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Solve literal equations and simultaneous literal equations

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Solve literal equations and simultaneous literal equations

A literal equation contains parameters whose values may vary. Treat the parameters as constants while isolating the target variable, and state any condition needed for division

\[Ax+B=C\Rightarrow x=\frac{C-B}{A},\quad A\ne0\]
  1. 1Treat parameters as constants Letters other than the target variable stay as fixed quantities during the rearrangement \(\displaystyle 3x+2a=5b\)
  2. 2Apply operations to both sides Move a non-target term by performing the same inverse operation on each side \(\displaystyle 3x+2a=5b\Rightarrow3x=5b-2a\)
  3. 3Divide by the whole coefficient The complete coefficient of the target variable divides the entire other side \(\displaystyle 3x=5b-2a\Rightarrow x=\frac{5b-2a}{3}\)
  4. 4State non-zero restrictions A denominator or coefficient used for division must not be zero \(\displaystyle x=\frac{C-B}{A},\quad A\ne0\)
Year 12, VCE Units 3 & 4, Chapter 1, 4 difficulty levels, No calculatorBuild a worksheet

Solve literal equations and simultaneous literal equations

A literal equation contains parameters whose values may vary. Treat the parameters as constants while isolating the target variable, and state any condition needed for division

\[Ax+B=C\Rightarrow x=\frac{C-B}{A},\quad A\ne0\]
  1. Treat parameters as constants Letters other than the target variable stay as fixed quantities during the rearrangement \(\displaystyle 3x+2a=5b\)
  2. Apply operations to both sides Move a non-target term by performing the same inverse operation on each side \(\displaystyle 3x+2a=5b\Rightarrow3x=5b-2a\)
  3. Divide by the whole coefficient The complete coefficient of the target variable divides the entire other side \(\displaystyle 3x=5b-2a\Rightarrow x=\frac{5b-2a}{3}\)
  4. State non-zero restrictions A denominator or coefficient used for division must not be zero \(\displaystyle x=\frac{C-B}{A},\quad A\ne0\)

Worked example

Make x the subject of the literal equation, assuming p is non-zero

\[px+2q=3r,\quad p\ne0\]
  1. \(\displaystyle px+2q=3r\)The target variable x is multiplied by p and the non-target term is 2q
  2. \(\displaystyle px=3r-2q\)Subtract 2q from both sides
  3. \(\displaystyle x=\frac{3r-2q}{p},\quad p\ne0\)Divide the whole right side by p and state the division restriction
  4. \(\displaystyle p\left(\frac{3r-2q}{p}\right)+2q=3r\)Substitution into the original equation verifies the rearranged formula when p is non-zero

Common mistakes

Cancelling across a sum
✕\(\displaystyle 3x+2a=5b\Rightarrow x=\frac{5b}{3}\)✓\(\displaystyle 3x+2a=5b\Rightarrow x=\frac{5b-2a}{3}\)
The term 2a is part of a sum and must be subtracted from the entire other side
Applying an operation to one side
✕\(\displaystyle 3x+2a=5b\Rightarrow3x=5b\)✓\(\displaystyle 3x+2a=5b\Rightarrow3x=5b-2a\)
Subtracting 2a must preserve equality by changing both sides of the equation
Ignoring a denominator restriction
✕\(\displaystyle x=\frac{c}{a-b}\)✓\(\displaystyle x=\frac{c}{a-b},\quad a-b\ne0\)
The expression is defined only when the denominator used for division is non-zero

Difficulty levels

introRearrange a single literal equation with one bracketed or fractional term
standardSolve a single equation with the target on both sides or solve a pair with one equation ready
challengeSolve a pair with symbolic coefficients and state the denominator restrictions
examSelect the correct solved pair and fill one labelled coefficient

Questions about this topic

What is a literal equation?+

It is an equation containing letters that represent quantities or parameters. A target letter is rearranged in terms of the other letters

Why must a non-zero restriction be stated?+

Division by zero is undefined. If a letter expression is used as a denominator, it must be stated as non-zero

How are simultaneous literal equations solved?+

Eliminate one variable or solve one equation for it, substitute into the other equation, then simplify both symbolic solutions and check them in the original equations

Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 1: Solving equations

In this chapter

Factorise, solve and reconstruct a rational quadraticSolve a quadratic exactly with the quadratic formulaSolve an equation that reduces to quadratic formDetermine parameter values from a discriminant conditionDivide polynomials and state quotient and remainderApply the remainder and factor theoremsMove between roots, factors, polynomial rules and equationsDetermine parameters in a polynomial identityUse unit-circle coordinates to evaluate circular functionsEvaluate exact circular-function values by symmetry and periodicityUse exact trigonometric identities and quadrant informationSolve circular equations over finite domainsGeneral solutions of circular equations and interval extractionLocate and bisect a circular-equation rootRearrange literal equations and recognise equivalent line formsSolve a pair of simultaneous linear equationsSolve a linear system with three or four unknownsVerify coordinate incidence and determine a parameterSolve literal equations and simultaneous literal equationsClassify and parameterise linear systems