Linear Equations in One Variable - 8th math
NCERT Exercise 2.6 Solution
Solve the following equations
Question (1) 
Solution
Given 
After cross multiplication, we get
⇒ 8x – 3 = 3x × 2
⇒ 8x – 3 = 6x
After transposing 6x to LHS we get:
⇒ 8x – 3 – 6x = 0
⇒ 2x – 3 = 0
After transposing –3 to RHS, we get
⇒ 2x = 3
After transposing 2 to RHS, we get
`x=3/2` Answer
Checking of Result
Given,
As calculated we have x = 3/2
Thus, after substituting the value of x in LHS, we get
= 2 = RHS Proved
Question (2) 
Solution
Given, 
After cross multiplication, we get
9x = (7 – 6x) × 15
⇒ 9x = 105 – 90 x
After transposing 90x to LHS, we get
⇒ 9x + 90x = 105
⇒ 99x = 105
After transposing 99 to RHS, we get
`x=105/99`
⇒ `x=35/33` Answer
Checking of Result
We have,
After substituting the value of x = 35/33, in LHS, we get
= 15 = RHS Proved
Question (3) 
Solution
Given, 
After cross multiplication, we get
9z = 4 (z + 15)
⇒ 9z = 4z + 60
After transposing 4z to LHS, we get
⇒ 9z – 4z = 60
⇒ 5z = 60
After transposing 5 to RHS, we get
z = 60/5 = 12
Thus, z = 12 Answer
Checking of Result
Given,
After substituting the value of z = 12 in LHS, we get
= 4/9 = RHS Proved
Question (4) 
Solution
Given, 
After cross multiplication, we get
⇒ 5(3y + 4) = –2(2 – 6y)
⇒ 15y + 20 = –4 + 12y
After transposing 12y to RHS, we get
⇒ 15y + 20 – 12y = –4
After transposing 20 to RHS, we get
⇒ 15y – 12y = –4 –20
⇒ 3y = –24
After transposing 3 to RHS, we get
⇒ y = –24/3 = –8
Thus, y = –8 Answer
Checking of Result
Given,
After substituting the value of y = –8 in LHS, we get
= –2/5 = RHS Proved
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