91Ó°ÊÓ

The substitution method simplifies complicated equations. Here, we replaced a complex expression with a single variable.
For instance, in the equation: \[9\frac{(x+2)^2}{(x+3)^2} - 6\frac{x+2}{x+3} + 1 = 0\]
we introduced a new variable, \(y\). We set \(y = \frac{x+2}{x+3}\).
So the original complex equation became: \[9y^2 - 6y + 1 = 0\]
This makes the problem simpler to handle because it's easier working with a quadratic equation in one variable. By substituting, the problem becomes clearer and lessens the chances of making mistakes.

The quadratic formula is a powerful tool for solving quadratic equations. Any quadratic equation of the form \(ax^2 + bx + c = 0\) can be solved using it. The general formula is: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
By plugging in the values of \(a\), \(b\), and \(c\) from our simplified equation, \[9y^2 - 6y + 1 = 0\],
we get: \[a = 9\], \[b = -6\], and \[c = 1\].So using the formula,
we substitute these values:\[y = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(9)(1)}}{2(9)}\]
This simplifies to: \[y = \frac{6 \pm \sqrt{36 - 36}}{18}\]
Further simplifying to: \[y = \frac{6 \pm 0}{18}\]
Ultimately, \[y = \frac{6}{18}\], which boils down to \[y = \frac{1}{3}\].
With the roots of the quadratic equation, we move to the next step.

Simplification is a crucial step in problem-solving. After finding \(y\), we replace our substitution back into the original context.
So if \(y = \frac{1}{3}\), and recall \(y = \frac{x+2}{x+3}\), we substitute it back:
\[\frac{x+2}{x+3} = \frac{1}{3}\]
To solve for \(x\) from here, we follow these steps:

• Cross multiply to get rid of the fraction:
\[3(x+2) = x+3\]
• Distribute the 3:
\[3x + 6 = x + 3\]
• Move \(x\) to one side of the equation:
\[3x - x = 3 - 6\]
• Simplify further:
\[2x = -3\]
• Divide both sides by 2 to find \(x\):
\[x = -\frac{3}{2}\].

Simplifying helps reduce errors and keeps your work organized. The final result, \(x = -\frac{3}{2}\), is much easier to understand.