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When solving linear equations, a crucial first step is isolating the variable. This simply means getting the variable (often represented by letters like \( x \), \( y \), etc.) on its own on one side of the equation. In our example, the equation is \(-3.5x + 2.8 = -11.2\). The term involving the variable, \(-3.5x\), is mixed with a constant number. Our job is to isolate \(x\), which can be achieved by removing the number from its side.

To do this, we use subtraction to eliminate \(+2.8\) by subtracting \(2.8\) from both sides. This is a basic property of equations: whatever you do to one side, you must do to the other. It simplifies our equation to:

• \(-3.5x = -14\)

This isolated the variable term, setting us up for the next steps of solving the equation.

Once the variable term is isolated, we use algebraic manipulation to rearrange and simplify the equation for further steps. Algebra involves manipulating these equations to make the 'unknown' or variable clear. We have already isolated \(-3.5x\) in the previous step. This algebraic manipulation helps us systematically move towards solving for the variable itself.

Previously, we removed \(+2.8\), which was our first manipulative step. However, algebraic manipulation doesn’t stop there. It often requires working with different arithmetic operations, like addition, subtraction, multiplication, and division, to simplify the equation in hand. As we go forward, these techniques will help solve the equation efficiently.

The final stage in solving certain types of linear equations, like in our given problem \(-3.5x = -14\), often involves division. Once you've isolated a variable term, you might notice a number attached to the variable. Here, \(-3.5\) is multiplied by \(x\). To isolate \(x\), we do the opposite operation, which is division.

By dividing each side of the equation by \(-3.5\), we are left with the variable \(x\) alone:

• \(\frac{-3.5x}{-3.5} = \frac{-14}{-3.5}\)
• this simplifies to \(x = 4\)

Division helps eliminate the number multiplying the variable, revealing the solution. Understanding this critical operation is key, as it unravels the answer and completes the solution of the equation. By executing these methods carefully, you can solve many linear equations confidently.