Algebra at the Class 6–8 level is the formal language that turns word problems into solvable puzzles. The central idea is simple: an algebraic expression is a combination of constants, variables (like x, y, a), and arithmetic operations. A linear equation constrains that expression to equal a specific value, while a polynomial is any expression built from non-negative integer powers of a variable.
Think of variables as labelled boxes. You don't know what's inside the box, but you know the rules the box must obey. The equation 2x + 3 = 11 is just saying: "Whatever is in box x, doubling it and adding 3 gives 11." Algebra is the process of figuring out what's in the box.
Here's a useful three-tier picture that maps to the NCERT progression from Class 6 to 8:
3x² − 5xy + 2y. You can simplify (combine like terms) but you cannot "solve" for anything.ax + b = c) has exactly one solution. Two equations in two variables define a unique point (usually) in the coordinate plane.The CTET Paper II questions on this topic almost always fall into one of four patterns: (1) add or subtract multi-variable expressions, (2) substitute values and evaluate, (3) solve simultaneous linear equations, or (4) factor a polynomial and divide. Recognising which pattern you're in — within the first 10 seconds of reading the question — is the single most valuable skill you can develop here.
The golden rule: only like terms can be combined. Like terms share exactly the same variable-and-power combination. 3x²y and −7x²y are like terms. 3x²y and 3xy² are not — the powers are switched.
When a question asks "what must be subtracted from A to get B", the answer is A − B. Flip the sign of every term in B, then add to A. This is where most exam errors happen: students add instead of subtract, or flip only some signs.
Subtraction template:
A − B = A + (−B) → negate every term in B, then group like terms.
When the question gives a = 1, b = −3, c = −2 and asks you to evaluate a complex expression, break the expression into named chunks. Compute b² = 9, a² = 1, c² = 4 first and write them down. Then evaluate each bracket separately before combining. Skipping intermediate steps is the primary source of sign errors in multi-substitution problems.
Watch the negative sign rules:
b = −3 means b² = 9 (positive)c = −2 means −c = 2 (the negative sign outside flips)A single equation like 2x + 3y = 6 has infinitely many solutions — it's a line. To pin down a unique solution, you need a second equation (the system). The two main techniques:
Elimination method: Multiply equations by suitable constants so one variable's coefficients become equal (or equal-and-opposite). Add or subtract the equations to eliminate that variable. This is the faster technique for most CTET patterns.
Substitution method: Isolate one variable from one equation and substitute into the other. More useful when one equation is already simple (e.g., x = 2y + 1).
A third "trick" question type: the paper gives you two equations and asks for x + y or x − y directly, without asking for individual values. Try to construct x + y algebraically from the equations before solving fully — sometimes it's faster.
For 2x + 3y = 12 and 3x − 2y = 5: multiply the first by 2 and the second by 3, add to eliminate y. 4x + 6y = 24 and 9x − 6y = 15 add to give 13x = 39, so x = 3. Then y = 2, and x + y = 5. Direct.
Factor theorem: If p(x) is a polynomial and p(a) = 0, then (x − a) is a factor. This is the doorway to every cubic/quadratic factoring question.
For a polynomial p(x) = 3x³ + 8x² + kx − 4 where (x + 2) is a factor, plug in x = −2 (because x + 2 = 0 means x = −2). Set p(−2) = 0, solve for k, and then divide the full polynomial by (x + 2) using polynomial long division or the grouping method.
Polynomial division (speed method — grouping): For (2x² + 7x + 6) ÷ (x + 2), factor the numerator. Look for two numbers that multiply to 2 × 6 = 12 and add to 7. Those are 3 and 4. Split the middle term: 2x² + 3x + 4x + 6 = x(2x + 3) + 2(2x + 3) = (x + 2)(2x + 3). Cancel (x + 2). Answer: (2x + 3). Much faster than long division.
For a cubic x³ − px² + qx − r with roots α, β, γ:
α + β + γ = pαβ + βγ + αγ = qαβγ = rThese are the direct relationships between coefficients and roots. When roots are given, compute sum, sum-of-products-of-pairs, and product — then read off the coefficients. No solving needed.
When asked "what is subtracted from A to get B", write the formula immediately: answer = A − B. Mechanically change the sign of every single term in B, write them out, then group like terms. Do not rearrange in your head — write each term with its new sign on paper first.
Example: A = 5y − 13x − 8a, B = 11x − 16y + 7a. Negate B: −11x + 16y − 7a. Now A + (−B) = (5y + 16y) + (−13x − 11x) + (−8a − 7a) = 21y − 24x − 15a.
Standard method (mental rearrangement): ~60 seconds, high error rate. This write-every-negated-term method: ~25 seconds, near-zero sign errors.
For multi-variable evaluation problems, pre-compute all powers first. Write a mini table: a=1, b=−3, b²=9, c=−2, c²=4, a²=1. Then evaluate each bracket separately and write the result. Finally, combine.
This turns a 5-step mental chain into 3 short independent calculations. For the expression a(a+b²+c) + b²(a²+b²+c²) − c(a+b²), the three brackets give 8, 126, and −20 respectively. Handling −c(...) carefully: c = −2, so −c = 2? No — the expression is −c(...) = −(−2)(10) = +20. Total: 8 + 126 + 20 = 154.
Standard mental-chain method: ~90 seconds with sign-error risk. Bracket-by-bracket with pre-computed table: ~40 seconds.
For any quadratic ax² + bx + c, instead of the quadratic formula, find two numbers m and n such that m × n = a × c and m + n = b. Split the middle term, group in pairs, factor out common factors from each pair.
