SSC CGL Preparation: Concepts and Practice

Learning outcome

Solve one-variable linear equations by equivalent operations, model a simple situation, and distinguish unique, absent and infinitely many solutions.

Concepts and assumptions

An equation asks which values make both sides equal. Unless the context requires positive lengths or another restriction, the domain here is all real numbers.

A one-variable linear equation has the form ax + b = 0 with a ≠ 0. More complicated-looking equations may simplify to this form. Others lose their variable entirely and need separate classification.

Adding or subtracting the same expression on both sides preserves equality. Multiplying or dividing both sides by the same nonzero constant is reversible and preserves the solution set. “Moving a term” is shorthand for such an operation.

After simplification, write Ax = B. If A ≠ 0, division gives the unique solution x = B/A. If A = 0 and B ≠ 0, no solution exists because zero cannot equal a nonzero number. If A = B = 0, every value in the original domain works.

Do not divide by an unknown expression without examining when it is zero; this can discard valid solutions. The nonzero constant denominators here exclude no x-values.

Clear numerical denominators using a common multiple, distribute brackets, collect variable terms and constants, and then check the answer in the original equation. Word problems also require a check against their physical domain.

Worked examples

Example 1 — Variables on both sides. Solve 5x − 7 = 3x + 9.

Subtract 3x: 2x − 7 = 9. Add 7: 2x = 16. Divide by 2: x = 8. Check: 5(8) − 7 = 33 and 3(8) + 9 = 33.

Example 2 — Fractional coefficients. Solve (x − 2)/3 + (x + 1)/4 = 5.

Multiply every term by 12: 4(x − 2) + 3(x + 1) = 60. Expand: 4x − 8 + 3x + 3 = 60. Thus 7x − 5 = 60, then 7x = 65. Therefore x = 65/7. Check: the two original fractions become 17/7 and 18/7; their sum is 35/7 = 5.

Example 3 — When the variable disappears. Classify both equations.

For 4(x − 2) = 4x − 5, expansion gives 4x − 8 = 4x − 5. Subtracting 4x leaves −8 = −5, which is false: no solution.

For 3(2x + 1) = 6x + 3, expansion gives identical sides. Subtraction leaves 0 = 0: infinitely many solutions, all real x.

Common mistakes

Changing only one side; forgetting to multiply every term when clearing fractions; dividing by zero; assuming variable cancellation always means no solution.

Practice questions

1. Solve 7x + 5 = 3x + 29.
2. A rectangle’s length exceeds its width by 5 cm, and its perimeter is 58 cm. Find both dimensions.
3. Solve or classify 5(x − 2) = 5x + 1.
4. Solve or classify 4(2x − 3) − 2x = 6x − 12.

Worked solutions

1. Subtract 3x and then 5: 4x = 24. Hence x = 6. Both original sides equal 47.
2. Let width be w cm, with w > 0; length is w + 5 cm. Then 2(w + w + 5) = 58, giving 4w + 10 = 58 and w = 12. Dimensions are 12 cm and 17 cm; 2(12 + 17) = 58.
3. Expanding gives 5x − 10 = 5x + 1. Subtraction leaves −10 = 1, a contradiction. Therefore no real solution.
4. The left side becomes 8x − 12 − 2x = 6x − 12, identical to the right side. Therefore every real x is a solution.