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When solving linear equations, one of the key steps is to isolate the variable. This means we want to get the variable by itself on one side of the equation. To do this, we perform operations that will keep the equation balanced but help move everything else to the other side. For example, consider the equation in the exercise: \(\frac{c}{-4} - 3 = 5\). Here, we isolate the fraction term by adding 3 to both sides of the equation: \(\frac{c}{-4} - 3 + 3 = 5 + 3\). This ensures that the -3 and +3 cancel each other out, leaving us with: \(\frac{c}{-4} = 8\). Now, the variable term is isolated, and we can focus on eliminating the fraction.

Fractions can make equations look complicated, but eliminating them can simplify the process. After we isolate the variable term, the next step is to clear the fraction. In our equation, \(\frac{c}{-4}\) is the fraction. To eliminate it, we need to multiply both sides by the denominator of the fraction. In this case, we multiply both sides by -4: \(\frac{c}{-4} \times -4 = 8 \times -4\). Multiplying by the denominator (-4) cancels out the fraction on the left side, which simplifies to: \c = -32\. Now, the fraction is gone, and the equation is much simpler.

Understanding inverse operations is crucial when solving equations. Inverse operations are operations that undo each other, like addition and subtraction or multiplication and division. We use these operations to solve for the variable. Going back to our example: \(\frac{c}{-4} - 3 = 5\), we initially used addition to cancel out the -3. Addition is the inverse operation of subtraction. Next, we multiplied by -4 to clear the fraction. Multiplication is the inverse operation of division. By using inverse operations, we systematically isolate the variable and simplify the equation. So, after isolating the variable and eliminating the fraction, we arrive at the solution: \c = -32\. Inverse operations help maintain the balance of the equation and are a fundamental concept in algebra.