(a) 2y + \(\frac{5}{2} = \frac{37}{2}\)
Substitute the value of y = 8 in the given equation
(b) 5t + 28 = 10
The given equation is 5t + 28 = 10
Transposing 28 from LHS to RHS
5t = 10 – 28 [on transposing + 28 becomes -28]
5t = -18
Dividing by 5 both the sides
Substitute the value of t = \(\frac{-18}{5}\) in the given equation
5t + 28 = 10
5 x \(\frac{-18}{5}\) + 28 = 10
18 + 28 = 10
10 = 10
∴ LHS = RHS (verified)
(c) \(\frac{a}{5} + 3 = 2\)
The given equation is \(\frac{a}{5} + 3 = 2\)
Transposing + 3 from LHS to RHS
Substitute the value of a = -5 in the given equation.
(d) \(\frac{q}{4} + 7 = 5\)
Substitute the value of q = -8 in the given equation.
(e) \(\frac{5}{2}\)x = - 10
Substitute the value of x = -4 in the given equation.
∴ LHS = RHS (verified)
(f) \(\frac{5}{2}\)x = \(\frac{25}{4}\)
∴ LHS = RHS (verified)
(g) 7m + \(\frac{19}{2}\) = 13
Substitute the value of m = \(\frac{1}{2}\) in the given equation
(h) 6z + 10 = -2
The given equation is 6z + 10 = -2
Transposing + 10 from LHS to RHS
6z = -2 – 10(on Transposing + 10 becomes – 10)
6z = -12
Divide by 6 both the sides
Substitute the value of z = -2 in the given equation
6z + 10 = -2
6(-2) + 10 = -2
-12 + 10 = -2
-2 = -2
∴ LHS = RHS (verified)
(i) \(\frac{3l}{2} = \frac{2}{3}\)
Substitute the value of l = 
in the given equation.
∴ LHS = RHS (verified)
(j) \(\frac{2b}{3} - 5 = 3\)
Substitute the value of b = 12 in the given equation
