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When solving linear equations, it's essential to first simplify the equation by combining like terms. Like terms are terms that have the same variable raised to the same power. In the given exercise, the equation starts with multiple x-terms on the left side: \(7x - 4x + 12 = 36 - 5y\). Combining like terms involves adding or subtracting the coefficients (numerical factors) of these terms.

For instance, \(7x\) and \(4x\) are like terms because they both contain the variable \(x\) to the first power. To combine them, subtract the coefficient of \(4x\) from the coefficient of \(7x\), resulting in \(3x\). No other like terms exist in the equation, so the simplified form becomes \(3x + 12 = 36 - 5y\). This step is a foundation for efficiently solving the equation as it reduces complexity and minimizes potential errors in later steps.

Rearranging equations is a crucial step in solving for a specific variable. After combining like terms, the next step in our problem is to isolate the variable \(y\) on one side of the equation. To achieve this, we rearrange the equation by performing the same operation on both sides, thus maintaining the equality.

In the equation \(3x + 12 = 36 - 5y\), we want to move all terms not containing \(y\) to the opposite side. This is done by subtracting \(36\) and \(3x\) from both sides. This step can be visualized as balancing a scale—whatever you do to one side, you must do to the other to keep the scale balanced. By performing these subtractions, we get \( -5y = -3x - 12 \), which is a neater arrangement where \(y\) is closer to being isolated.

The final step to solve for \(y\) is to isolate the variable, meaning to get \(y\) by itself on one side of the equality. In this case, we want to isolate \(y\) from the equation \( -5y = -3x - 12\). This is accomplished by dividing all terms in the equation by the coefficient of \(y\), which is \( -5\).

Dividing each term by \( -5 \) results in \(y = \frac{3}{5}x + \frac{12}{5}\). It's important to apply the division across all terms to preserve the equation's balance. Now \(y\) is isolated and expressed as a function of \(x\), achieving our goal. This form is useful for graphing the equation or for identifying how changes in \(x\) will affect \(y\). The process of isolating variables is fundamental in algebra and can be applied to various types of equations to solve for unknowns.