Agroup of children, adults, and senior citizens attended three different art exhibits that have different ticket prices for each age group. let c represent the number of children, a represent the number of adults, and s represent the number of senior citizens. the system represents the cost of each type of ticket and the total cost of the tickets for all three exhibits. what are the number of children, adults, and seniors that attended these three exhibits? 6c + 7a + 2s = 990 3c + 3a + 4s = 430 2c + 4a + 4s = 480 a) c = 120, a = 30, s = 30 b) c = 20, a = 40, s = 70 c) c = 100, a = 50, s = 5 d) c = 100, a = 50, s = 20

 (D)  c = 100, a = 50, s = 20

Solution:

If you solve the equation by putting in the values given in the option you will deduce that only option d matches LHS to the RHS

Lets see how,

6c + 7a + 2s = 990                 where, c = 100, a = 50, s = 20

6(100) + 7(50) + 2(20) = 990

600 + 750 + 40 = 990

990 = 990 hence true

2c + 4a + 4s = 480                     where, c = 100, a = 50, s = 20

2 (100) + 4 ( 50) + 4 (20)  = 480

200 + 250 + 80  =  480

See for the other option too with the same steps done above  to have a more clear understanding :)

if x decreases by 4 then y increases by 10

step-by-step explanation:

$6.42

step-by-step explanation: