Let a = adult price, c=child price, s= senior price,

(eqn1) 2a + 2c + 2s = $46

(eqn2) s + 2c = $19.50

(eqn3) a + s + 3c = $36 => 2a + 2s +6c = $72 (multiply both sides by 2)

So: 2x(eqn3) – (eqn1)

=> 2a + 2s +6c – 2a – 2c – 2s = $72 – $46

=> 4c = 26 => c = $26/4= $6.50

By (eqn2)

s = $19.50 – 2c = $19.50 – 2 x $6.50= $19.50 – $13.00 = $6.50

(eqn1) – (eqn2)

=> 2a + 2c + 2s – s – 2c = $46 – $19.50

=> 2a +s = $26.50

=> 2a = $26.50 – s = $26.50 – $6.50 = $20

=> a = $20 / 2 = $10

So Adults $10, Children $6.50 and Seniors $6.50

Check

(eqn1) 2 x $10 + 2 x $6.50 + 2 x $6.50 = $20 + $13 + $13 = $46

(eqn2) s + 2c = $6.50 + 2 x $$6.50 = $6.50 + $13.50 = $19.50

(eqn3) a + s + 3c = $10 + $6.50 + 3 x $6.50 = $10 + $6.50 + $19.50 = $36

From what the problem gives:

2a + 2s + 2c = 46

s + 2c = 19.5

a + s + 3c = 36

So now we have 3 variables and 3 equations (recall systems of equations with 3 variables). There’s several ways to go about solving them, the easiest probably being matrices (if you have a calculator) or by solving for each variable and substituting.

2s+2a+2c=46

2s+ 2(36-s-3c)+2c=46

2s+72-2s-6c+2c=46

72-4c=46

-4c=-26

c=6.5

s+2c=19.50

s+2(6.5)=19.5

s+13=19.5

s=6.5

2(6.5)+2a+2(6.5)=46

13+2a+13=46

2a=20

a=10