5.4 Substituting into Formulae, Linear Equations, and Number Patterns

Substituting into Real-World Word and Symbolic Formulae

LANTITE frequently presents a practical rule—either in words or in algebraic symbols—and asks you to evaluate the output for given inputs, or work backward from a known output to find an unknown input. Accurate substitution relies on two habits: wrapping substituted values (especially negative numbers and decimals) in brackets and applying the order of operations (BODMAS/BIMDAS).

Common formula structures include:

1. Linear cost models ($C = F + rn$): For a school excursion with fixed bus hire of $450 and student entry of $18.50 per student, the total cost for $n$ students is $C = 450 + 18.5n$. For $n = 36\text{ students}$, multiply before adding: $C = 450 + (18.5 \times 36) = 450 + 666 = 1{,}116$ ($1,116).
2. Temperature conversions ($F = 1.8C + 32$ and $C = \frac{5(F - 32)}{9}$): When substituting a sub-zero Celsius temperature such as $C = -5^\circ\text{C}$, preserve the negative sign: $F = 1.8(-5) + 32 = -9 + 32 = 23^\circ\text{F}$. Conversely, if $F = 68^\circ\text{F}$, subtract $32$ inside the brackets first: $C = \frac{5(68 - 32)}{9} = \frac{5 \times 36}{9} = 20^\circ\text{C}$.
3. Multi-variable and squared formulas (BMI, medication dosages, and stopping distance):
   • Body Mass Index: $\text{BMI} = \frac{m}{h^2}$, where $m$ is mass in $\text{kg}$ and $h$ is height in metres (convert $160\text{ cm}$ to $1.6\text{ m}$ first, then square $1.6^2 = 2.56$ before dividing: for $m = 64\text{ kg}$, $\text{BMI} = \frac{64}{2.56} = 25$).
   • Pediatric dosage (Young's Rule): $D = \frac{A \times Y}{Y + 12}$, where $A$ is the adult dose in $\text{mg}$ and $Y$ is the child's age in years. Always evaluate the entire denominator $(Y + 12)$ before dividing: for an $8\text{-year-old}$ child when the adult dose is $A = 250\text{ mg}$, $D = \frac{250 \times 8}{8 + 12} = \frac{2{,}000}{20} = 100\text{ mg}$.
   • Vehicle stopping distance: $d = 0.25v + 0.008v^2$. For $v = 50\text{ km/h}$, square $v$ before multiplying by $0.008$: $d = 0.25(50) + 0.008(2{,}500) = 12.5 + 20 = 32.5\text{ m}$.

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Forming and Solving Linear Equations ($ax + b = c$ and $ax + b = cx + d$)

To find an unknown input when the output or comparison condition is known, translate the word problem into a linear equation and isolate the pronumeral $n$ using inverse operations:

• One-target equations ($ax + b = c$): Subtract the fixed term $b$ from both sides first, then divide by the per-unit rate $a$: $x = \frac{c - b}{a}$.
• Equating two competing quotes ($ax + b = cx + d$): Collect the variable terms on one side and the fixed constants on the other: $(a - c)x = d - b$.

Worked Example 1: Comparing Two School Formal Venue Quotes

A Year 12 coordinator compares two venues for the graduation dinner:

• Venue A: $600 fixed room hire plus $12 per student ($C_A = 600 + 12n$).
• Venue B: $240 fixed room hire plus $20 per student ($C_B = 240 + 20n$).

For what number of students $n$ do both venues charge the exact same total price, and which venue is cheaper for $60\text{ students}$?

1. Set the two cost expressions equal:

$$600 + 12n = 240 + 20n$$

1. Subtract $12n$ and $240$ from both sides:

$$600 - 240 = 20n - 12n \implies 360 = 8n$$

1. Divide by $8$:

$$n = \frac{360}{8} = 45\text{ students}$$

At $n = 45$, both venues cost $600 + 12(45) = 1{,}140$ ($1,140). Notice how each extra student costs $8 less at Venue A ($12 vs. $20), which takes $360 \div 8 = 45\text{ students}$ to offset Venue A's $360 higher fixed fee. 4. Compare for $60\text{ students}$: Because $60 > 45$, Venue A is cheaper ($600 + 12(60) = 1{,}320$, or $1,320, vs. Venue B at $240 + 20(60) = 1{,}440$, or $1,440).

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Linear Inequalities and Context-Driven Rounding

When a school budget has an upper ceiling $K$, the linear model forms an inequality:

$$F + rn \le K \implies n \le \frac{K - F}{r}$$

Suppose an excursion costs $C = 450 + 18.5n$ and the total budget cannot exceed $2,000:

$$450 + 18.5n \le 2{,}000 \implies 18.5n \le 1{,}550 \implies n \le \frac{1{,}550}{18.5} \approx 83.78$$

Even though $83.78$ is closer to $84$ than $83$ under standard rounding, you must round down to $83\text{ students}$ because $84\text{ students}$ would cost $450 + 18.5(84) = 2{,}004$ ($2,004), violating the $2,000 budget limit.

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Linear Number Patterns and Tables of Values ($T_n = mn + c$)

A sequence or table of values is linear (arithmetic) when the output changes by a constant first difference $m$ every time the step number $n$ increases by $1$. The general rule for the $n\text{th}$ term is:

$$T_n = mn + c$$

where:

• $m$ is the constant step change ($T_2 - T_1$, or $\frac{\Delta T}{\Delta n}$ when table entries skip steps), and
• $c = T_1 - m$ is the starting constant at $n = 0$ (the zero term).

Critical Pattern Trap: Never plug the first term $T_1$ directly in as $c$ without subtracting $m$ first!

Worked Example 2: Hexagonal Reading Tables Pattern

A school library places hexagonal tables end-to-end in a single row. $1\text{ table}$ seats $6\text{ students}$, $2\text{ joined tables}$ seat $10\text{ students}$, and $3\text{ joined tables}$ seat $14\text{ students}$. Find the rule for the number of seats $S_n$ at $n$ joined tables, and determine how many tables are required to seat $50\text{ students}$.

1. Find the constant first difference $m$: Each new table adds $10 - 6 = 4\text{ seats}$, so $m = 4$.
2. Find the zero term $c$: Step back from $n = 1$ ($S_1 = 6$) to $n = 0$: $c = 6 - 4 = 2$ (physically representing the $2$ end seats at the far left and far right of the row). Thus:

$$S_n = 4n + 2$$

1. Solve for $S_n = 50$: $4n + 2 = 50 \implies 4n = 48 \implies n = 12\text{ tables}$.