Question

In each of the following equations is the number given in the bracket a solution of the equation?

(1) 5y = 16 [8]
(2) $5=\frac{35}{x}\left[7\right]$
(3) $5=\frac{x}{35}\left[7\right]$
(4) 5m − 1 = 19 [4]
(5) 2p = p + 3 [2]
(6) 3t = 7t [0]

Answer 1

(1) If we put y = 8 in the given equation, we get:
L.H.S. = 5y
= 5 × 8
= 40
R.H.S. = 16
∵ 40 ≠ 16,
∴ L.H.S. ≠ R.H.S.
Hence, 8 is not the solution of the given equation.

(2) If we put x = 7 in the given equation, we get:
$$\begin{gathered} \mathrm{L}.\mathrm{H}.\mathrm{S}.=5 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{35}{x} \\ =\frac{35}{7} \\ =5 \end{gathered}$$
∵ 5 = 5
∴ L.H.S. = R.H.S.
Hence, 7 is the solution of the given equation.

(3) If we put x = 7 in the given equation, we get:

$$\begin{gathered} \mathrm{L}.\mathrm{H}.\mathrm{S}.=5 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{x}{35} \\ =\frac{7}{35} \\ =\frac{1}{5} \\ \because 5\ne \frac{1}{5} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}. \end{gathered}$$
Hence, 7 is not the solution of the given equation.

(4) If we put m = 4 in the given equation, we get:
L.H.S. = 5m − 1
= 5 × 4 – 1
= 20 – 1 (Performing multiplication first)
= 19
R.H.S. = 19
∵19 = 19
∴ L.H.S. = R.H.S.
Hence, 4 is the solution of the given equation.

(5) If we put p = 2 in the given equation, we get:
L.H.S. = 2p
= 2 × 2
= 4
R.H.S. = p+3
= 2 + 3
= 5
∵ 4 ≠ 5,
∴ L.H.S. ≠ R.H.S.
Hence, 2 is not the solution of the given equation.

(6) If we put t = 0 in the given equation, we get:
LHS = 3t
= 3 × 0
= 0
RHS = 7t
=7 × 0 = 0
∵ 0 = 0
∴ LHS = RHS
Hence, 0 is the solution of the given equation.