Linear Algebra Problem Solving

Question

A group of 14 people spent P560.00 for admission tickets to a certain show. Gate A charges P50.00 for adults, P30.00 for students and P20.00 for children. If the same people could have entered through Gate B which charges P80.00 for adults, P40.00 for students and P20.00 for children, they could have spent P840.00. How many adults, students and children are there?

Answer 1

gates adult student kid total price

Gate A 50 30 20 560

Gate B 80 40 20 840

let us say number of adults is x , students is y and kids is z

putting in equation

50x + 30y + 20z = 560

80x + 40y + 20z = 840

solving equations 1 and 2 ( subtract first one from the second ) we get

30x + 10y = 280

eliminating z we get

50x + 30y = 560

80x + 40y = 840

to eliminate y multiply first equation by 4 and second eq by 3

4(50x+30y) = 560x4

3(80x+40y) = 840x3

= > 240x + 120y = 2520

200x + 120y = 2240

40x = 280

x = 7

solving for y

30x +10y = 280

30x7 +10y = 280

10y = 280 - 210 = 70

y = 7

solving for z

50x+30y+20z = 560

50x7 + 30 x 7 + 20z = 560

350 + 210 + 20z = 560

560 +20z= 560

z = 0

so the number of adults = 7, students =7 and kids = 0