Linear Equations in One Variable - 8th math

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NCERT Solution Exercise 2.5

Solve the following linear equations.

Question (1)

linear equation ncert exercise 2.3 math

Solution

Given,

`x/2-1/5=x/3+1/4`

After transposing –1/5 to RHS, we get

`x/2=x/3+1/4+1/5`

After transposing x/3 to LHS, we get

linear equation ncert exercise 2.5_1a math

After cross multiplication, we get

20x = 9 × 6

⇒ 20x = 54

After transposing 20 to RHS, we get

linear equation ncert exercise 2.5_1b math

Checking of Result

Given,

`x/2-1/5=x/3+1/4`

Now, LHS `=x/2-1/5`

Now, after substituting the value of `x=27/10`, we get

linear equation ncert exercise 2.5_1c math

Thus, LHS `23/20`

Now, RHS = ` x/3+1/4`

Now, after substituting the value of `x=27/10` in RHS, we get

linear equation ncert exercise 2.5_1d math

Thus, RHS `23/20`

Thus, LHS = RHS Proved

Thus, `x=27/10` Answer

Question (2)

Solution:

Given

`n/2-(3n)/4+(5n)/6=21`

linear equation ncert exercise 2.5_2a math

After cross multiplication, we get

⇒ 7n = 21 × 12

⇒ 7n = 252

After transposing 7 to RHS, we get

`n=252/7`

⇒ n = 36

Checking of Result

Given

`n/2-(3n)/4+(5n)/6=21`

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

linear equation ncert exercise 2.5_2b math

= 18 – 27 + 30

= 21 = RHS

Thus, LHS = RHS Proved

Thus, n = 36 Answer

Question (3)

Solution

Given,

`x+7-(8x)/3=17/6-(5x)/2`

After transposing `-(5x)/2` to LHS, we get

`x+7-(8x)/3+(5x)/2=17/6`

Again after transposing 7 to RHS, we get

linear equation ncert exercise 2.5_3a math

After multiplying both sides by 6 we get

linear equation ncert exercise 2.5_3b math

⇒ 5x = –25

After transposing 5 to RHS, we get

`x=-25/5`

⇒ x = –5

Checking of Result

Given,

`x+7-(8x)/3=17/6-(5x)/2`

Now, LHS = `x+7-(8x)/3`

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

linear equation ncert exercise 2.5_3c math

Thus, LHS = `46/3`

Now, RHS = `17/6-(5x)/2`

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

linear equation ncert exercise 2.5_3d math

Thus, RHS = `46/3`

Thus, LHS = RHS Proved

Thus, x = –5 Answer

Question (4)

Solution

Given,

`(x-5)/3=(x-3)/5`

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,

`(x-5)/3=(x-3)/5`

We have LHS = `(x-5)/3`

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

`(8-5)/3`

`3/3=1`

Thus, LHS = 1

Now, RHS = ` (x-3)/5`

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

`(8-3)/5`

`5/5=1`

Thus, RHS = 1

Thus, LHS = RHS Proved

Question (5)

Solution:

Given,

After transposing –t to LHS, we get

linear equation ncert exercise 2.5_5b math

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

`(3t-2)/4-(2t+3)/3`

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

linear equation ncert exercise 2.5_5c math

Thus, LHS = `-4/3`

Now, RHS = `2/3-t`

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

`2/3-2`

`=(2-6)/3`

`=-4/3`

Thus, RHS `-4/3`

Thus, LHS = RHS Proved

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