For 2x² + 7x + 6: target m × n = 12, m + n = 7. That's 3 and 4. Split: 2x² + 3x + 4x + 6. Group: x(2x+3) + 2(2x+3) = (x+2)(2x+3). Done in 4 steps.
Quadratic formula: 6–7 steps with square root computation. Split-middle method: 4 steps, no square roots, faster for integer-root quadratics. Applied here for the train-speed PYQ — saves at least 30 seconds.
When a polynomial question gives you the roots and asks for coefficients (or vice versa), use Vieta's formulas as a direct read-off — no algebra required.
For roots 1, 2, 3 in x³ − px² + qx − 6: sum = 1+2+3 = 6 = p. Sum-of-pairs = (1)(2)+(2)(3)+(1)(3) = 2+6+3 = 11 = q. Product = 6 (matches constant term). Answer: p=6, q=11 in about 20 seconds.
Solving three equations from substituting each root: ~3 minutes. Vieta's direct read-off: ~20 seconds. Use this whenever roots are explicitly stated.
For simultaneous equations 2x + 3y = 12 and 3x − 2y = 5, when asked for x + y rather than individual values, don't stop at finding x and y separately if you can construct the answer more directly. But if you must solve fully, always eliminate the variable with coefficients that are easiest to match. Here the y coefficients are 3 and −2 — LCM is 6. Multiply first equation by 2, second by 3, and add (the y terms cancel). This is 2 multiplications + 1 addition = 3 operations to isolate x.
Substitution from the first equation would require isolating y = (12 − 2x)/3 and substituting — that's 5–6 operations with fraction risk. Elimination: 3 operations, no fractions. Save ~30 seconds.
When you see an algebra question in the exam hall, run this decision tree in about 10 seconds:
Step 1 — Is there an equals sign?
Step 2 — How many variables and equations?
Step 3 — Does the question involve polynomials with degree ≥ 2?
k given a factor → Plug in the root, set to zero, solve for k.Step 4 — Does the answer involve a ratio, speed, or rate?
If you spend more than 90 seconds on any single algebra question, skip and return. These questions have clean integer answers — if your arithmetic is getting messy, you've likely made a sign error early.
Why this question: Tests whether you know the subtraction-from-expression setup and can handle three-variable sign management without collapsing.
Solving path: Identify the formula: answer = (5y − 13x − 8a) − (11x − 16y + 7a). Negate every term of the second expression: −11x + 16y − 7a. Add to the first: (5y + 16y) + (−13x − 11x) + (−8a − 7a) = 21y − 24x − 15a. Match with option C.
Why this question: Multi-variable substitution with negative values — the dominant source of sign errors in the algebra section.
Solving path: Pre-compute: b² = 9, a² = 1, c² = 4. Bracket 1: a(a + b² + c) = 1(1 + 9 − 2) = 8. Bracket 2: b²(a² + b² + c²) = 9(1 + 9 + 4) = 9(14) = 126. Bracket 3: c(a + b²) = −2(1 + 9) = −20. The expression is B1 + B2 − B3 = 8 + 126 − (−20) = 8 + 126 + 20 = 154.
Why this question: Polynomial addition and subtraction across three expressions, with the trap of a subtracted parenthesis that flips signs.
Solving path: P + Q first: collect xy terms: 12xy + 9xy = 21xy; y² terms: −10y² + 12y² = 2y²; x² terms: −18x² + 14x² = −4x². So P + Q = 21xy + 2y² − 4x². Now subtract R = 5y² − x² + xy: negate R → −5y² + x² − xy. Add: (21xy − xy) + (2y² − 5y²) + (−4x² + x²) = 20xy − 3y² − 3x². This matches option D.
Why this question: Applied algebra — speed-distance framed as polynomial division, which CTET Paper II tests by dressing up pure factoring as a word problem.
Solving path: Speed = Distance ÷ Time = (2x² + 7x + 6) ÷ (x + 2). Factor the numerator: find m × n = 12, m + n = 7 → m = 3, n = 4. Split: 2x² + 3x + 4x + 6 = x(2x + 3) + 2(2x + 3) = (2x + 3)(x + 2). Cancel (x + 2). Answer: (2x + 3) km/hr.
Why this question: Tests simultaneous equations using a "find x + y" framing — must solve both variables cleanly, then combine.
Solving path: Eliminate y. Multiply 2x + 3y = 12 by 2 → 4x + 6y = 24. Multiply 3x − 2y = 5 by 3 → 9x − 6y = 15. Add: 13x = 39 → x = 3. Substitute: 6 + 3y = 12 → y = 2. Therefore x + y = 5.
Forgetting to negate all terms when subtracting. When you subtract an expression in parentheses, every single term inside flips sign. Missing even one term gives a wrong answer. Write out −(...) explicitly before combining.
Confusing b² with (−b)² during substitution. If b = −3, then b² = (−3)² = 9, which is positive. Students sometimes write b² = −9. Pre-compute powers and write them down before substituting into the full expression.
Applying the factor theorem with the wrong root sign. If (x + 2) is a factor, the root is x = −2, not x = +2. Substituting the wrong sign gives a non-zero result and leads to an incorrect k.
Stopping at finding x without computing the asked quantity. Many questions ask for x + y, x − y, or some combined expression — not x alone. After solving the system, re-read the question before writing the answer.
Trying polynomial long division when split-middle-term is available. Long division works but takes 2–3 minutes. If the numerator is a quadratic or can be factored by grouping, factor first and cancel — it's 4 steps versus 10+.
Misapplying Vieta's sign conventions. For x³ − px² + qx − r, the sum of roots equals p (not −p). The signs alternate in the standard form. Write out x³ − (sum)x² + (sum-of-pairs)x − (product) = 0 as a template before reading off coefficients.