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Linear Equations

From a single unknown to a pair of equations — the fast, error-free methods examiners reward in CDS & OTA Maths.

13 min read Graduate / CDS level Exam-ready notes By The Cavalier
🎯 What you'll learn
  • Solve linear equations in one and two variables accurately
  • Pick the fastest method: substitution, elimination or cross-multiplication
  • Translate word problems into equations without errors
  • Judge whether a pair of equations has one, none or infinite solutions

Linear equations are the backbone of CDS Algebra. Almost every paper carries questions on equations in one or two variables, often disguised as age, money, or train-fare word problems. Master substitution, elimination and cross-multiplication, and you can crack these in under a minute. This Cavalier guide builds the topic from first principles to exam-ready speed.

What a Linear Equation Really Is

An equation is a statement that two expressions are equal. It becomes linear when the highest power of every variable is exactly one — no squares, no products of two variables, no variable under a root.

The general form in one variable is ax + b = 0, where a ≠ 0. In two variables it is ax + by + c = 0. Because the power is one, the graph of such an equation is always a perfectly straight line — that is why we call it “linear”.

Key point

One variable: ax + b = 0 → one unique solution, x = −b/a.
Two variables: ax + by + c = 0 → needs a second equation to pin down a unique (x, y).

So 3x − 7 = 0 is linear, and 2x + 5y = 12 is linear. But x2 + 3 = 0 and xy = 6 are not linear, because a variable carries power 2 (or two variables multiply together). The same idea holds for three or more variables: as long as every power is one, the equation stays linear.

In CDS, you are rarely asked to draw the line; you are asked to find the value of the unknown that makes the statement true. That value is called the solution or root of the equation. Checking a candidate value is simple — put it back in and see whether both sides become equal.

Solving an Equation in One Variable

The golden rule: whatever you do to one side, do to the other. Add, subtract, multiply or divide both sides equally and the balance stays true. Your goal is to isolate the variable on one side.

  1. Remove brackets and clear fractions (multiply through by the LCM of denominators).
  2. Bring all variable terms to one side, all constants to the other — a term changes sign when it crosses the equals sign (transposition).
  3. Combine like terms.
  4. Divide by the coefficient of the variable.
Worked example

Solve: 3(x + 2) − 5 = 2x + 7

3x + 6 − 5 = 2x + 7 3x + 1 = 2x + 7 3x − 2x = 7 − 1 x = 6
Exam tip

Always substitute your answer back into the original equation. Plugging x = 6 above gives 3(8) − 5 = 19 and 2(6) + 7 = 19. Both sides match — the answer is confirmed in five seconds.

Equations With Fractions and Decimals

Fractions slow people down and cause sign slips. The cure is to clear them at the very start by multiplying every term by the LCM of all denominators.

Worked example

Solve: (x/2) + (x/3) = 10

LCM of 2 and 3 = 6 Multiply each term by 6: 3x + 2x = 60 5x = 60 x = 12

For decimals, multiply through by 10, 100 or 1000 to make every coefficient a whole number. So 0.2x + 0.05 = 1.05 becomes 20x + 5 = 105 (multiply by 100), giving x = 5.

Common mistake

When you multiply by the LCM, multiply every single term, including the constant on the right. Forgetting the lone number on the other side is the most common slip in fraction problems.

A Pair of Linear Equations in Two Variables

A single equation like 2x + y = 7 has endless solutions — (1,5), (2,3), (3,1) and so on. To fix a unique pair (x, y) you need a second independent equation. Together they form a simultaneous (or pair of) linear equations.

Geometrically each equation is a line. The solution is the single point where the two lines cross. CDS gives you three standard tools to find that point: substitution, elimination, and cross-multiplication.

Remember

Two variables need two equations; three variables need three equations. In general, you need as many independent equations as you have unknowns.

Method 1: Substitution

Substitution works best when one variable already has a coefficient of 1, so it is easy to express in terms of the other.

  1. From one equation, write one variable in terms of the other.
  2. Substitute that expression into the second equation — now only one variable remains.
  3. Solve for it, then back-substitute to find the second variable.
Worked example

Solve: x + y = 9 and 2x − y = 3

From the first: y = 9 − x Substitute into the second: 2x − (9 − x) = 3 2x − 9 + x = 3 3x = 12 → x = 4 y = 9 − 4 = 5

So x = 4, y = 5. Check: 4 + 5 = 9 and 2(4) − 5 = 3. Both hold.

Method 2: Elimination

Elimination is usually the fastest method in the exam. You make the coefficient of one variable equal in both equations, then add or subtract to cancel it.

  1. Multiply one or both equations so a chosen variable has the same coefficient.
  2. Add the equations if the signs are opposite; subtract if they are the same. That variable vanishes.
  3. Solve the single remaining equation, then back-substitute.
Worked example

Solve: 3x + 2y = 16 and 5x − 2y = 8

The +2y and −2y already match in size. Add the two equations: (3x + 5x) + (2y − 2y) = 16 + 8 8x = 24 → x = 3 Put x = 3 in 3x + 2y = 16: 9 + 2y = 16 → y = 3.5
Exam tip

Scan the two equations first. If a variable already has equal or opposite coefficients (like ±2y above), elimination needs no extra multiplication — that is your quickest route.

Method 3: Cross-Multiplication

For the standard pair a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, the solution comes from a single formula. It is handy when the numbers are awkward for elimination.

