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When faced with equations involving fractions, many students feel a sense of dread. However, there's a systematic approach that can turn these pesky fractions into more manageable whole numbers. The key is to find a common denominator and use it to clear the fractions.

Clearing Fractions by Multiplication

Consider the given equation \(\frac{3}{5}+\frac{7}{x+2}=2\). You can see two terms with fractions where denominators are \(5\) and \(x+2\). To get rid of the fractions, multiply every term by the least common multiple (LCM) of the denominators, which, in this case, is simply \(5(x+2)\). This multiplication cancels out the denominators, leaving you with an equation with integers and no fractions, which is often easier to solve.

The objective of solving an algebraic equation is to find the value of the unknown variable, often referred to as 'isolating the variable.' This process involves moving all terms containing the variable to one side of the equation and constant terms to the opposite side.

To efficiently isolate the variable, balance the equation by performing the same operations on both sides. For instance, if the variable is being subtracted by a number, add that number to both sides to cancel it out. In our exercise, after simplifying, we arrive at \(3x - 29 = 10x - 20\). To isolate \(x\), we subtract \(3x\) from and add \(20\) to both sides, resulting in a simpler expression \(-9 = 7x\) that can then be resolved by dividing both sides by the coefficient of \(x\).

Simplifying equations is a critical skill in algebra that involves reducing expressions to their most basic form. This simplification process often includes distributing multiplication over addition, combining like terms, and performing basic arithmetic operations.

In our example, after multiplying to eliminate fractions, we have \(\frac{3(x+2)}{5} + 7 = 2x + 4\). The next step is to distribute the \(\frac{3}{5}\) across \((x+2)\), combine like terms, and perform arithmetic operations to streamline the equation. This simplification is vital as it transforms a complex-looking equation into one that is straightforward to solve. It's akin to tidying up a room so you can clearly see and reach what you need - in this case, the solution for the variable \(x\).