Solve the equation 3x + 2(x + 2) = 20 – (2x – 5). Check your resul

1.	Solve the equation 3x + 2(x + 2) = 20 – (2x – 5). Check your result.
Answer» The above question can be written as,= 3x + 2x + 4 = 20 – 2x + 5 Now, find the value of x by considering the above equation, = 5x + 4 = 20 – 2x + 5 Transposing 4 to RHS and it becomes -4 and -2x to LHS it becomes 2x = 5x + 2x = 20 – 4 + 5 = 7x = 25 – 4 = 7x = 21 Multiplying both side by (1/7) = 7x × (1/7) = 21 × (1/7) = x = 21/7 = x = 3 By substituting 3 in the place of x in given equation, we get LHS, = 5x + 4 = 5 × (3) + 4 = 15 + 4 = 19 RHS, = 20 – 2x + 5 = 20 – 2 × (3) + 5 = 20 – 6 + 5 = 25 – 6 = 19 By comparing LHS and RHS = 19 = 19 ∴ LHS = RHS Hence, the result is verified.

Solve the equation 3x + 2(x + 2) = 20 – (2x – 5). Check your result.

Answer»

The above question can be written as,= 3x + 2x + 4 = 20 – 2x + 5

Now, find the value of x by considering the above equation,

= 5x + 4 = 20 – 2x + 5

Transposing 4 to RHS and it becomes -4 and -2x to LHS it becomes 2x

= 5x + 2x = 20 – 4 + 5

= 7x = 25 – 4

= 7x = 21

Multiplying both side by (1/7)

= 7x × (1/7) = 21 × (1/7)

= x = 21/7

= x = 3

By substituting 3 in the place of x in given equation, we get

LHS,

= 5x + 4

= 5 × (3) + 4

= 15 + 4

= 19

RHS,

= 20 – 2x + 5

= 20 – 2 × (3) + 5

= 20 – 6 + 5

= 25 – 6

= 19

By comparing LHS and RHS

= 19 = 19

∴ LHS = RHS

Hence, the result is verified.