Simple Equations

Solving,

$$\begin{aligned}\Rightarrow x - 4 = 8 \\[1em] \text{ [Adding 4 on each side] } \\[1em] \Rightarrow x - 4 + 4 = 8 + 4 \\[1em] \Rightarrow x + 0 = 12 \\[1em] \Rightarrow x = 12\end{aligned}$$

Hence, $\bm{x = 12}$.

Solving,

$$\begin{aligned}\Rightarrow x + 7 = 2 \\[1em] \text{ [Subtracting 7 from each side] } \\[1em] \Rightarrow x + 7 - 7 = 2 - 7 \\[1em] \Rightarrow x + 0 = -5 \\[1em] \Rightarrow x = -5\end{aligned}$$

Hence, $\bm{x = -5}$.

Solving,

$$\begin{aligned}\Rightarrow x - 6 = -3 \\[1em] \text{ [Adding 6 on each side] } \\[1em] \Rightarrow x - 6 + 6 = -3 + 6 \\[1em] \Rightarrow x + 0 = 3 \\[1em] \Rightarrow x = 3\end{aligned}$$

Hence, $\bm{x = 3}$.

Solving,

$$\begin{aligned}\Rightarrow y + 2 = -6 \\[1em] \text{ [Subtracting 2 from each side] } \\[1em] \Rightarrow y + 2 - 2 = -6 - 2 \\[1em] \Rightarrow y + 0 = -8 \\[1em] \Rightarrow y = -8\end{aligned}$$

Hence, $\bm{y = -8}$.

Solving,

$$\begin{aligned}\Rightarrow x + 3 = -1 \\[1em] \text{ [Subtracting 3 from each side] } \\[1em] \Rightarrow x + 3 - 3 = -1 - 3 \\[1em] \Rightarrow x + 0 = -4 \\[1em] \Rightarrow x = -4\end{aligned}$$

Hence, $\bm{x = -4}$.

Solving,

$$\begin{aligned}\Rightarrow z - 5 = -8 \\[1em] \text{ [Adding 5 on each side] } \\[1em] \Rightarrow z - 5 + 5 = -8 + 5 \\[1em] \Rightarrow z + 0 = -3 \\[1em] \Rightarrow z = -3\end{aligned}$$

Hence, $\bm{z = -3}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{x}{3} = 6 \\[1em] \text{ [Multiplying each side by 3] } \\[1em] \Rightarrow \frac{x}{3} \times 3 = 6 \times 3 \\[1em] \Rightarrow x = 18\end{aligned}$$

Hence, $\bm{x = 18}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{y}{2} = -4 \\[1em] \text{ [Multiplying each side by 2] } \\[1em] \Rightarrow \frac{y}{2} \times 2 = -4 \times 2 \\[1em] \Rightarrow y = -8\end{aligned}$$

Hence, $\bm{y = -8}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{2x}{4} = 3 \\[1em] \Rightarrow \frac{x}{2} = 3 \\[1em] \text{ [Multiplying each side by 2] } \\[1em] \Rightarrow \frac{x}{2} \times 2 = 3 \times 2 \\[1em] \Rightarrow x = 6\end{aligned}$$

Hence, $\bm{x = 6}$.

Solving,

$$\begin{aligned}\Rightarrow 2x = 10 \\[1em] \text{ [Dividing each side by 2] } \\[1em] \Rightarrow \frac{2x}{2} = \frac{10}{2} \\[1em] \Rightarrow x = 5\end{aligned}$$

Hence, $\bm{x = 5}$.

Solving,

$$\begin{aligned}\Rightarrow 3y = -12 \\[1em] \text{ [Dividing each side by 3] } \\[1em] \Rightarrow \frac{3y}{3} = \frac{-12}{3} \\[1em] \Rightarrow y = -4\end{aligned}$$

Hence, $\bm{y = -4}$.

