A linear equation is any equation where the highest power of the variable is 1. No x², no x³ — just x (or y, or whatever letter you pick). The word "linear" comes from the fact that if you plot these equations on a graph, they make straight lines. For the exam, you don't need to worry about graphs — you need to solve them fast.
Think of it like a weighing scale (तराजू). Whatever you do to one side, you do to the other side. The equation stays balanced. Your only job is to isolate the variable on one side.
One variable: 3x + 7 = 22 — one unknown, one equation, one answer.
Two variables (simultaneous equations): 2x + y = 10 and x + 3y = 12 — two unknowns, two equations, unique solution.
Here is the real-world analogy that makes this click: Imagine you run a tea stall. Chai costs ₹10 and samosa costs ₹5. A customer pays ₹45. Another pays ₹35. Two transactions (two equations), two unknowns (how many chai, how many samosa) — that is a simultaneous equation problem.
In UP Police Constable papers, linear equations appear as:
3x - 5 = 13"0.2x + 0.04 = 0.5x - 0.11"The word problem type is the most common. The real skill is not solving the equation — it is translating the Hindi or English sentence correctly into an equation. Once you have the equation written, a 12th-pass student can solve it in under 30 seconds.
The standard form is ax + b = c where a ≠ 0.
Rule: Move all x terms to the left, all constants to the right. Then divide.
Example: 3.5x - 1.5x = 9.25 - 2.75
Step 1 — Collect like terms on each side: 2x = 6.5
Step 2 — Divide both sides by 2: x = 3.25
That is it. Two steps. The UP Police paper frequently uses decimal coefficients to slow you down — but the method is identical to integer coefficients. Multiply both sides by 10 or 100 to clear decimals if that is faster for you.
Decimal-clearing trick: For 0.2x + 0.04 = 0.5x - 0.11, multiply every term by 100:
20x + 4 = 50x - 11
Now it is clean integers. Rearrange: 4 + 11 = 50x - 20x → 15 = 30x → x = 0.5.
You need exactly two independent equations to solve for two unknowns. Two methods work:
Best when one equation is easy to rearrange (i.e., one variable has coefficient 1).
Step 1 — From the simpler equation, express one variable in terms of the other. Step 2 — Plug that expression into the second equation. Step 3 — Solve the resulting one-variable equation. Step 4 — Back-substitute to find the first variable.
Example: 2c + t = 170 and 5c + 4t = 530
From equation 1: t = 170 - 2c
Substitute into equation 2: 5c + 4(170 - 2c) = 530
5c + 680 - 8c = 530
-3c = -150
c = 50
Then t = 170 - 2(50) = 70.
Best when both equations have the same coefficient for one variable (or you can make it so by multiplying).
Step 1 — Multiply one or both equations so that one variable has the same coefficient in both. Step 2 — Add or subtract the equations to eliminate that variable. Step 3 — Solve for the remaining variable.
Example: Same problem above.
Multiply equation 1 by 4: 8c + 4t = 680
Subtract equation 2: (8c + 4t) - (5c + 4t) = 680 - 530
3c = 150
c = 50
For two-variable word problems, elimination tends to be faster when both equations have similar-looking coefficients. Substitution is faster when one variable is already isolated or has coefficient 1.
Sometimes the problem gives you a ratio instead of a full equation. For example: y/x = 4/3 means y = 4x/3. Substitute this directly into the second equation. You will end up with a one-variable equation.
This is where most candidates lose marks — not in the algebra, but in the translation. Here is a systematic approach:
x = correct answers, let y = mistakes.y = 2x.x + y = 48.Watch for trigger phrases:
M/3 - E/2 = 30When a problem gives you y/x = a/b, immediately write y = (a/b)x and substitute. You skip one full equation and reduce a two-variable problem to one variable in a single step.
Worked example: y/x = 4/3 and 7x + 6y = 29.
Write y = 4x/3. Plug in: 7x + 6(4x/3) = 29 → 7x + 8x = 29 → 15x = 29 → x = 29/15.
Then y = 4/3 × 29/15 = 116/45 ≈ 2.58 ≈ 2.5.
Standard method (treat as full simultaneous system): 5-6 steps. This method: 3 steps.
Any equation with decimals can be converted to integers instantly. Count the maximum decimal places in any coefficient, then multiply every term by 10^n.
0.2x + 0.04 = 0.5x - 0.11 — maximum 2 decimal places, so multiply by 100:
20x + 4 = 50x - 11
Now you have integers and the problem becomes trivial.
Standard approach (working with 0.2, 0.04 etc.): error-prone, ~45s. After multiplying by 100: ~15s, zero rounding errors.
When the options are neat numbers, substitute directly into both original equations instead of solving algebraically. Check which option satisfies both simultaneously.
