Linear Equations in One Variable - 8th math

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

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

⇒ 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