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The distributive property is a fundamental concept in algebra that helps simplify expressions. It involves multiplying each term inside a bracket by a factor outside the bracket. This property helps to break down complicated expressions into simpler parts.
Consider the expression \(2(x-1)\). Applying the distributive property means multiplying 2 by both \(x\) and \(-1\), resulting in \(2x - 2\). Similarly, for \(-3(x+1)\), you would distribute \(-3\) to both \(x\) and \(1\), giving us \(-3x - 3\).
By expanding each term like this, you can see all elements of the expression, which is useful for further operations like combining terms.

Once you expand an equation using the distributive property, the next step is combining like terms. Like terms are terms that have the same variable raised to the same power. This allows you to simplify the equation by adding or subtracting the coefficients of these terms.
For example, in the equation \(2x - 2 + 3 = x - 3x - 3\), \(2x\) and \(x\) are like terms because they both have the variable \(x\). Also, \(-2\) and \(3\) on the left, as well as \(-3\) on the right, are constants that can be combined.
On the left side of this equation, combining \(2x\) and the constant terms \(-2 + 3\) results in \(2x + 1\). On the right side, combining \(x - 3x\) and \(-3\) yields \(-2x - 3\). This simplification helps in solving the equation as it organizes terms in a way that makes isolating the variable easier.

Isolating the variable is the step where we solve for the unknown, usually denoted as \(x\). This involves operations that move all terms containing the variable to one side of the equation and constant terms to the other.
In our example, the equation is \(2x + 1 = -2x - 3\). To isolate \(x\), we try to gather all \(x\) terms on one side. We add \(2x\) to both sides to get all \(x\) terms on the left. This action gives us \(4x + 1 = -3\).
Next, subtract 1 from both sides to further isolate the term with \(x\):\(4x = -4\). Finally, divide each side by 4 to find \(x = -1\). Isolating the variable is crucial because it reveals the solution to the equation.

After finding a solution, always verify that it satisfies the original equation. Checking solutions ensures that calculations were accurate and that the found value is indeed a solution.
To check \(x = -1\), substitute it back into the original equation: \[2(x-1)+3=x-3(x+1)\].Substituting \(-1\) for \(x\) yields:\[2(-1-1)+3=(-1)-3(-1+1)\].Simplified, both sides equal to \(-1\), confirming our solution is correct.
This step is important as it verifies that no mistakes were made during earlier calculations, thus solidifying your confidence in the solution.