Exercise 122 Page 626

Let's introduce a few variables.

a c = price per adult = price per child

The first family bought tickets for

2

adults and

3

children and paid

$ 27.75 .

With this information, we can write an equation.

2 a + 3 c = 27.75

The second family bought tickets for

3

adults and

2

children, for which they paid

$ 32.25 .

With this information, we can write a second equation.

3 a + 2 c = 32.25

If we combine these equations, we get a system of equations which we can solve by using the Elimination Method.

{2 a + 3 c = 27.75 3 a + 2 c = 32.25 (I) (II)

▼

Solve by substitution

$(I):$ $LHS \cdot 2 = RHS \cdot 2$

{4 a + 6 c = 55.5 3 a + 2 c = 32.25

$(II):$ $LHS \cdot 3 = RHS \cdot 3$

{4 a + 6 c = 55.5 9 a + 6 c = 96.75

$(I):$ Subtract $(II)$

{4 a + 6 c - (9 a + 6 c) = 55.5 - 96.75 9 a + 6 c = 96.75

$(I):$ Distribute $- 1$

{4 a + 6 c - 9 a - 6 c = 55.5 - 96.75 9 a + 6 c = 96.75

$(I):$ Subtract terms

{- 5 a = - 41.25 9 a + 6 c = 96.75

$(I):$ $LHS / (- 5) = RHS / (- 5)$

{a = 8.25 9 a + 6 c = 96.75

Having solved for

a,

we can substitute this value into the second equation and solve for

c .

{a = 8.25 9 a + 6 c = 96.75

$(II):$ $a = 8.25$

{a = 8.25 9 (8.25) + 6 c = 96.75

▼

(II):

Solve for

c

$(II):$ Multiply

{a = 8.25 74.25 + 6 c = 96.75

$(II):$ $LHS - 74 = RHS - 74$

{a = 8.25 6 c + 0.25 = 22.75

$(II):$ $LHS - 0.25 = RHS - 0.25$

{a = 8.25 6 c = 22.5

$(II):$ $LHS / 6 = RHS / 6$

{a = 8.25 c = 3.75

An adult ticket costs

$ 8.25,

and a child's ticket costs

$ 3.75 .