Linear Equations in One Variable - 8th math

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NCERT Solution Exercise 2.3 part2


 ncert math exercise 2.3 part 2 Linear Equations in one varible

Question (6) 8x + 4 = 3 (x – 1) + 7

Solution:

Given, 8x + 4 = 3(x – 1) + 7

8x + 4 = 3x – 3 + 7

8x + 4 = 3x + 4

After transposing 4 to RHS, we get

⇒ 8x = 3x + 4 –4

After transposing 3x to LHS, we get

8x – 3x = 4 – 4

5x = 0

After transposing 5 to RHS, we get

⇒ x = 0/5 = 0

Thus, x = 0 Answer

CHECKING OF RESULT:

Given, 8x + 4 = 3x + 4

Now, LHS = 8x + 4

After substituting the value of x = 0 in LHS we get

8 × 0 + 4

⇒ LHS = 4

Now, RHS = 3x + 4

After substituting the value of x = 0 in RHS we get

3 × 0 + 4

⇒ RHS = 4

Thus, LHS = RHS proved.

Question (7) x = 4/5 (x + 10)

Solution:

Given, x = 4/5 (x + 10)

linear equation ncert math question 7

After transposing to LHS, we get

linear equation ncert math question 7a

After transposing 5 to RHS, we get, or by cross multiplication, we get

x = 8 × 5

⇒ x = 40 Answer

CHECKING OF RESULT:

Given, x = 4/5 (x + 10)

RHS 4/5(x + 10)

After substituting the value of x = 40 in RHS we get

= 4 × 10 = 40

⇒ RHS = 40

Now, since LHS = x = 40

Thus, LHS = RHS Proved

Question (8) linear equation ncert math question 8

Solution

Given, linear equation ncert math question 8a

After transposing 1 to RHS, we get

linear equation ncert math question 8b

Now, after transposing to LHS, we get

linear equation ncert math question 8c

After transposing 15 to RHS, we get

3x = 2 × 15

3x = 30

After transposing 3 to RHS, we get

x = 30/3

x = 10 Answer

CHECKING OF RESULT:

Given, linear equation ncert math question 8d

LHS

After substituting the value of x = 10 in LHS, we get

linear equation ncert math question 8e

LHS

Now, RHS

After substituting the value of x = 10 in RHS, we get

linear equation ncert math question 8f

RHS

Thus, LHS = RHS Proved

Question (9) 2y + 5/3 = 26/3 – y

Solution:

Given, 2y + 5/3 = 26/3 – y

After transposing 5/3 to RHS we get

2y = 26/3 – y – 5/3

After transposing –y to LHS, we get

linear equation ncert math question 9

3y = 7

After transposing 3 to RHS, we get

y = 7/3 Answer

CHECKING OF RESULT

Given, 2y + 5/3 = 26/3 – y

LHS

After substituting the value of y = 7/3 in LHS, we get

linear equation ncert math question 9a

Thus, LHS

Now, RHS

After substituting the value of y = 7/3 in RHS, we get

linear equation ncert math question 9b

Thus, RHS

Thus, LHS = RHS. Proved

Question (10) 3m = 5m – 8/5

Solution:

Given 3m = 5m – 8/5

After transposing 5m to LHS, we get

linear equation ncert math question 10

After cross multiplication, we get

2m × 5 = 8

⇒ 10m = 8

After transposing 10 to RHS, we get

m = 8/10

⇒ m = 4/5 Answer

CHECKING OF RESULT

Given, 3m = 5m – 8/5

Now, LHS = 3m

By substituting the value of m = 4/5 in LHS, we get

linear equation ncert math question 10a

Thus, LHS

Now, RHS

By substituting the value of m = 4/5 in RHS, we get

linear equation ncert math question 10b

Thus, RHS

Thus, LHS = RHS. Proved

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