Theory & Concepts

SSC CGL Linear Equations Questions, Formulas & Tricks

Get comprehensive theory, expert shortcuts, and hand-picked practice questions for Linear Equations (1 & 2 variables) specifically designed for the SSC CGL 2025-26 pattern.

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25 min readDifficulty: Easy-Intermediate

Linear equations are the foundation of word problems in arithmetic (Ages, Time & Work, Speed). In SSC CGL, solving two-variable equations quickly and understanding the conditions for infinite or no solutions are guaranteed marks.

Learning path

  • One Variable Isolation
  • Two Variables (Elimination)
  • Conditions for Solutions
  • 10 Exam-Level Examples

1. Solving Linear Equations

Use these two primary methods to quickly find the values of unknown variables.

1-Variable (Isolation)

Move all terms with the variable to one side, and constants to the other. Change signs when crossing the equal sign.

3x+5=14  ⟹  3x=9  ⟹  x=33x + 5 = 14 \implies 3x = 9 \implies x = 3

2-Variable (Elimination)

Multiply the equations by constants to make the coefficients of one variable equal, then subtract or add to eliminate it.

2. Conditions for Solutions

For a system of two equations: a1x+b1y=c1a_1x + b_1y = c_1 and a2x+b2y=c2a_2x + b_2y = c_2.

The Coefficient Ratios

Check the ratio of coefficients to determine the graph lines:

1. Unique Solution (Intersecting Lines): a1a2≠b1b2\frac{a_1}{a_2} \neq \frac{b_1}{b_2}

2. No Solution (Parallel Lines): a1a2=b1b2≠c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}

3. Infinite Solutions (Coincident Lines): a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

Examiners love giving "No Solution" and "Infinite Solution" conditions and asking you to find the missing variable 'k'.

3. 10 Solved examples

Question 01Exam Pattern

Solve for \( x \): \( 3x + 5 = 14 \).

2
3
4
5
Correct answer: b) 3

Solution

Step 1: Isolate the term with \( x \) by moving 5 to the other side.
Step 2: \( 3x = 14 - 5 \).
Step 3: \( 3x = 9 \).
Final calculation: Divide by 3. \( x = 3 \).
Question 02Exam Pattern

Solve for \( x \): \( 2(x - 3) + 4x = 24 \).

5
6
4
7
Correct answer: a) 5

Solution

Step 1: Expand the bracket: \( 2x - 6 + 4x = 24 \).
Step 2: Combine like terms: \( 6x - 6 = 24 \).
Step 3: Move the constant to the right: \( 6x = 30 \).
Final calculation: \( x = 5 \).
Question 03Exam Pattern

Solve the system of equations: \( x + y = 10 \) and \( x - y = 4 \).

x=7, y=3
x=6, y=4
x=8, y=2
x=5, y=5
Correct answer: a) x=7, y=3

Solution

Step 1: Add the two equations together.
Step 2: \( (x + y) + (x - y) = 10 + 4 \).
Step 3: \( 2x = 14 \implies x = 7 \).
Final calculation: Substitute \( x = 7 \) into the first equation: \( 7 + y = 10 \implies y = 3 \).
Question 04Exam Pattern

Solve for \( x \) and \( y \): \( 2x + 3y = 12 \) and \( 3x - y = 7 \).

x=2, y=3
x=3, y=2
x=4, y=1
x=1, y=4
Correct answer: b) x=3, y=2

Solution

Step 1: Multiply the second equation by 3 to match the \( y \) coefficients: \( 9x - 3y = 21 \).
Step 2: Add this to the first equation: \( (2x + 3y) + (9x - 3y) = 12 + 21 \).
Step 3: \( 11x = 33 \implies x = 3 \).
Final calculation: Substitute \( x = 3 \) into \( 3x - y = 7 \). \( 9 - y = 7 \implies y = 2 \).
Question 05Exam Pattern

For what value of \( k \) will the system \( 2x + 3y = 5 \) and \( 4x + ky = 10 \) have infinite solutions?

4
5
6
8
Correct answer: c) 6

Solution

Step 1: For infinite solutions, lines must be coincident. Condition: \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
Step 2: Set up the ratios: \( \frac{2}{4} = \frac{3}{k} = \frac{5}{10} \).
Step 3: The ratio is \( \frac{1}{2} \).
Final calculation: \( \frac{3}{k} = \frac{1}{2} \implies k = 6 \).
Question 06Exam Pattern

Find the value of \( k \) for which \( 3x + 2y = 8 \) and \( 6x + ky = 11 \) has no solution.

2
4
6
8
Correct answer: b) 4

Solution

Step 1: For no solution, lines must be parallel. Condition: \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
Step 2: Set up the ratios: \( \frac{3}{6} = \frac{2}{k} \neq \frac{8}{11} \).
Step 3: Evaluate the left side: \( \frac{1}{2} = \frac{2}{k} \).
Final calculation: Cross multiply to find \( k \). \( k = 4 \).
Question 07Exam Pattern

The sum of two numbers is 25 and their difference is 9. Find the larger number.

17
16
18
15
Correct answer: a) 17

Solution

Step 1: Let the numbers be \( x \) and \( y \).
Step 2: Set up the equations: \( x + y = 25 \) and \( x - y = 9 \).
Step 3: Add the equations: \( 2x = 34 \).
Final calculation: \( x = 17 \). The numbers are 17 and 8. The larger is 17.
Question 08Exam Pattern

A father is 3 times as old as his son. After 10 years, he will be twice as old. Find the father's present age.

30
40
45
50
Correct answer: a) 30

Solution

Step 1: Let the son's present age be \( S \) and father's be \( F \). Eq 1: \( F = 3S \).
Step 2: After 10 years, \( F + 10 = 2(S + 10) \).
Step 3: Substitute \( F = 3S \) into Eq 2: \( 3S + 10 = 2S + 20 \).
Step 4: Solve for \( S \): \( 3S - 2S = 20 - 10 \implies S = 10 \).
Final calculation: Father's age \( F = 3(10) = 30 \).
Question 09Exam Pattern

A fraction becomes 1/2 when 1 is added to the numerator. It becomes 1/3 when 1 is added to the denominator. Find the fraction.

3/8
4/9
2/5
3/7
Correct answer: a) 3/8

Solution

Step 1: Let fraction be \( \frac{x}{y} \). Eq 1: \( \frac{x+1}{y} = \frac{1}{2} \implies 2x + 2 = y \).
Step 2: Eq 2: \( \frac{x}{y+1} = \frac{1}{3} \implies 3x = y + 1 \).
Step 3: Substitute \( y \) from Eq 1 into Eq 2: \( 3x = (2x + 2) + 1 \).
Step 4: Solve for \( x \): \( 3x = 2x + 3 \implies x = 3 \).
Final calculation: Substitute \( x = 3 \) into \( y = 2x + 2 \implies y = 8 \). Fraction is \( \frac{3}{8} \).
Question 010Exam Pattern

The cost of 3 pens and 2 pencils is 24. The cost of 2 pens and 3 pencils is 21. Find the cost of 1 pen.

5
6
4
7
Correct answer: b) 6

Solution

Step 1: Let pen = \( x \), pencil = \( y \). \( 3x + 2y = 24 \) and \( 2x + 3y = 21 \).
Step 2: Shortcut when coefficients are swapped: Add the equations. \( 5x + 5y = 45 \implies x + y = 9 \).
Step 3: Subtract the equations. \( (3x+2y) - (2x+3y) = 24 - 21 \implies x - y = 3 \).
Step 4: Now solve the simple system: \( x + y = 9 \) and \( x - y = 3 \).
Final calculation: Add them: \( 2x = 12 \implies x = 6 \). The cost of a pen is 6.

4. Strategy errors to avoid

!

The Sign Swap Oversight

The most common mistake is forgetting to change the sign when moving a term to the other side of the equal sign. \\( 2x + 5 = 10 \\) becomes \\( 2x = 10 - 5 \\), not \\( 2x = 10 + 5 \\). Slow down during the transposition step.

!

Confusing the Solution Ratios

In "No Solution" and "Infinite Solution" problems, students often mix up the constant ratio \\( c_1/c_2 \\). Remember: No Solution requires \\( a_1/a_2 = b_1/b_2 \neq c_1/c_2 \\). If the third ratio is also equal, you have Infinite Solutions.

Master the next modules

Quadratic EquationsPolynomials & FactorizationInequalities