The above question can be written as,= 13y – 52 – 3y + 27 – 5y – 20 = 0
Now, find the value of y by considering the above equation,
= 13y – 3y – 5y – 52 + 27 – 20 = 0
= 13y – 8y – 72 + 27 = 0
= 5y – 45 = 0
Transposing -45 to RHS and it becomes 45
= 5y = 45
Multiplying both side by (1/5)
= 5y × (1/5) = 45 × (1/5)
= y = 9
By substituting 9 in the place of y in given equation, we get
LHS,
= 5y – 45
= 5 × (9) – 45
= 45 – 45
= 0
RHS,
= 0
By comparing LHS and RHS
= 0 = 0
∴ LHS = RHS
Hence, the result is verified.
