# 1.03 Equations arising from substitution

Lesson

A formula is like an equation. It usually has an equals sign and describes a relationship between variables. Formulas are often used to solve specific real world problems. They tend to have multiple variables represented by uppercase and lowercase letters.

As an example, the formula for the area of a trapezium is given by  $A=\frac{h}{2}\left(a+b\right)$A=h2(a+b).

This formula uses four variables $a$a, $b$b, $h$h and $A$A, where $A$A represents the area, $a$a and $b$b are the lengths of the parallel sides and $h$h is the perpendicular distance between the parallel sides.

Note that the plural of formula can be written as either formulas or formulae.

### The subject of a formula

All formulas have a single variable by itself on one side of the equals sign (usually the left-hand side). This is known as the subject of the formula. In the area of a trapezium example above, the variable $A$A is the subject.

In evaluating algebraic expressions, we substituted values for the variables in a formula and evaluated the subject.

### Solving equations resulting from substitution into formulae

In some situations we know the values for all of the variables in a formula, except for one, which is not the subject. To find this unknown, we substitute all known values into the formula, and simplify any calculations. This will result in an equation with one unknown variable, which we solve just like any other equation.

#### Worked examples

##### example 1

The perimeter $P$P of a rectangle with side lengths $a$a and $b$b, is given by the formula $P=2a+2b$P=2a+2b.

If $P=480$P=480 and $a=150$a=150, find the value of $b$b.

Think: Begin by replacing all known variables in the formula with their values. After simplifying, solve the resulting equation for $b$b.

Do:

 $P$P $=$= $2a+2b$2a+2b $480$480 $=$= $2\times150+2b$2×150+2b (Substitute $P=480$P=480 and $a=150$a=150) $480$480 $=$= $300+2b$300+2b (This is our equation to solve)

This is a linear equation, which we solve in the normal way:

 $480-300$480−300 $=$= $300+2b-300$300+2b−300 (Subtract $300$300 from both sides) $180$180 $=$= $2b$2b $\frac{180}{2}$1802​ $=$= $\frac{2b}{2}$2b2​ (Divide both sides by $2$2) $90$90 $=$= $b$b

So the shorter side length of the rectangle is $b=90$b=90 units.

##### example 2

The volume $V$V of a cylinder is given by the formula $V=\pi r^2h$V=πr2h, where $r$r is the radius of the cylinder and $h$h is its height.

If $V=1000$V=1000 and $r=4.5$r=4.5, find $h$h. Round your answer to two decimal places.

Think: Once again, we begin by replacing all known variables in the formula with their values. After simplifying, solve the resulting equation to find $h$h.

Do:

 $V$V $=$= $\pi r^2h$πr2h $1000$1000 $=$= $\pi\times4.5^2\times h$π×4.52×h (Substitute $r=4.5$r=4.5) $1000$1000 $=$= $20.25\pi h$20.25πh (This is our equation to solve) $\frac{1000}{20.25\pi}$100020.25π​ $=$= $\frac{20.25\pi h}{20.25\pi}$20.25πh20.25π​ (Divide both sides by $20.25\pi$20.25π) $15.719\ldots$15.719… $=$= $h$h $h$h $=$= $15.72$15.72 (2 d.p.)

Reflect: Notice that it is often easiest to keep $\pi$π as a symbol in our working, until the final step. In this way we avoid rounding errors in our calculations.

##### example 3

The formula for converting a temperature, $F$F, in degrees Fahrenheit to a temperature, $C$C, in degrees Celsius, is given by $C=\frac{5}{9}\left(F-32\right)$C=59(F32).

If $C=35$C=35, find $F$F.

Solution

 $C$C $=$= $\frac{5}{9}\left(F-32\right)$59​(F−32) $35$35 $=$= $\frac{5}{9}\left(F-32\right)$59​(F−32) (Substitute $C=35$C=35. This gives our equation to solve) $35\times9$35×9 $=$= $\frac{5}{9}\left(F-32\right)\times9$59​(F−32)×9 (Multiply both sides by $9$9) $315$315 $=$= $5\left(F-32\right)$5(F−32) $\frac{315}{5}$3155​ $=$= $\frac{5\left(F-32\right)}{5}$5(F−32)5​ (Divide both sides by $5$5) $63$63 $=$= $F-32$F−32 $63+32$63+32 $=$= $F-32+32$F−32+32 (Add $32$32 to both sides) $95$95 $=$= $F$F

So the temperature corresponding to $35$35$^\circ$°C is $95$95$^\circ$°F.

#### Practice questions

##### Question 1

The surface area of the rectangular prism is given by formula $S=2\left(ab+bh+ah\right)$S=2(ab+bh+ah).

Find the value of $h$h when $a=4$a=4, $b=8$b=8 and $S=304$S=304.

##### Question 2

Newton's second equation of motion is $s=ut+\frac{1}{2}at^2$s=ut+12at2.

Find the value of $a$a if $s=4000$s=4000, $u=300$u=300 and $t=20$t=20.

##### Question 3

The resistance of a parallel circuit is given by the formula $\frac{1}{a}=\frac{1}{b}+\frac{1}{c}$1a=1b+1c. Find the value of $b$b (correct to two decimal places) if $a=12$a=12 and $c=27$c=27.

### Outcomes

#### MS11-1

uses algebraic and graphical techniques to compare alternative solutions to contextual problems