Example (chairs and tables): Options give c = 50, 40, 60, 70. Plug c = 50 into 2c + t = 170: t = 70. Check 5(50) + 4(70) = 250 + 280 = 530. Correct.
You verified in ~20 seconds without solving the simultaneous system. Full algebraic method: ~60s.
Use this when: the answer options are specific numbers (not expressions), and the equations are easy to evaluate.
For any equation of the form ax + b = cx + d, the one move you need is: collect x terms left, constants right.
3.5x - 1.5x = 9.25 - 2.75 → 2x = 6.5 → x = 3.25
The pattern: (coefficient of x on left) - (coefficient of x on right) = (constant on right) - (constant on left). Write it as a formula in your head: (a - c)x = (d - b). This cuts out intermediate rearrangement steps — 4 steps reduced to 2.
After finding your answer, spend 5 seconds plugging it back into the original equation. If both sides match, you are done. If not, you caught an error before marking the wrong option — zero extra time wasted.
Example: x = 0.5. Check: 0.2(0.5) + 0.04 = 0.10 + 0.04 = 0.14. Right side: 0.5(0.5) - 0.11 = 0.25 - 0.11 = 0.14. Match confirmed.
This takes ~5s and saves you from a misread. Do it for every linear equation question.
In the exam hall, look at the question and make one decision first:
Is it one variable or two?
One variable (ax + b = cx + d form):
10ⁿ to clear them.x terms left, constants right.Two variables (word problem or pair of equations):
Word problem translation checklist:
Total time budget per question: 60-90 seconds for one-variable, 90-120 seconds for two-variable word problems.
Why this question: Tests your ability to set up and solve a one-variable equation from a real-world context — the most common linear equation format in UP Police papers.
Solving path: Let correct answers = x. Mistakes = 2x (twice as many). Total attempted = x + 2x = 3x = 48. So x = 16. Done in 2 steps.
Why this question: Decimal linear equations appear consistently. Candidates lose time trying to work with 0.2, 0.04 etc. The decimal-clearing trick turns this into a trivial problem.
Solving path: Multiply by 100: 20x + 4 = 50x - 11. Rearrange: 15 = 30x. So x = 0.5. Verify: LHS = 0.1 + 0.04 = 0.14, RHS = 0.25 - 0.11 = 0.14. Match.
Why this question: Tests a two-equation simultaneous system from a word problem — the hardest linear equation format in UP Police papers. Also tests equation-building from English text.
Solving path: Let chair = c, table = t. Equations: 2c + t = 170 and 5c + 4t = 530. From equation 1: t = 170 - 2c. Substitute: 5c + 4(170 - 2c) = 530 → 5c + 680 - 8c = 530 → -3c = -150 → c = 50.
Why this question: The ratio-to-substitution pattern. y/x is given, not a full second equation. Most candidates try to treat this as two separate equations and waste time. The fast path is immediate substitution.
Solving path: y/x = 4/3 → y = 4x/3. Substitute: 7x + 6(4x/3) = 29 → 7x + 8x = 29 → 15x = 29 → x = 29/15. Then y = 4/3 × 29/15 = 116/45 ≈ 2.58. Closest option is 2.5.
Why this question: Tests equation-building from a two-condition word problem involving fractions — "a third of" and "a half of" are common triggers that candidates misread.
Solving path: Let English = E, Maths = M. Total: M + E = 240. Condition: M/3 = E/2 + 30 → M = 3E/2 + 90. Substitute: 3E/2 + 90 + E = 240 → 5E/2 = 150 → E = 60.
Flipping the "exceeds by" equation. "A exceeds B by 30" means A - B = 30, not B - A = 30. Get the direction wrong and your equation produces a wrong answer that still looks like a clean number.
Not clearing decimals before rearranging. Working with 0.3x = 0.15 is fine, but most candidates make arithmetic errors under time pressure. Multiply by 10 or 100 first — it takes 3 seconds and prevents miscalculation.
Treating a ratio as a second equation incorrectly. When y/x = 4/3, some candidates write two equations: y = 4k and x = 3k. This works but adds an extra variable k and extra steps. Just write y = 4x/3 directly.
Sign errors when moving terms across the equals sign. 3x - 7 = x + 5 becomes 3x - x = 5 + 7, not 5 - 7. The sign flips when you cross the equality. Write it out — do not do this in your head under pressure.
Substituting into the same equation you derived from. When using substitution, derive t from equation 1 and substitute into equation 2. If you substitute back into equation 1, you get 0 = 0 — a true but useless result.
Not verifying the answer when options are close decimals. Questions like the y/x = 4/3 problem give options 2.2, 2.5, 2.4, 2.3 — all within 0.3 of each other. If you round at an intermediate step, you may land on the wrong option. Keep fractions exact until the final step, then compare with options.