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Expanding brackets in algebra is a fundamental skill, as it allows us to simplify expressions and solve equations. When you come across a problem like \(7x + 6(2 - x) + 3 = 4 + 3(6 + x)\), your first step is to apply the distributive property. This property states that for any numbers a, b, and c, the expression \(a(b + c)\) is equal to \(ab + ac\).

Let’s apply it to our example. On the left side of the equation \(6(2 - x)\), the distributive property tells us to multiply 6 by each term inside the brackets: \(6 \times 2\) and \(6 \times (-x)\), which results in 12 and -6x respectively. Repeat the same process on the right side with \(3(6 + x)\) to get 18 and \(3x\). Ensuring you carefully handle positive and negative signs during this process is crucial for getting the correct result.

Once the brackets are expanded, our next move is to simplify the expressions. This means combining like terms, which are terms that have the same variables raised to the same power. In this case, we're working with simple linear terms, involving 'x' and constants.

In our equation, after expanding the brackets, we’re left with terms involving 'x' and standalone numbers. Simplify the expression by combining the like terms. For instance, on the left side, we can combine \(7x - 6x\) into just 'x' and \(12 + 3\) into 15. This step turns complex-looking equations into simpler ones that are much easier to solve. It's like tidying up: group together similar items so you can see clearly what you're working with.

After simplifying the expressions, the goal is to isolate the variable, in this case, 'x'. To do this, you want to move all the terms with 'x' to one side and the constants to the opposite side. You can subtract or add terms to both sides to maintain the balance of the equation.

In our problem, subtract 'x' from both sides to get the terms with 'x' on one side: \(3x - x\). Then subtract 22 from both sides to move the constants to the other side: \(15 - 22\). You're essentially removing distractions from both sides of the equation, concentrating all the 'x's on one side and the constants on the other, making the path to 'x' much clearer. The end result shows you what 'x' equals, entirely on its own, without any extra baggage. In other words, you’ve successfully isolated 'x', leading to a simple equation that could be solved to find the value of 'x'.