Equations, Identities and Equivalence

Proving Algebraic Equivalence

Two algebraic expressions are equivalent if they produce the same value for every possible value of the variables. This equivalence is recorded using the identity symbol \( \equiv \). Proving an identity requires algebraic manipulation — typically expanding, factorising or simplifying one side (or both sides independently) until both sides are shown to be identical in form. The key discipline is to work on one side at a time rather than treating the statement as an equation to solve.

A related skill is finding unknown constants when two expressions are told to be equivalent: set up equations by comparing coefficients.

Method for Proving an Identity

Choose the more complex side. Apply algebraic operations (expanding, collecting like terms, factorising, combining fractions) until it equals the simpler side. End the proof with "= RHS" (or "= LHS") and a tick or "as required".

Do not cross-multiply or rearrange as if the identity were an equation — this assumes what you are trying to prove.

Avoid: starting with the identity as written and performing the same operation to both sides. This approach relies on the identity being true (circular reasoning). Instead, manipulate one side independently.

Worked Examples

Show that \( (x+4)^2 - (x-2)^2 \equiv 12(x+1) \).

Take the LHS (more complex) and expand:

\[ (x+4)^2 - (x-2)^2 = (x^2+8x+16) - (x^2-4x+4) \] \[ = x^2+8x+16-x^2+4x-4 = 12x+12 = 12(x+1) = \text{RHS} \checkmark \]
Show that \( \dfrac{x^2-9}{x-3} \equiv x+3 quad (x \neq 3) \).
\[ \frac{x^2-9}{x-3} = \frac{(x+3)(x-3)}{x-3} = x+3 = \text{RHS} \checkmark \]
Show that \( n^2 + 3n + 2 \equiv (n+1)(n+2) \).

Take the LHS and factorise:

\[ n^2+3n+2 \equiv (n+1)(n+2) = \text{RHS} \checkmark \]

Or expand the RHS: \( (n+1)(n+2) = n^2+3n+2 = \text{LHS} \checkmark \)

Show that \( (a+b)^2 - (a-b)^2 \equiv 4ab \).
\[ (a+b)^2 - (a-b)^2 = [(a+b)+(a-b)][(a+b)-(a-b)] quad \text{(difference of two squares)} \] \[ = [2a][2b] = 4ab = \text{RHS} \checkmark \]
Find the values of \( a \) and \( b \) such that \( ax^2 + bx - 6 \equiv (2x+3)(x-2) \).

Expand the RHS and compare coefficients:

\[ (2x+3)(x-2) = 2x^2-4x+3x-6 = 2x^2 - x - 6 \]

Comparing with \( ax^2 + bx - 6 \): \( a = 2, b = -1 \).

Find \( p \), \( q \) and \( r \) such that \( x^2 + 6x + 11 \equiv (x+p)^2 + r \).

Expand: \( (x+p)^2 + r = x^2 + 2px + p^2 + r \). Comparing coefficients:

\[ 2p = 6 \implies p = 3 \qquad p^2 + r = 11 \implies r = 11-9 = 2 \] \[ x^2 + 6x + 11 \equiv (x+3)^2 + 2 \]

 Key Takeaways

  • To prove an identity: manipulate the more complex side using algebraic operations until it equals the simpler side. Work on one side at a time.
  • Never treat the identity as an equation to rearrange — that approach assumes the truth of what you are proving.
  • Useful techniques: expanding brackets, collecting like terms, factorising (especially difference of two squares), combining fractions.
  • To find unknown constants: expand any brackets, then compare coefficients of like powers of \( x \) on both sides.
  • When simplifying a fraction, factor the numerator and denominator first — common factors then cancel cleanly.