91Ó°ÊÓ

To solve linear equations, it's essential to isolate the variable, which means to get the variable by itself on one side of the equation. For the given equation, \(7 + \frac{2}{3} x = -1\) the first step involves getting rid of constants on the same side as the variable. This is done by performing the inverse operation. Here, we subtract 7 from both sides, resulting in the term with \(x\) on one side, \(\frac{2}{3} x = -8\).
This step sets the stage for the next crucial move: to isolate the variable completely. Remember, our goal is a clean equation with the variable on one side and a number on the other, looking something like \( x = \text{number} \) to denote a solution.

Simplification of an equation makes it easier to solve. After isolating the variable term on one side, we work towards simplifying the coefficient of the variable to unity. In our example, after subtracting 7, we reduced the equation to \(\frac{2}{3} x = -8\) which is simpler than the initial equation. However, the variable \(x\) still has a coefficient \(\frac{2}{3}\) that we need to eliminate.

Removing fractional coefficients involves multiplying by the inverse of the fraction, which you can think of as flipping the fraction over. Simplification is like cleaning up your workspace; it gives you a clear view of what you're working with and makes the subsequent steps feel much more manageable.

Dividing by a fraction can be a tricky concept, but it's actually based on a simple rule: dividing by a fraction is the same as multiplying by its reciprocal. A reciprocal is what you get when you flip a fraction's numerator and denominator. So for the fraction \(\frac{2}{3}\) in our equation, the reciprocal is \(\frac{3}{2}\).

To isolate \(x\) from \(\frac{2}{3} x = -8\) we need to 'cancel out' the \(\frac{2}{3}\) by dividing it from both sides, which is the same as multiplying by \(\frac{3}{2}\) — its reciprocal. This gives us \( x = -8 \times \frac{3}{2} = -12 \) as our final answer. Always remember, when dividing by a fraction, flip the divider and multiply. This method dramatically simplifies the solving process, turning what is seemingly a more complicated division into an easy multiplication.