Linear Equations and Inequalities: Question 2

Worked solution

Part (a): write the two equations

Each sale gives one equation. The total cost is (number of adult tickets) $\times a$ plus (number of child tickets) $\times c$.

• Friday: $7$ adult and $5$ child tickets for $109: $7a + 5c = 109. \qquad (1)$
• Saturday: $4$ adult and $9$ child tickets for $93: $4a + 9c = 93. \qquad (2)$

Part (b): solve the simultaneous equations

Method: elimination. Make the $c$-terms match. Multiply $(1)$ by $9$ and $(2)$ by $5$ so both have $45c$:

$9 \times (1): \quad 63a + 45c = 981, \qquad (3)$ $5 \times (2): \quad 20a + 45c = 465. \qquad (4)$

Subtract $(4)$ from $(3)$ to eliminate $c$:

$63a - 20a = 981 - 465 \;\Rightarrow\; 43a = 516 \;\Rightarrow\; a = \frac{516}{43} = 12.$

Substitute $a = 12$ into $(1)$:

$7(12) + 5c = 109 \;\Rightarrow\; 84 + 5c = 109 \;\Rightarrow\; 5c = 25 \;\Rightarrow\; c = 5.$

Check in $(2)$: $\;4(12) + 9(5) = 48 + 45 = 93$ ✓.

So an adult ticket costs $12 and a child ticket costs $5:

$\boxed{a = 12, \quad c = 5}$

Part (c): cost for the family

The family buys $2$ adult and $3$ child tickets:

$2a + 3c = 2(12) + 3(5) = 24 + 15 = 39.$

The family pays a total of $\boxed{39}$ dollars, i.e. $39.