CDS Mathematics · Algebraic Identities and Simplification

Conditional Identities

When a question gives a relation between the letters, use it to rewrite each piece of the expression until the pieces match; the answer is usually a constant.

All Algebraic Identities and Simplification notes CDS Mathematics strategy

Why this matters

Sixteen PYQs, eleven of them HARD — the hardest page in the chapter by count. Almost all of them have a small constant answer (0, 1, −1, 2). That makes a numeric test the fastest check: pick numbers that satisfy the condition, evaluate, and strike out options.

Concept 1 of 3: Rewrite each piece with the condition

The condition is a machine for replacing one expression by another. Apply it to each denominator or bracket separately; the pieces usually turn into multiples of one common factor, which then cancels.

Definition

$ab + bc + ca = 0$ gives $bc = -a(b + c)$ , so $a^2 - bc = a(a + b + c)$ , and likewise for $b$ and $c$ .
$a + b = 2c$ gives $b - c = c - a = -(a - c)$ .
$x^2 = y + z$ gives $x^2 + x = x + y + z$ , so $\dfrac{1}{x + 1} = \dfrac{x}{x + y + z}$ .
A relation like $\left(x + \dfrac{1}{yz}\right) - \left(y + \dfrac{1}{zx}\right)$ regroups as $(x - y)\left(1 + \dfrac{1}{xyz}\right)$ .

Check the result on numbers that satisfy the condition, avoiding values that make a denominator zero.

A typical rewrite

ab + bc + ca = 0 \;\Rightarrow\; a^2 - bc = a(a + b + c)

Worked example

a + b = 2c

, find

\dfrac{a}{a - c} + \dfrac{b}{b - c}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2017 · CDS (II) 2017 — Elementary Mathematics · Q32Hard

Example 1 · Algebraic Identities and Simplification · Conditional Identities

ab + bc + ca = 0

, then what is the value of

\frac{a^2}{a^2 - bc} + \frac{b^2}{b^2 - ca} + \frac{c^2}{c^2 - ab}

Test with numbers that fit

When

ab + bc + ca = 0

a = b = 1

does not fit unless

c = -\dfrac12

. Build the test triple from the condition, then evaluate; a single wrong-looking value rules out options quickly.

Drill 7 more on rewrite each piece with the condition

Concept 2 of 3: Equal ratios

If several fractions are equal, each is also equal to the fraction you get by adding all the numerators and all the denominators. That gives the common value at once — unless the denominators add to zero, which is a separate case.

Definition

If $\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = k$ , then $k = \dfrac{a + c + e}{b + d + f}$ whenever $b + d + f \ne 0$ .
Write each numerator as $k$ times its denominator and add; if the added denominators are zero, solve that case directly.
$\dfrac{p + q}{q + r} = \dfrac{r + s}{s + p}$ : cross-multiply and factor, giving $(p - r)(p + q + r + s) = 0$ — either one or the other.

Adding equal ratios

\dfrac ab = \dfrac cd = \dfrac ef = k \;\Rightarrow\; k = \dfrac{a + c + e}{b + d + f}

Worked example

\dfrac{a + b}{c} = \dfrac{b + c}{a} = \dfrac{c + a}{b} = k

, find the possible values of

k

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2019 · CDS (II) 2019 — Elementary Mathematics · Q32Moderate

Example 2 · Algebraic Identities and Simplification · Conditional Identities

\frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b}

, then which one of the following statements is correct ?

Do not drop the zero-sum case

Adding numerators and denominators divides by their sum. When that sum can be zero, the ratio takes a second value, and the options usually include both the full answer and the half answer.

Drill 4 more on equal ratios

Concept 3 of 3: Factor first, then substitute

If a quadratic expression is given a value and a linear one is also known, the quadratic almost always factors with the linear one as a factor. Divide and the rest drops out.

Definition

Factor the expression so the given combination appears: $4p^2 + 4pq - 3q^2 = (2p + 3q)(2p - q)$ .
Differences of squares hide in products like $(u + v)(u - v)$ with $u$ , $v$ themselves sums.
Surds: if $x - 2 = u + u^2$ with $u^3 = 2$ , cube both sides; the $u^3$ terms become numbers.
Divide only by factors known to be non-zero, and say so.

Divide out the known factor

x^2 + xy - 2y^2 = (x + 2y)(x - y)

Worked example

x + 2y = 5

and

x^2 + xy - 2y^2 = 15

, find

x - y

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2016 · CDS (II) 2016 — Elementary Mathematics · Q31Moderate

Example 3 · Algebraic Identities and Simplification · Conditional Identities

2p + 3q = 12

and

4p^2 + 4pq - 3q^2 = 126

, then what is the value of

p + 2q

Dividing by something that could be zero

From

x^2(m - 1) = mab(m - 1)

you may cancel

m - 1

only if

m \ne 1

. From

(x - y)k = (y - z)k

you may not conclude

x - y = y - z

unless

k \ne 0

— and sometimes

k = 0

is the answer the question wants.

Drill 2 more on factor first, then substitute

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Rewrite each piece with the condition
A typical rewrite
$ab + bc + ca = 0 \;\Rightarrow\; a^2 - bc = a(a + b + c)$
Equal ratios
Adding equal ratios
$\dfrac ab = \dfrac cd = \dfrac ef = k \;\Rightarrow\; k = \dfrac{a + c + e}{b + d + f}$
Factor first, then substitute
Divide out the known factor
$x^2 + xy - 2y^2 = (x + 2y)(x - y)$

Watch out for (3)

Test with numbers that fit
→ Rewrite each piece with the condition
Do not drop the zero-sum case
→ Equal ratios
Dividing by something that could be zero
→ Factor first, then substitute

Test yourself on Algebraic Identities and Simplification

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