NCERT Solutions for Class 7th Maths Chapter 7 .2 Solving Equations Systematically — Figure it Out

Book page 1727 Updated on2026-09-19

Q1.

Solve these equations and check the solutions. (a) 3x – 10 = 35 (b) 5s = 3s (c) 3u – 7 = 2u + 3 (d) 4 (m + 6) – 8 = 2m – 4 (e) u/15 = 6

Answer

(a) x = 15 (b) s = 0 (c) u = 10 (d) m = −10 (e) u = 90

(a) 3x − 10 = 35
3x = 35 + 10 = 45 (remove the term −10)
x = 45 ÷ 3 = 15
Check: 3 × 15 − 10 = 45 − 10 = 35 ✓

(b) 5s = 3s
5s − 3s = 3s − 3s (subtract 3s from both sides)
2s = 0
s = 0 ÷ 2 = 0
Check: LHS = 5 × 0 = 0, RHS = 3 × 0 = 0 ✓

(c) 3u − 7 = 2u + 3
3u − 7 − 2u = 2u + 3 − 2u (subtract 2u from both sides)
u − 7 = 3
u = 3 + 7 = 10
Check: 3 × 10 − 7 = 23 and 2 × 10 + 3 = 23 ✓

(d) 4(m + 6) − 8 = 2m − 4
4m + 24 − 8 = 2m − 4 (open the bracket)
4m + 16 = 2m − 4
4m + 16 − 2m = 2m − 4 − 2m
2m + 16 = −4
2m = −4 − 16 = −20
m = −20 ÷ 2 = −10
Check: LHS = 4(−10 + 6) − 8 = 4 × (−4) − 8 = −24. RHS = 2 × (−10) − 4 = −24 ✓

(e) u ÷ 15 = 6
Remove the divisor 15 by multiplying both sides by 15:
u = 6 × 15 = 90
Check: 90 ÷ 15 = 6 ✓

Why it happens: notice how part (b) surprises people. "5s = 3s" looks impossible until you remember that s = 0 makes both sides zero. And in (d) the bracket must be opened first, because 4 multiplies the whole of (m + 6), not just the m.

Tip: in (e) the unknown is being divided, so you undo it by multiplying — the opposite of what you do when the unknown is multiplied.

Q2.

Frame an equation that has no solution. [Hint: 4 more than a number, and 5 more than a number can never be equal!]

Answer

One such equation is x + 4 = x + 5.

x + 4 = x + 5
Subtract x from both sides:
x + 4 − x = x + 5 − x
4 = 5 — which is false

No value of x can make this true, so the equation has no solution.

Some more equations with no solution:

2y + 1 = 2y + 9 (subtracting 2y gives 1 = 9)
3(t + 2) = 3t + 10 (this becomes 3t + 6 = 3t + 10, so 6 = 10)
5 − p = 8 − p (adding p gives 5 = 8)

Why it happens: in every one of these the unknown cancels completely, leaving a statement about numbers alone. If that statement is false, no value of the unknown can rescue it. Picture it as a balance: both pans hold the same sack, but one pan also has 4 kg and the other 5 kg. Whatever is in the sack, that scale can never balance.

Try This: what if the leftover statement is true? Try 2(x + 3) = 2x + 6. Cancelling gives 6 = 6, which is always true — so every number is a solution.

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