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To solve linear equations, the first step is often to isolate the variable. This means getting the variable, in this case, 'r', by itself on one side of the equation. We do this by performing the same operation on both sides of the equation to keep it balanced. For instance, start with the equation:

\[2r + 8 = 6r + 10\].

We want 'r' on one side, so subtract 2r from both sides:

\[2r + 8 - 2r = 6r + 10 - 2r\]

This simplifies to:

\[8 = 4r + 10\].

Now, the variable terms are isolated on one side.

Once the variable is isolated on one side, the next step is to solve for it by performing operations that will get the variable completely by itself. From our previous step, we have:

\[8 = 4r + 10\].

To isolate 'r', subtract 10 from both sides:

\[8 - 10 = 4r + 10 - 10\].

This simplifies to:

\[-2 = 4r\].

Finally, divide both sides by 4 to solve for 'r':

\[r = \frac{-2}{4}\]. This simplifies to:

\[r = -\frac{1}{2}\].

Now, we have our value for 'r'.

After finding the solution, it is important to check if it satisfies the original equation. Substitute \(r = -\frac{1}{2}\) back into the original equation:

\[2(-\frac{1}{2}) + 8 = 6(-\frac{1}{2}) + 10\].

Simplify both sides:

\[-1 + 8 = -3 + 10\].

This results in:

\[7 = 7\].

Since both sides of the equation are equal, the solution \(r = -\frac{1}{2}\) is correct.

Identities in equations occur when the solving process reveals that both sides of the equation are identical for any value of the variable. For example, an equation like \[2(x - 3) = 2x - 6\] simplifies to \[2x - 6 = 2x - 6\]. Both sides are the same regardless of what value x takes, making it an identity. This indicates that every value of x is a solution. It is important to recognize identities because it means checking each instance of the variable is redundant; the equation is true universally. Conversely, contradictions are equations that have no solutions because they simplify to false statements (e.g., \[0 = 5\]). In our case, there is no contradiction or identity, but it's crucial to understand these concepts while solving equations.