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A solution of an equation is a number substituted for an unknown variable which makes the equality in the equation true.

Example:

Consider the equation \(2x + 3 = 7\). Solve the equation.

We know that the equality shows that the unknown variable makes the equation true. That is, when the variable \(x = 2\), we can see that the LHS and RHS are same. Hence, \(x = 2\) is the solution of the given equation.

**1**. The DO - UNDO method:

Let us understand the concept with an example.

Find the solution for \(2\) times \(x\) is more than \(3\) equals \(7\).

**Solution**:

First, let us frame the equation.

\(2x + 3 = 7\)

We have framed the equation from \(x\). To determine the value of \(x\), we need to undo what we did to arrive at the solution. Hence, we do to frame the equation and undo to find the solution.

Now, let us undo the equation.

\(2x +3 - 3 = 7 - 3\) (Subtract \(3\) from both sides)

\(2x = 4\) (Simplify)

\(\frac{2x}{2}=\frac{4}{2}\) (Divide by \(2\) on both sides)

\(x = 2\) (Simplify)

Therefore, \(x = 2\) is the solution of the equation.

**2**. Transposition method:

Transferring the number from one side to the other side of the equation is called a transposition method.

Example:

**1**. Solve the equation \(2x + 3 = 7\) using transposition method.

**Solution**:

Transfer the number \(3\) to the other side of the equation and changing its sign.

Thus, we have:

\(2x = 7 - 3\)

\(2x = 4\)

Now, instead of dividing both sides of the equation by \(2\), let us transfer the number \(2\) to the other side of the equation and reciprocating, we get:

\(x = \frac{4}{2}\)

\(x = 2\)

Thus, the solution of the equation is \(2\).

**2**. Find the solution of \(\frac{x}{2} - 1 = 13\).

**Solution**:

Transfer the number \(1\) to the other side of the equation and changing its sign.

Thus, we have:

\(\frac{x}{2} = 13 + 1\)

\(\frac{x}{2} = 14\)

Now, transfer \(2\) to the RHS and reciprocate the number. Then, we have:

\(x = 14 \times 2\)

\(x = 28\)

Thus, the solution is \(x = 28\).