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In algebra, combining like terms is a critical step towards simplifying expressions and equations. Like terms are terms that have the same variables raised to the same power, allowing them to be added or subtracted from each other.

For example, when facing an equation such as \(\frac{3x}{2} + \frac{3x}{4} = \frac{x}{4} - 4\), the first step is to identify and combine the terms that have the same variable, in this case, 'x'. Combining \(\frac{3x}{2} + \frac{3x}{4}\) leads to \(\frac{5x}{2}\), since you are essentially adding the coefficients of 'x' while the variable part remains unchanged.

To master this concept, focus on understanding that you're grouping like factors together, just as you would combine identical items into one pile. This not only simplifies equations but also lays the groundwork for solving them.

Solving linear equations is a foundational skill in algebra. A linear equation is one in which the highest power of the variable is one. The general form is \(ax + b = 0\), where 'a' and 'b' are constants. To solve for 'x', you must isolate the variable on one side of the equation.

In the given exercise, after combining like terms, you are left with \(10x = x - 16\). To solve for 'x', subtract 'x' from both sides to isolate the terms containing 'x'. This leads to the equation \(9x = -16\), which you can then solve by dividing both sides by '9', yielding \(x = -\frac{16}{9}\).

Remember, the goal is to get 'x' by itself, so reverse the operations that are being applied to 'x', maintaining the balance by performing each operation on both sides of the equation. Practice this concept by working through a variety of linear equations to develop a clear understanding of the solving process.

Substitution is a technique used in algebra to find the value of expressions. Once you find the value of the variable, you 'substitute' it into another expression to evaluate it. This is particularly important when solving a set of equations or evaluating expressions where the variable appears in more complex forms.

In our example, after solving the linear equation, you find that \(x = -\frac{16}{9}\). To evaluate \(x^2 - x\), you substitute the value of 'x' into the expression, resulting in \(\left(-\frac{16}{9}\right)^2 - \left(-\frac{16}{9}\right) = \frac{256}{81} + \frac{16}{9} = \frac{400}{81}\).

Substitution is a powerful tool that can be used for a variety of purposes in algebra, such as solving systems of equations or evaluating functions. It is a straightforward process, but requires careful attention to ensure that the substitution is done correctly. Precise substitution is key to obtaining the correct results in algebraic computations.