Key point

x / (b1c2 − b2c1) = y / (c1a2 − c2a1) = 1 / (a1b2 − a2b1)

Worked example

Solve: 2x + 3y − 13 = 0 and 3x − 2y − 0 = 0

a1=2, b1=3, c1=−13; a2=3, b2=−2, c2=0 x / (b1c2 − b2c1) = x / (0 − (−2)(−13)) = x / (−26) y / (c1a2 − c2a1) = y / ((−13)(3) − 0) = y / (−39) 1 / (a1b2 − a2b1) = 1 / ((2)(−2) − (3)(3)) = 1 / (−13) x = −26 / −13 = 2, y = −39 / −13 = 3

So x = 2, y = 3. Write the constant term carefully with its correct sign before applying the formula.

One Solution, None, or Infinitely Many

Before solving, you can predict how many solutions a pair has just by comparing coefficients. This is a favourite CDS one-mark question.

Key point

For a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0:

  • a1/a2 ≠ b1/b2unique solution (lines intersect).
  • a1/a2 = b1/b2 ≠ c1/c2no solution (parallel lines).
  • a1/a2 = b1/b2 = c1/c2infinite solutions (same line).

Example: 2x + 3y = 5 and 4x + 6y = 9. Here 2/4 = 3/6 = 1/2 but 5/9 ≠ 1/2, so the ratios match for a and b but not c — the lines are parallel and there is no solution.

Turning Word Problems Into Equations

Most CDS marks come from word problems, not bare equations. The skill is translation: name the unknowns, then convert each English statement into one equation.

  • “sum of two numbers is 50” → x + y = 50
  • “A is 6 more than B” → A = B + 6
  • “twice the smaller equals the larger” → 2s = l
  • “ages 5 years ago” → subtract 5 from each current age
Worked example

The sum of two numbers is 35 and their difference is 13. Find the numbers.

Let the numbers be x and y. x + y = 35 x − y = 13 Add: 2x = 48 → x = 24 y = 35 − 24 = 11
Common mistake

Read what the question actually asks. After solving you may have x and y, but the question might want their product, their ratio, or just the larger one. Answer the exact demand.

Age, Money and Two-Digit Number Problems

Three problem types recur in nearly every CDS paper. Learn their set-up once.

Age problems

If a present age is x, then n years ago it was (x − n) and n years hence it will be (x + n). Apply the same shift to every person mentioned.

Money / coin problems

Value = number of coins × denomination. So x coins of ₹2 and y coins of ₹5 give a total value of (2x + 5y) rupees.

Two-digit number problems

Key point

A two-digit number with tens digit t and units digit u equals 10t + u. Reversing the digits gives 10u + t.

Worked example

A father is 3 times as old as his son. After 12 years he will be twice as old. Find their present ages.

Let son = x, father = 3x. After 12 years: 3x + 12 = 2(x + 12) 3x + 12 = 2x + 24 x = 12 Son = 12 years, Father = 36 years

Previous-Year Style Question

Previous-year style question

Q. The sum of the digits of a two-digit number is 9. When the digits are reversed, the new number is 27 greater than the original number. What is the original number?

Answer: Let the tens digit be t and units digit u. Then t + u = 9, and the original number is 10t + u. The reversed number is 10u + t. Given (10u + t) − (10t + u) = 27 → 9u − 9t = 27 → u − t = 3. Solving u + t = 9 and u − t = 3 gives u = 6, t = 3. So the original number is 36 (check: reversed 63 = 36 + 27).

Exam tip

For “difference of digits” clues, the constant always works out to a multiple of 9, because reversing a two-digit number changes its value by 9 × (difference of digits). Spotting this saves time.

Quick Revision

60-second recap
  • Linear means highest power of each variable is 1; its graph is a straight line.
  • One variable: transpose terms, isolate x, divide by its coefficient.
  • Clear fractions and decimals first by multiplying through by the LCM.
  • Two variables need two equations: use substitution, elimination or cross-multiplication.
  • Compare a1/a2, b1/b2, c1/c2 to judge unique, no, or infinite solutions.
  • For word problems: name unknowns, translate each sentence, then answer the exact question asked.
  • Two-digit number = 10t + u; reversed = 10u + t.

Practise five mixed word problems daily and time yourself. With elimination as your default and cross-multiplication as backup, linear-equation questions become guaranteed marks in CDS and OTA Maths.

Frequently asked questions

Which method is fastest in the CDS exam?

Elimination is usually quickest, especially when a variable already has equal or opposite coefficients. Use substitution when one variable has a coefficient of 1, and cross-multiplication when the numbers are awkward.

How do I know if a pair of equations has no solution?

Compare the coefficient ratios. If a1/a2 = b1/b2 but this ratio is not equal to c1/c2, the lines are parallel and there is no solution.

What is the most common mistake with fraction equations?

Forgetting to multiply the constant term on the other side of the equals sign by the LCM. Always multiply every single term, not just the fractions.

How do I represent a two-digit number in equations?

A two-digit number with tens digit t and units digit u equals 10t + u. Reversing the digits gives 10u + t. This set-up solves almost all digit problems.

Can a linear equation have more than one solution?

A single linear equation in one variable has exactly one solution. A pair in two variables has one unique solution if the lines intersect, none if parallel, and infinitely many if they are the same line.

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