NCERT Solutions Class 8th Math Chapter – 2 एक चर वाले रैखिक समीकरण (Linear Equations in One Variable) प्रश्नावली 2.5

NCERT Solutions Class 8th Math Chapter – 2 एक चर वाले रैखिक समीकरण (Linear Equations in One Variable) प्रश्नावली 2.5

TextbookNCERT
Class 8th
Subject गणित (Mathematics)
Chapter2nd
Chapter Name एक चर वाले रैखिक समीकरण (Linear Equations in One Variable)
CategoryClass 8th गणित
Medium Hindi
SourceLast Doubt

NCERT Solutions Class 8th Math Chapter – 2 एक चर वाले रैखिक समीकरण (Linear Equations in One Variable) प्रश्नावली 2.5 जसमें हम समीकरणों,  एक चर वाले रैखिक समीकरण, रैखिक, समीकरण, LHS, RHS, LHS = RHS, आदि के बारे में हम बिस्तार से पढ़ेंगे।

NCERT Solutions Class 8th Math Chapter – 2 एक चर वाले रैखिक समीकरण (Linear Equations in One Variable) प्रश्नावली 2.5

Chapter – 2

एक चर वाले रैखिक समीकरण

प्रश्नावली 2.5

निम्नलिखित रैखिक समीकरणों को हल कीजिए।

1. x/2 – 1/5 = x/3 + 1/4
हल:
x/2 – 1/5 = x/3 +
x/2 – x/3 = ¼+ 1/5
(3x – 2x) /6 = (5 + 4)/20
3x – 2x = 9/20 × 6
⇒ x = 54/20
⇒ x = 27/10
2. n/2 – 3n/4 + 5n/6 = 21
हल:
n/2 – 3n/4 + 5n/6 = 21
(6n – 9n + 10n)/12 = 21
7n/12 = 21
⇒ 7n = 21 x 12
n = 252/7
n = 36
3. x + 7 – 8x/3 = 17/6 – 5x/2
हल:
x + 7 – 8x/3 = 17/6 – 5x/2
x – 8x/3 + 5x/2 = 17/6 – 7
⇒ (6x – 16x + 15x)/6 = (17 – 42)/6
⇒ 5x/6 = – 25/6
⇒ 5x = – 25
⇒ x = – 5
4. (x – 5)/3 = (x – 3)/5
हल:
(x – 5)/3 = (x – 3)/5
⇒ 5(x-5) = 3(x-3)
⇒ 5x -25 = 3x-9
⇒ 5x – 3x = -9+25
⇒ 2x = 16
⇒ x = 8
5. (3t – 2) / 4 – (2t + 3) / 3 = 2/3 – t
हल:
(3t – 2) / 4 – (2t + 3) / 3 = 2/3 – t
⇒ (3t – – 2) / 4) × 12 – (2t + 3) / 3) × 12 3
(3t – 2) × 3 – (2t + 3) × 4 = 2 × 4 – 12t
⇒ 9t – 6 – 8t – 12 = 8 – 12t
9t – 6 – 8t – 12 = 8 – 12t
t – 18 = 8 – 12t
t + 12t = 8 + 18
⇒ 13t = 26
⇒ t = 2
6. m – (m – 1)/2 = 1 – (m – 2)/3
हल:
m – (m – 1)/2 = 1 – (m – 2)/3
m – m/2 – 1 /2 = 1 – (m/3 – 2/3)
m – m/2 + ½ = 1 – m/3 + 2/3
m – m/2 + m/3 = 1 + 2/3 – ½
⇒ m/2 + m/3 = ½ + 2/3
(3m + 2m)/6 = (3 + 4)/6
⇒ 5m/6 = 7/6
⇒ m = 7/6 × 6/5
m = 7/5
7. 3(t – 3) = 5(2t + 1)
हल:
3(t – 3) = 5(2t + 1)
⇒ 3t – 9 = 10t + 5
⇒ 3t – 10t = 5 + 9
⇒ -7t = 14
t = 14/-7
t = -2

निम्न समीकरणों को सरल रूप में बदलते हुए हल कीजिए।

8.  15 (y – 4) -2 (y – 9) + 5 (y + 6) = 0
हल:
15 (y – 4) -2 (y – 9) + 5 (y + 6) = 0
15y – 60 -2y + 18 + 5y + 30 = 0
15y – 2y + 5y = 60 – 18 – 30
⇒ 18y = 12
y = 12/18
⇒ y = 2/3
9. 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
हल:
3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
⇒ 15z – 21 – 18z + 22 = 32z – 52 – 17
15z – 18z – 32z = -52 – 17 + 21 – 22
⇒ -35z = -70
z = -70/-35
⇒ z = 2
10. 0.25(4f – 3) = 0.05(10f – 9)
हल:
0.25(4f – 3) = 0.05(10f – 9) ⇒
f – 0.75 = 0.5f – 0.45
f – 0.5f = -0.45 + 0.75
0.5f = 0.30
f = 0.30/0.5
f = 3/5
⇒ f = 0.6

NCERT Solution Class 8th Math all Chapter in Hindi Medium