Linear Algebra and Linear Equation

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

Let a = adults, s=students and c = children. Three unknowns, need 3 equations.

a + s + c = 14

50a+30s+20c=560

80a+40s+20c=840

Divide the 2nd equation by 10 and the third equation by 20 and put into a matrix

[1 1 1 14]

[5 3 2 56]

[4 2 1 42]

R2 = R2-5R1

R3 = R3-4R1

[1 1 1 14]

[0 -2 -3 -14]

[0 -2 -3 -14]

R2 and R3 are redundant, so set R3 to all zeroes.

R2=-R2/2

[1 1 1 14]

[ 0 1 3/2 7]

[0 0 0 0]

c is a free variable. Let c = 2 and reverse solve

s + 3 = 7 --> s = 4

a +4 + 2 = 14 --> a = 8

8 adults, 4 students and 2 children is one solution

Checking:

50(8) + 30(4) + 20(2) = 400 + 120 + 40 =560 √

80(8) + 40(4) + 20(2) = 640 + 160 + 40 =840 √