Solving,

$$\begin{aligned}\Rightarrow 2x - 3 = 7 \\[1em] \text{ [Adding 3 on each side] } \\[1em] \Rightarrow 2x - 3 + 3 = 7 + 3 \\[1em] \Rightarrow 2x = 10 \\[1em] \text{ [Dividing each side by 2] } \\[1em] \Rightarrow \frac{2x}{2} = \frac{10}{2} \\[1em] \Rightarrow x = 5\end{aligned}$$

Hence, $\bm{x = 5}$.

Solving,

$$\begin{aligned}\Rightarrow 3x - 5 = 4 \\[1em] \text{ [Adding 5 on each side] } \\[1em] \Rightarrow 3x - 5 + 5 = 4 + 5 \\[1em] \Rightarrow 3x = 9 \\[1em] \text{ [Dividing each side by 3] } \\[1em] \Rightarrow \frac{3x}{3} = \frac{9}{3} \\[1em] \Rightarrow x = 3\end{aligned}$$

Hence, $\bm{x = 3}$.

Solving,

$$\begin{aligned}\Rightarrow 4x - 1 = 7 \\[1em] \text{ [Adding 1 on each side] } \\[1em] \Rightarrow 4x - 1 + 1 = 7 + 1 \\[1em] \Rightarrow 4x = 8 \\[1em] \text{ [Dividing each side by 4] } \\[1em] \Rightarrow \frac{4x}{4} = \frac{8}{4} \\[1em] \Rightarrow x = 2\end{aligned}$$

Hence, $\bm{x = 2}$.

Solving,

$$\begin{aligned}\Rightarrow 5z - 3 = 2 \\[1em] \text{ [Adding 3 on each side] } \\[1em] \Rightarrow 5z - 3 + 3 = 2 + 3 \\[1em] \Rightarrow 5z = 5 \\[1em] \text{ [Dividing each side by 5] } \\[1em] \Rightarrow \frac{5z}{5} = \frac{5}{5} \\[1em] \Rightarrow z = 1\end{aligned}$$

Hence, $\bm{z = 1}$.

Solving,

$$\begin{aligned}\Rightarrow 2x - 3 = -5 \\[1em] \text{ [Adding 3 on each side] } \\[1em] \Rightarrow 2x - 3 + 3 = -5 + 3 \\[1em] \Rightarrow 2x = -2 \\[1em] \text{ [Dividing each side by 2] } \\[1em] \Rightarrow \frac{2x}{2} = \frac{-2}{2} \\[1em] \Rightarrow x = -1\end{aligned}$$

Hence, $\bm{x = -1}$.

Solving,

$$\begin{aligned}\Rightarrow 4y + 5 = -3 \\[1em] \text{ [Subtracting 5 from each side] } \\[1em] \Rightarrow 4y + 5 - 5 = -3 - 5 \\[1em] \Rightarrow 4y = -8 \\[1em] \text{ [Dividing each side by 4] } \\[1em] \Rightarrow \frac{4y}{4} = \frac{-8}{4} \\[1em] \Rightarrow y = -2\end{aligned}$$

Hence, $\bm{y = -2}$.

Solving,

$$\begin{aligned}\Rightarrow 6z - 8 = 1 \\[1em] \text{ [Adding 8 on each side] } \\[1em] \Rightarrow 6z - 8 + 8 = 1 + 8 \\[1em] \Rightarrow 6z = 9 \\[1em] \text{ [Dividing each side by 6] } \\[1em] \Rightarrow \frac{6z}{6} = \frac{9}{6} \\[1em] \Rightarrow z = \frac{3}{2}\end{aligned}$$

Hence, $\bm{z = \frac{3}{2}}$.

Solving,

$$\begin{aligned}\Rightarrow 5x + 4 = 19 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow 5x = 19 - 4 \\[1em] \Rightarrow 5x = 15 \\[1em] \text{ [Dividing each side by 5] } \\[1em] \Rightarrow x = \frac{15}{5} \\[1em] \Rightarrow x = 3.\end{aligned}$$

Hence, $\bm{x = 3}$.

Solving,

$$\begin{aligned}\Rightarrow 3x - 8 = 10 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow 3x = 10 + 8 \\[1em] \Rightarrow 3x = 18 \\[1em] \text{ [Dividing each side by 3] } \\[1em] \Rightarrow x = \frac{18}{3} \\[1em] \Rightarrow x = 6.\end{aligned}$$

Hence, $\bm{x = 6}$.

Solving,

$$\begin{aligned}\Rightarrow 5 - 2y = 9 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow -2y = 9 - 5 \\[1em] \Rightarrow -2y = 4 \\[1em] \text{ [Dividing each side by −2] } \\[1em] \Rightarrow y = \frac{4}{-2} \\[1em] \Rightarrow y = -2.\end{aligned}$$

Hence, $\bm{y = -2}$.

Solving,

$$\begin{aligned}\Rightarrow -6 + 3x = -5 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow 3x = -5 + 6 \\[1em] \Rightarrow 3x = 1 \\[1em] \text{ [Dividing each side by 3] } \\[1em] \Rightarrow x = \frac{1}{3}.\end{aligned}$$

Hence, $\bm{x = \frac{1}{3}}$.

Solving,

$$\begin{aligned}\Rightarrow 2x + 12 = 3 - x \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow 2x + x = 3 - 12 \\[1em] \Rightarrow 3x = -9 \\[1em] \text{ [Dividing each side by 3] } \\[1em] \Rightarrow x = \frac{-9}{3} \\[1em] \Rightarrow x = -3.\end{aligned}$$

Hence, $\bm{x = -3}$.

Solving,

$$\begin{aligned}\Rightarrow 2 - 5y = 2y - 5 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow -5y - 2y = -5 - 2 \\[1em] \Rightarrow -7y = -7 \\[1em] \text{ [Dividing each side by −7] } \\[1em] \Rightarrow y = \frac{-7}{-7} \\[1em] \Rightarrow y = 1.\end{aligned}$$

Hence, $\bm{y = 1}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{x}{5} + 5 = 2 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow \frac{x}{5} = 2 - 5 \\[1em] \Rightarrow \frac{x}{5} = -3 \\[1em] \text{ [Multiplying each side by 5] } \\[1em] \Rightarrow x = -3 \times 5 \\[1em] \Rightarrow x = -15.\end{aligned}$$

Hence, $\bm{x = -15}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{4x}{3} + 3 = x - 2 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow \frac{4x}{3} - x = -2 - 3 \\[1em] \text{ [Taking L.C.M.] } \\[1em] \Rightarrow \frac{4x - 3x}{3} = -5 \\[1em] \Rightarrow \frac{x}{3} = -5 \\[1em] \text{ [Multiplying each side by 3] } \\[1em] \Rightarrow x = -5 \times 3 \\[1em] \Rightarrow x = -15.\end{aligned}$$

Hence, $\bm{x = -15}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{3x + 2}{5} = 4 \\[1em] \text{ [Multiplying each side by 5] } \\[1em] \Rightarrow 3x + 2 = 4 \times 5 \\[1em] \Rightarrow 3x + 2 = 20 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow 3x = 20 - 2 \\[1em] \Rightarrow 3x = 18 \\[1em] \text{ [Dividing each side by 3] } \\[1em] \Rightarrow x = \frac{18}{3} \\[1em] \Rightarrow x = 6.\end{aligned}$$

Hence, $\bm{x = 6}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{x}{2} - 6 = \frac{x}{3} - 5 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow \frac{x}{2} - \frac{x}{3} = -5 + 6 \\[1em] \text{ [Taking L.C.M.] } \\[1em] \Rightarrow \frac{3x - 2x}{6} = 1 \\[1em] \Rightarrow \frac{x}{6} = 1 \\[1em] \text{ [Multiplying each side by 6] } \\[1em] \Rightarrow x = 6.\end{aligned}$$

Hence, $\bm{x = 6}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{2y}{3} - 8 = 2 - y \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow \frac{2y}{3} + y = 2 + 8 \\[1em] \text{ [Taking L.C.M.] } \\[1em] \Rightarrow \frac{2y + 3y}{3} = 10 \\[1em] \Rightarrow \frac{5y}{3} = 10 \\[1em] \text{ [Multiplying each side by 3] } \\[1em] \Rightarrow 5y = 30 \\[1em] \text{ [Dividing each side by 5] } \\[1em] \Rightarrow y = \frac{30}{5} \\[1em] \Rightarrow y = 6.\end{aligned}$$

Hence, $\bm{y = 6}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{x}{2} + 2 = 5 - x \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow \frac{x}{2} + x = 5 - 2 \\[1em] \text{ [Taking L.C.M.] } \\[1em] \Rightarrow \frac{x + 2x}{2} = 3 \\[1em] \Rightarrow \frac{3x}{2} = 3 \\[1em] \text{ [Multiplying each side by 2] } \\[1em] \Rightarrow 3x = 6 \\[1em] \text{ [Dividing each side by 3] } \\[1em] \Rightarrow x = \frac{6}{3} \\[1em] \Rightarrow x = 2.\end{aligned}$$

Hence, $\bm{x = 2}$.

Solving,

$$\begin{aligned}\Rightarrow 6(2x - 1) - 5(x - 3) = 2(x + 2) \\[1em] \text{ [Removing brackets] } \\[1em] \Rightarrow 12x - 6 - 5x + 15 = 2x + 4 \\[1em] \Rightarrow 7x + 9 = 2x + 4 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow 7x - 2x = 4 - 9 \\[1em] \Rightarrow 5x = -5 \\[1em] \text{ [Dividing each side by 5] } \\[1em] \Rightarrow x = \frac{-5}{5} \\[1em] \Rightarrow x = -1.\end{aligned}$$

Hence, $\bm{x = -1}$.

Solving,

$$\begin{aligned}\Rightarrow 7(3x - 1) - 6(2x + 3) = 8(x - 2) + 1 \\[1em] \text{ [Removing brackets] } \\[1em] \Rightarrow 21x - 7 - 12x - 18 = 8x - 16 + 1 \\[1em] \Rightarrow 9x - 25 = 8x - 15 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow 9x - 8x = -15 + 25 \\[1em] \Rightarrow x = 10.\end{aligned}$$

Hence, $\bm{x = 10}$.

Thrice $x = 3x$.

5 less than thrice $x$ means 5 is subtracted from $3x$, i.e., $3x - 5$.

Hence, option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 5x + 8 = 4 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow 5x = 4 - 8 \\[1em] \Rightarrow 5x = -4 \\[1em] \text{ [Dividing each side by 5] } \\[1em] \Rightarrow x = \frac{-4}{5}.\end{aligned}$$

∴ $3x = 3 \times \frac{-4}{5} = \frac{-12}{5}$

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 3m - 8 = -4 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow 3m = -4 + 8 \\[1em] \Rightarrow 3m = 4 \\[1em] \text{ [Dividing each side by 3] } \\[1em] \Rightarrow m = \frac{4}{3}.\end{aligned}$$

Hence, option 4 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 4x + 3 = 15 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow 4x = 15 - 3 \\[1em] \Rightarrow 4x = 12 \\[1em] \text{ [Dividing each side by 4] } \\[1em] \Rightarrow x = \frac{12}{4} = 3.\end{aligned}$$

∴ $2x - 3 = 2 \times 3 - 3 = 6 - 3 = 3$

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{m}{3} + 2 = 5 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow \frac{m}{3} = 5 - 2 \\[1em] \Rightarrow \frac{m}{3} = 3 \\[1em] \text{ [Multiplying each side by 3] } \\[1em] \Rightarrow m = 3 \times 3 = 9.\end{aligned}$$

Hence, option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{x}{2} + 3 = 5 - x \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow \frac{x}{2} + x = 5 - 3 \\[1em] \text{ [Taking L.C.M.] } \\[1em] \Rightarrow \frac{x + 2x}{2} = 2 \\[1em] \Rightarrow \frac{3x}{2} = 2 \\[1em] \text{ [Multiplying each side by 2] } \\[1em] \Rightarrow 3x = 4 \\[1em] \text{ [Dividing each side by 3] } \\[1em] \Rightarrow x = \frac{4}{3}.\end{aligned}$$

Hence, option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{x}{2} - 5 = \frac{x}{3} - 6 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow \frac{x}{2} - \frac{x}{3} = -6 + 5 \\[1em] \text{ [Taking L.C.M.] } \\[1em] \Rightarrow \frac{3x - 2x}{6} = -1 \\[1em] \Rightarrow \frac{x}{6} = -1 \\[1em] \text{ [Multiplying each side by 6] } \\[1em] \Rightarrow x = -6.\end{aligned}$$

Hence, option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{x}{3} - \frac{x}{4} = 5 \\[1em] \text{ [Taking L.C.M.] } \\[1em] \Rightarrow \frac{4x - 3x}{12} = 5 \\[1em] \Rightarrow \frac{x}{12} = 5 \\[1em] \text{ [Multiplying each side by 12] } \\[1em] \Rightarrow x = 60.\end{aligned}$$

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 4x - 1 = 7 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow 4x = 7 + 1 \\[1em] \Rightarrow 4x = 8 \\[1em] \text{ [Dividing each side by 4] } \\[1em] \Rightarrow x = \frac{8}{4} = 2.\end{aligned}$$

∴ $3x - 2 = 3 \times 2 - 2 = 6 - 2 = 4$

Hence, option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{2x}{2} - 8 = 2 - x \\[1em] \Rightarrow x - 8 = 2 - x \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow x + x = 2 + 8 \\[1em] \Rightarrow 2x = 10 \\[1em] \text{ [Dividing each side by 2] } \\[1em] \Rightarrow x = \frac{10}{2} = 5.\end{aligned}$$

Hence, option 1 is the correct option.

(i) Let the number be $x$.

$$\begin{aligned}\Rightarrow x + 6 = 15 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow x = 15 - 6 \\[1em] \Rightarrow x = 9.\end{aligned}$$

Hence, the number is 9.

(ii) Solving,

$$\begin{aligned}\Rightarrow \frac{5x}{7} - \frac{3x}{2} = \frac{11}{14} \\[1em] \text{ [Taking L.C.M.] } \\[1em] \Rightarrow \frac{10x - 21x}{14} = \frac{11}{14} \\[1em] \Rightarrow \frac{-11x}{14} = \frac{11}{14} \\[1em] \text{ [Equating numerators] } \\[1em] \Rightarrow -11x = 11 \\[1em] \text{ [Dividing each side by −11] } \\[1em] \Rightarrow x = -1.\end{aligned}$$

Hence, $\bm{x = −1}$.

(iii) Solving,

$$\begin{aligned}\Rightarrow \frac{5x}{6} - \frac{2x}{3} = 9 \\[1em] \text{ [Taking L.C.M.] } \\[1em] \Rightarrow \frac{5x - 4x}{6} = 9 \\[1em] \Rightarrow \frac{x}{6} = 9 \\[1em] \text{ [Multiplying each side by 6] } \\[1em] \Rightarrow x = 54.\end{aligned}$$

Hence, $\bm{x = 54}$.

(iv) Solving,

$$\begin{aligned}\Rightarrow 5(3 - 2x) - 3(5 - 3x) = 6 \\[1em] \text{ [Removing brackets] } \\[1em] \Rightarrow 15 - 10x - 15 + 9x = 6 \\[1em] \Rightarrow -x = 6 \\[1em] \text{ [Multiplying each side by −1] } \\[1em] \Rightarrow x = -6.\end{aligned}$$

Hence, $\bm{x = −6}$.

(v) Solving,

$$\begin{aligned}\Rightarrow \frac{x}{3} - 3 = 3 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow \frac{x}{3} = 3 + 3 \\[1em] \Rightarrow \frac{x}{3} = 6 \\[1em] \text{ [Multiplying each side by 3] } \\[1em] \Rightarrow x = 18.\end{aligned}$$

Hence, $\bm{x = 18}$.

(vi) Solving,

Let the number be $x$.

One-third of the number = $\frac{x}{3}$

3 more than one-third of the number = $\frac{x}{3} + 3$

∴ The statement is written as $\frac{x}{3} + 3 = 5$.

Hence, the required equation is $\bm{\frac{x}{3} + 3 = 5}$.

(i) Solving,

$$\begin{aligned}\Rightarrow \frac{4x}{3} + 3 = x - 2 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow \frac{4x}{3} - x = -2 - 3 \\[1em] \text{ [Taking L.C.M.] } \\[1em] \Rightarrow \frac{4x - 3x}{3} = -5 \\[1em] \Rightarrow \frac{x}{3} = -5 \\[1em] \text{ [Multiplying each side by 3] } \\[1em] \Rightarrow x = -15.\end{aligned}$$

Since $x$ = −15 and not 15, the given statement is false.

Hence, the given statement is false (F).

(ii) Solving,

$$\begin{aligned}\Rightarrow \frac{3y}{2} - 5 = -7 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow \frac{3y}{2} = -7 + 5 \\[1em] \Rightarrow \frac{3y}{2} = -2 \\[1em] \text{ [Multiplying each side by 2] } \\[1em] \Rightarrow 3y = -4 \\[1em] \text{ [Dividing each side by 3] } \\[1em] \Rightarrow y = \frac{-4}{3}.\end{aligned}$$

Since $y = \frac{-4}{3}$ and not $\frac{4}{3}$, the given statement is false.

Hence, the given statement is false (F).

(iii) Solving,

$$\begin{aligned}\Rightarrow 2x + 3x + 4x = 5x + 1 \\[1em] \Rightarrow 9x = 5x + 1 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow 9x - 5x = 1 \\[1em] \Rightarrow 4x = 1 \\[1em] \text{ [Dividing each side by 4] } \\[1em] \Rightarrow x = \frac{1}{4}.\end{aligned}$$

Since $x = \frac{1}{4}$, the given statement is true.

Hence, the given statement is true (T).

(iv) Solving,

$$\begin{aligned}\Rightarrow \frac{x}{5} + 4 = 5 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow \frac{x}{5} = 5 - 4 \\[1em] \Rightarrow \frac{x}{5} = 1 \\[1em] \text{ [Multiplying each side by 5] } \\[1em] \Rightarrow x = 5.\end{aligned}$$

Since $x$ = 5 and not −5, the given statement is false.

Hence, the given statement is false (F).

(v) Solving,

$$\begin{aligned}\Rightarrow -6 + 3x = -5 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow 3x = -5 + 6 \\[1em] \Rightarrow 3x = 1 \\[1em] \text{ [Dividing each side by 3] } \\[1em] \Rightarrow x = \frac{1}{3}.\end{aligned}$$

Since $x = \frac{1}{3}$, the given statement is true.

Hence, the given statement is true (T).

In the equation $3x − 4 = 2 + x$, on transposing −4 to R.H.S. it becomes +4 and on transposing $x$ to L.H.S. it becomes −$x$.

$$\begin{aligned}\Rightarrow 3x - 4 = 2 + x \\[1em] \Rightarrow 3x - x = 2 + 4.\end{aligned}$$

So, Assertion (A) is true.

Any term of an equation can indeed be shifted to the other side with change in its sign. This is the rule of transposition, which is exactly the rule used above. So, Reason (R) is true and it is the correct explanation of Assertion (A).

Hence, option 1 is the correct option.

Putting $x$ = 2 in the given equation,

$$\begin{aligned}\text{L.H.S.} = \frac{6 \times 2}{2} + 3 = 6 + 3 = 9 \\[1em] \text{R.H.S.} = 3 \times 2 = 6.\end{aligned}$$

Since L.H.S. ≠ R.H.S., $x$ = 2 is not the solution of the given equation.

In fact, on simplifying the equation we get $3x + 3 = 3x$, i.e., 3 = 0, which is not possible. So, the equation has no solution. Assertion (A) is false.

The solution (root) of an equation is the value of the variable which makes L.H.S. = R.H.S., i.e., it satisfies the equation. So, Reason (R) is true.

Hence, option 4 is the correct option.