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

Book page 1817 Updated on2026-09-19

Q1.
Write 5 equations whose solution is x = – 2.
Answer

Build each one by doing the same thing to both sides of x = − 2.

EquationHow it was builtCheck with x = − 2
x + 2 = 0Add 2 to both sides− 2 + 2 = 0 ✓
3x = − 6Multiply both sides by 33 × (− 2) = − 6 ✓
5x + 4 = − 6Multiply by 5, then add 4− 10 + 4 = − 6 ✓
x ÷ 4 = − 0.5Divide both sides by 4− 2 ÷ 4 = − 0.5 ✓
7x + 9 = 4x + 3Multiply by 7 and add 9; the same value also gives 4x + 3− 14 + 9 = − 5 and − 8 + 3 = − 5 ✓
Why it happens: the value x = − 2 satisfies x = − 2, and every operation applied to both sides carries that value forward. So you can make as many equations as you like, and they will all have − 2 as their solution.
Tip: to make the last kind, choose any two expressions that are equal at x = − 2. Here 7x + 9 and 4x + 3 both give − 5.
Q2.
Find the value of each unknown: (a) 2y = 60 (b) – 8 = 5x – 3 (c) – 53w = –15 (d) 13 – z = 8 (e) k + 8 = 12 – k (f) 7m = m – 3 (g) 3n = 10 + n
Answer

(a) y = 30   (b) x = − 1   (c) w = 15/53   (d) z = 5   (e) k = 2   (f) m = − 1/2   (g) n = 5

(a) 2y = 60 → y = 60 ÷ 2 = 30   Check: 2 × 30 = 60 ✓
(b) − 8 = 5x − 3
− 8 + 3 = 5x → − 5 = 5x → x = − 5 ÷ 5 = − 1
Check: 5(− 1) − 3 = − 5 − 3 = − 8 ✓
(c) − 53w = − 15
w = (− 15) ÷ (− 53) = 15/53
Check: − 53 × 15/53 = − 15 ✓
(d) 13 − z = 8
13 − 8 = z → z = 5   Check: 13 − 5 = 8 ✓
(e) k + 8 = 12 − k
k + k + 8 = 12  (add k to both sides)
2k = 12 − 8 = 4 → k = 2   Check: 2 + 8 = 10 and 12 − 2 = 10 ✓
(f) 7m = m − 3
7m − m = − 3 → 6m = − 3 → m = − 3 ÷ 6 = − 1/2
Check: 7 × (− 1/2) = − 3.5 and − 1/2 − 3 = − 3.5 ✓
(g) 3n = 10 + n
3n − n = 10 → 2n = 10 → n = 5   Check: 15 = 10 + 5 ✓
Why it happens: parts (e), (f) and (g) all have the unknown on both sides, so the first move is always the same — bring the unknown terms together by adding or subtracting one of them from both sides. In (b) the unknown happens to sit on the right; that changes nothing, since an equation reads equally well in either direction.
Tip: in (c) the answer is a fraction and that is perfectly fine. Dividing a negative by a negative gives a positive.
Q3.
I am a 3-digit number. My hundred's digit is 3 less than my ten's digit. My ten's digit is 3 less than my unit's digit. The sum of all the three digits is 15. Who am I?
Answer

The number is 258.

Let the units digit be u.

Tens digit = u − 3
Hundreds digit = (tens digit) − 3 = u − 3 − 3 = u − 6

The three digits add up to 15:
(u − 6) + (u − 3) + u = 15
3u − 9 = 15
3u = 15 + 9 = 24
u = 24 ÷ 3 = 8

Tens digit = 8 − 3 = 5    Hundreds digit = 8 − 6 = 2
Check: 2 + 5 + 8 = 15 ✓; 2 is 3 less than 5 ✓; 5 is 3 less than 8 ✓
Why it happens: naming the units digit is the smart choice, because the other two are described in terms of it. Had we named the hundreds digit h, we would get h, h + 3, h + 6 and the same equation 3h + 9 = 15, giving h = 2 — the same number.
Tip: always check that the digits are single digits between 0 and 9, and that the hundreds digit is not 0. Here 2, 5 and 8 pass both tests.
Q4.
The weight of a brick is 1 kg more than half its weight. What is the weight of the brick?
Answer

The brick weighs 2 kg.

Let the weight of the brick be w kg.
"1 kg more than half its weight" is w/2 + 1, and that equals w:
w = w/2 + 1
w − w/2 = 1  (subtract w/2 from both sides)
w/2 = 1
w = 1 × 2 = 2 kg
Check: half of 2 kg is 1 kg, and 1 kg more than that is 2 kg. ✓
Why it happens: a whole brick minus half a brick is half a brick. So that missing half must be exactly the 1 kg mentioned in the question — and a brick is two halves, so 2 kg. Picture a balance: on one pan the whole brick, on the other half a brick and a 1 kg weight. Remove half a brick from each pan and you are left with half a brick against 1 kg.
Tip: read such sentences slowly. "1 kg more than half its weight" is w/2 + 1, not (w + 1)/2.
Q5.
One quarter of a number increased by 9 gives the same number. What is the number?
Answer

The number is 12.

Let the number be x.
x/4 + 9 = x
9 = x − x/4  (subtract x/4 from both sides)
9 = 3x/4  (because x − x/4 = 4x/4 − x/4 = 3x/4)
9 × 4 = 3x → 36 = 3x
x = 36 ÷ 3 = 12
Check: one quarter of 12 is 3, and 3 + 9 = 12 ✓
Why it happens: the number is made up of four quarters. Taking one quarter away leaves three quarters, and those three quarters are worth the 9 that was added. So one quarter is 3, and the whole number is 4 × 3 = 12.
Q6.
Given 4k + 1 = 13, find the values of: (a) 8k + 2 (b) 4k (c) k (d) 4k – 1 (e) – k – 2
Answer

(a) 26   (b) 12   (c) 3   (d) 11   (e) − 5

From 4k + 1 = 13:   4k = 13 − 1 = 12, so k = 12 ÷ 4 = 3
ExpressionQuick wayValue
(a) 8k + 2= 2 × (4k + 1) = 2 × 1326
(b) 4k= 13 − 112
(c) k= 12 ÷ 43
(d) 4k − 1= 12 − 111
(e) − k − 2= − 3 − 2− 5
Why it happens: part (a) needs no value of k at all. 8k + 2 is exactly twice 4k + 1, so it is twice 13. Spotting a relationship like this is faster than solving, and it is the same idea used in Examples 1 and 2 — using what is given instead of recomputing.
Check it yourself: with k = 3, 8k + 2 = 24 + 2 = 26 ✓, 4k − 1 = 12 − 1 = 11 ✓, − k − 2 = − 3 − 2 = − 5 ✓
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