Linear Literal Equations and Simultaneous Linear Literal Equations Revision Notes for VCE SSCE Mathematical Methods

Linear Literal Equations and Simultaneous Linear Literal Equations

What are literal equations?

When we solve a literal equation for $x$ , our answer contains pronumerals (letters representing unknown values) rather than specific numbers. This makes literal equations different from the numerical equations you might be more familiar with.

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Numerical vs Literal Equations

For the numerical equation $2x + 5 = 7$ , we get a specific number as our answer: $x = 1$ .

For the literal equation $ax + b = c$ , we get an answer expressed in terms of pronumerals:

x = \frac{c - b}{a}

The key difference is that literal equations contain letters in their solutions rather than numerical values.

The good news is that we solve literal equations using exactly the same algebraic techniques as numerical equations. We transpose the equation to make $x$ the subject, following the same rules of algebra you already know.

Solving linear literal equations

Let's work through some examples to see how this works in practice.

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Worked Example: Solving for x

Solve the following equations for $x$ :

Part a: $px - q = r$

Add $q$ to both sides:

px = r + q

Divide both sides by $p$ :

x = \frac{r + q}{p}

Part b: $ax + b = cx + d$

Collect the terms containing $x$ on the left side:

ax - cx = d - b

Factorise the left side:

x(a - c) = d - b

Divide both sides by $(a - c)$ :

x = \frac{d - b}{a - c}

Part c: $\frac{a}{x} = \frac{b}{2x} + c$

Multiply every term by $2x$ to eliminate the fractions:

2a = b + 2xc

Subtract $b$ from both sides:

2a - b = 2xc

Divide both sides by $2c$ :

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

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Exam Tip: Working with Fractions

When solving literal equations with fractions, multiply through by the lowest common denominator to clear the fractions first. This makes the algebra much easier to manage.

Simultaneous linear literal equations

Just as we can solve pairs of simultaneous equations with numerical coefficients, we can also solve simultaneous literal equations. We use the same two methods you're already familiar with: substitution and elimination.

Worked example using substitution

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Worked Example: Substitution Method

Solve the following simultaneous equations for $x$ and $y$ :

y = ax + c

y = bx + d

Since both expressions equal $y$ , we can equate them:

ax + c = bx + d

Collect the $x$ terms on the left:

ax - bx = d - c

Factorise:

x(a - b) = d - c

Therefore:

x = \frac{d - c}{a - b}

Now substitute this value of $x$ into the first equation to find $y$ :

y = a\left(\frac{d - c}{a - b}\right) + c

Expand the bracket:

y = \frac{ad - ac}{a - b} + c

Express $c$ as a fraction with denominator $(a - b)$ :

y = \frac{ad - ac}{a - b} + \frac{c(a - b)}{a - b}

y = \frac{ad - ac + ac - bc}{a - b}

Simplify:

y = \frac{ad - bc}{a - b}

Worked example using elimination

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Worked Example: Elimination Method

Solve the simultaneous equations $ax - y = c$ and $x + by = d$ for $x$ and $y$ .

Label the equations:

ax - y = c \text{ ... (1)}

x + by = d \text{ ... (2)}

To eliminate $y$ , multiply equation (1) by $b$ :

abx - by = bc \text{ ... (1')}

Add equations (1') and (2):

abx - by + x + by = bc + d

The $y$ terms cancel:

abx + x = bc + d

Factorise the left side:

x(ab + 1) = bc + d

Therefore:

x = \frac{bc + d}{ab + 1}

Now substitute this value into equation (1) to find $y$ . From equation (1), we can write:

y = ax - c

Substitute our expression for $x$ :

y = a\left(\frac{bc + d}{ab + 1}\right) - c

y = \frac{abc + ad}{ab + 1} - c

Express $c$ with the common denominator:

y = \frac{abc + ad}{ab + 1} - \frac{c(ab + 1)}{ab + 1}

y = \frac{abc + ad - abc - c}{ab + 1}

y = \frac{ad - c}{ab + 1}

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Exam Tip: Choosing Your Method

When solving simultaneous literal equations, choose the method that seems most straightforward for the given equations. If one variable already has a coefficient of 1, elimination might be easiest. If an equation is already solved for one variable, substitution is often quicker.

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Key Points to Remember:

A literal equation is one where the solution is expressed using pronumerals rather than numbers.
Literal equations are solved using exactly the same algebraic techniques as numerical equations - transpose to make the required variable the subject.
For simultaneous literal equations, use either the substitution method or the elimination method, just as you would with numerical equations.
When working with fractions in literal equations, multiply through by the lowest common denominator to simplify your work.
Always factorise when you have multiple terms containing your variable - this allows you to isolate and solve for that variable.

Linear Literal Equations and Simultaneous Linear Literal Equations (VCE SSCE Mathematical Methods): Revision Notes