🏡 Home
    1. Linear Equations in One Variable
    2. Understanding Quadrilaterals
    3. Practical Geometry
    4. Mensuration
    1. Math
    2. Chemistry
    3. Chemistry Hindi
    4. Biology
    5. Exemplar Solution
    1. 11th physics
    2. 11th physics-hindi
    1. Science 10th (English)
    2. Science 10th (Hindi)
    3. Mathematics
    4. Math (Hindi)
    5. Social Science
    1. Science (English)
    2. 9th-Science (Hindi)
    1. 8th-Science (English)
    2. 8th-Science (Hindi)
    3. 8th-math (English)
    4. 8th-math (Hindi)
    1. 7th Math
    2. 7th Math(Hindi)
    1. Sixth Science
    2. 6th Science(hindi)
    1. Five Science
    1. Science (English)
    2. Science (Hindi)
    1. Std 10 science
    2. Std 4 science
    3. Std two EVS
    4. Std two Math
    5. MCQs Math
    6. एमoसीoक्यूo गणित
    7. Civil Service
    1. General Math (Hindi version)
    1. About Us
    2. Contact Us
10upon10.com

Linear Equations in One Variable - 8th math

8th-math-home

8th-math hindi-home

NCERT Solution Exercise 2.5


Solve the following linear equations.

Question (1)

Solution

Given,

After transposing –1/5 to RHS, we get

After transposing x/3 to LHS, we get

After cross multiplication, we get

20x = 9 × 6

⇒ 20x = 54

After transposing 20 to RHS, we get

Checking of Result

Given,

Now, LHS

Now, after substituting the value of , we get

Thus, LHS

Now, RHS =

Now, after substituting the value of in RHS, we get

Thus, RHS

Thus, LHS = RHS Proved

Thus, Answer

Question (2)

Solution:

Given

After cross multiplication, we get

⇒ 7n = 21 × 12

⇒ 7n = 252

After transposing 7 to RHS, we get

⇒ n = 36

Checking of Result

Given

Now, after substituting the value of n = 36 in LHS, we get

= 18 – 27 + 30

= 21 = RHS

Thus, LHS = RHS Proved

Thus, n = 36 Answer

Question (3)

Solution

Given,

After transposing to LHS, we get

Again after transposing 7 to RHS, we get

After multiplying both sides by 6 we get

⇒ 5x = –25

After transposing 5 to RHS, we get

⇒ x = –5

Checking of Result

Given,

Now, LHS =

After substituting the value of x = –5 in this LHS, we get

Thus, LHS =

Now, RHS =

After substituting the value of x = –5 in this RHS, we get

Thus, RHS =

Thus, LHS = RHS Proved

Thus, x = –5 Answer

Question (4)

Solution

Given,

After cross multiplication, we get

5(x – 5) = 3(x – 3)

⇒ 5x – 25 = 3x – 9

After transposing –25 to RHS, we get

5x = 3x – 9 + 25

After transposing 3x to LHS, we get

5x – 3x = – 9 + 25

⇒ 2x = 16

After transposing 2 to RHS, we get

x = 16/2 = 8

⇒ x = 8 Answer

Checking of Result

Given,

We have LHS =

Now, after substituting the value of x = 8 in the LHS, we get

Thus, LHS = 1

Now, RHS =

Now, after substituting the value of x = 8 in the RHS, we get

Thus, RHS = 1

Thus, LHS = RHS Proved

Question (5)

Solution:

Given,

After transposing –t to LHS, we get

After cross multiplication, we get

3(13t – 18) = 2 × 12

⇒ 39t – 54 = 24

After transposing –54 to RHS, we get

39t = 24 + 54

⇒ 39t = 78

After transposing 39 to RHS, we get

t = 78/39 = 2

⇒ t = 2 Answer

Checking of Result

Given,

Now, LHS

After substituting the value of t = 2 in this LHS, we get

Thus, LHS =

Now, RHS =

After substituting the value of t = 2 in this LHS, we get

Thus, RHS

Thus, LHS = RHS Proved




Reference: