Question

Solve the following equation by Trial & Error Method.
$x+3=5$
[2 Marks]
[Trial and Error Method]

Answer 1

Solution:


$$x+3=5$$
Means,
$$\begin{array}{cc}& \text{LHS}=x+3,\\ & \text{RHS}=5.\end{array}$$
1. First, let's try by substituting $x=0$ in eq, (1), then,
$$\begin{array}{cc}& \text{LHS}=x+3\\ & =0+3\\ & =3\end{array}$$
Since RHS $=5$
$\Rightarrow$ LHS $\ne$ RHS.
Thus, $x=0$ is not a solution to the given equation.
2. Now, let's substitute $x=1$ in eq. (1), then,
$$\begin{array}{cc}& \text{LHS}=x+3\\ & =1+3\\ & =4\end{array}$$
Again, LHS $\ne$ RHS.
Thus, $x=1$ is not a solution to the given equation.
[1 Mark]
3. Now, let $x=2$, then, putting in eq. (1), we get,
$$\begin{array}{cc}& \text{LHS}=x+3\\ & =2+3\\ & =5\end{array}$$
We can see that when $x=2$, we get,
$$LHS=5=RHS$$
$$\Rightarrow \text{LHS}=\text{RHS}$$
Thus, $\underset{–}{x}=2$ is a solution to the given equation.
Therefore, using the trial and error method, the solution of the given equation $x+3=5$, is $x=2$.
[1 Mark]