Using and Transposing Formulas Revision Notes for VCE SSCE Mathematical Methods

Using and Transposing Formulas

What is a formula?

A formula is a mathematical relationship that uses symbols to connect two or more quantities. Formulas allow us to calculate unknown values when we know the values of other variables in the relationship.

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For example, the formula $A = \ell w$ expresses the relationship between the area of a rectangle and its dimensions. Here, $A$ represents area, $\ell$ represents length, and $w$ represents width.

The subject of a formula

The subject of a formula is the variable that we are calculating. It is typically written alone on the left-hand side of the equation. In the formula $A = \ell w$ , the variable $A$ is the subject because we can find its value by substituting known values for $\ell$ and $w$ .

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Worked Example: Finding the area of a rectangle

Find the area of a rectangle with length $\ell = 10$ cm and width $w = 4$ cm.

Solution:

A = \ell w

A = 10 \times 4

Substitute $\ell = 10$ and $w = 4$ .

A = 40 \text{ cm}^2

This example demonstrates basic substitution - replacing variables with their given numerical values to calculate the subject of the formula.

Transposing formulas

Sometimes we need to rearrange a formula to make a different variable the subject. This process is called transposing the formula. Transposing uses the same algebraic techniques you learned for solving linear equations - you perform inverse operations to both sides of the equation to isolate the desired variable.

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The key principle is maintaining balance: whatever operation you perform on one side of the equation, you must perform on the other side as well.

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Worked Example: Transposing a kinematic equation

Transpose the formula $v = u + at$ to make $a$ the subject.

Solution:

v = u + at

v - u = at

Subtract $u$ from both sides.

\frac{v - u}{t} = a

Divide both sides by $t$ .

Notice how we reversed the operations: $a$ was being multiplied by $t$ and then $u$ was being added, so we first subtracted $u$ , then divided by $t$ .

Evaluating unknowns: Two methods

When you need to find the value of a variable that is not currently the subject of the formula, you have two approaches:

Method 1: Substitute then solve - Replace all known variables with their values first, then solve the resulting equation for the unknown.
Method 2: Transpose then substitute - Rearrange the formula to make the unknown variable the subject first, then substitute the known values.

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Both methods will give you the same answer. Choose the method that feels most comfortable or that makes the calculation simpler.

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Worked Example: Comparing both methods

Evaluate $p$ if $2(p + q) - r = z$ , and $q = 2$ , $r = -3$ , and $z = 11$ .

Method 1: Substitute then solve

2(p + 2) - (-3) = 11

First substitute $q = 2$ , $r = -3$ , and $z = 11$ .

2p + 4 + 3 = 11

Then solve for $p$ .

2p = 4

p = 2

Method 2: Transpose then substitute

First, we rearrange the formula to make $p$ the subject:

2(p + q) - r = z

2(p + q) = z + r

p + q = \frac{z + r}{2}

p = \frac{z + r}{2} - q

Now substitute the known values:

p = \frac{11 + (-3)}{2} - 2

p = \frac{8}{2} - 2

p = 4 - 2

p = 2

Both methods produce the same answer. Method 1 can be quicker for simple calculations, while Method 2 gives you a general formula you can reuse.

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Worked Example: Path around a rectangular lawn

A path $x$ metres wide surrounds a rectangular lawn. The lawn is $\ell$ metres long and $b$ metres wide. The total area of the path is $A$ m².

a) Find $A$ in terms of $\ell$ , $b$ , and $x$ .

b) Find $b$ in terms of $\ell$ , $A$ , and $x$ .

c) Find the value of $b$ if $\ell = 6$ , $A = 72$ , and $x = 1.5$ .

Solution:

a) First, visualise the problem:

The overall dimensions are $(b + 2x)$ by $(\ell + 2x)$ because the path adds $x$ metres on each side.

The area of the path equals the total area minus the lawn area:

A = (b + 2x)(\ell + 2x) - b\ell

Expanding the brackets:

A = b\ell + 2x\ell + 2xb + 4x^2 - b\ell

The $b\ell$ terms cancel:

A = 2x\ell + 2xb + 4x^2

b) To make $b$ the subject, we need to isolate terms containing $b$ :

A - (2x\ell + 4x^2) = 2xb

Therefore:

b = \frac{A - (2x\ell + 4x^2)}{2x}

c) Substitute the given values into the expression for $b$ :

b = \frac{A - (2x\ell + 4x^2)}{2x}

b = \frac{72 - (2(1.5)(6) + 4(1.5)^2)}{2(1.5)}

b = \frac{72 - (18 + 9)}{3}

b = \frac{72 - 27}{3}

b = \frac{45}{3}

b = 15

Therefore, the width of the lawn is 15 metres.

Advanced transposing techniques

More complex formulas may involve square roots, fractions, or reciprocals. The key is to identify which operations have been applied to the variable you want to isolate, then undo those operations in reverse order.

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Worked Example: Transposing formulas with square roots and fractions

For each of the following, make $c$ the subject of the formula:

a) $e = \sqrt{3c - 7a}$

b) $\frac{1}{a} - \frac{1}{b} = \frac{1}{c - 2}$

Solution:

a) $e = \sqrt{3c - 7a}$

To eliminate the square root, square both sides of the equation:

e^2 = 3c - 7a

Now isolate $c$ :

3c = e^2 + 7a

c = \frac{e^2 + 7a}{3}

b) $\frac{1}{a} - \frac{1}{b} = \frac{1}{c - 2}$

First, establish a common denominator on the left-hand side:

\frac{b - a}{ab} = \frac{1}{c - 2}

Take the reciprocal of both sides (flip both fractions):

\frac{ab}{b - a} = c - 2

Therefore:

c = \frac{ab}{b - a} + 2

This example shows how reciprocals can be useful when dealing with fractions. When you have a fraction equal to another fraction, taking reciprocals of both sides can simplify the equation.

Using technology for complex transposing

While understanding the manual process of transposing formulas is essential, technology can help verify your work or assist with particularly complex rearrangements.

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Exam tip: CAS calculators can verify your work or help with particularly complex rearrangements.

On the TI-Nspire:

Use the solve() function from the Algebra menu to make a variable the subject of a formula.

On the Casio ClassPad:

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Use the solve function followed by the variable you want to isolate (e.g., , c). The expand function under Interactive > Transformation can produce neater answers.

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Key Points to Remember:

A formula is a mathematical relationship between two or more variables expressed using symbols
The subject of a formula is the variable being calculated, typically isolated on one side of the equation
Transposing means rearranging a formula to make a different variable the subject
When transposing, use inverse operations in reverse order: if something was added, subtract it; if something was multiplied, divide by it
You can evaluate unknowns using two methods: substitute then solve, or transpose then substitute - both give the same answer
For complex formulas involving square roots, square both sides; for fractions, consider using common denominators or reciprocals

Using and Transposing Formulas (VCE SSCE Mathematical Methods): Revision Notes