 10
there are 10 tables of 6 seats and 10 tables of 4 seats
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