NCERT Solutions for Class 7th Maths Chapter 7 .2 Solving Equations Systematically — Math Talk

Book page 1687 Updated on2026-09-19

Q1.

Can this equation have any other solution? [The equation 2n + 1 = 99, solved by trial and error to give n = 49.]

Answer

No. n = 49 is the only solution.

If n = 48, LHS = 2 × 48 + 1 = 97 (too small)
If n = 49, LHS = 2 × 49 + 1 = 99 (correct)
If n = 50, LHS = 2 × 50 + 1 = 101 (too big)

Why it happens: every time n increases by 1, the LHS increases by exactly 2 — it never goes back down. So the LHS passes through 99 once and only once. Solving it systematically shows the same thing: 2n + 1 = 99 forces 2n = 98, and 98 has only one half, namely 49.

Tip: not every equation behaves like this. x + 4 = x + 5 has no solution, and 5s = 3s has exactly one (s = 0). Equations you will meet later can have more than one. It is always worth asking the question.

Q2.

Try solving 5x – 4 = 7 using trial and error.

Answer

Trial and error works, but this equation shows why it is a poor tool — the answer is not a whole number.

Try x =	LHS = 5x − 4	Compare with RHS = 7
1	5 − 4 = 1	too small
2	10 − 4 = 6	a little too small
3	15 − 4 = 11	too big
2.1	10.5 − 4 = 6.5	still small
2.2	11 − 4 = 7	correct

So x = 2.2 = 11/5
Check: 5 × (11/5) − 4 = 11 − 4 = 7 ✓

Why it happens: the answer lies between 2 and 3, so whole-number guessing can never reach it. You have to start guessing decimals, and there are infinitely many of those. The systematic method has no such trouble: 5x − 4 = 7 gives 5x = 11 and then x = 11/5 straight away.

Tip: use trial and error to understand an equation and to check an answer. Use the balance method to find the answer.

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