91Ó°ÊÓ

The substitution method is a powerful tool for solving complex equations by transforming them into simpler ones. For the given equation, we are starting with \(10 x^{-2} + 33 x^{-1} - 7 = 0\). Solving this directly can be difficult. So, we use substitution to convert it into an easier form.

We set \(y = x^{-1}\). This means everywhere we see \(x^{-1}\), we replace it with \(y\). Doing this turns our original equation into \(10y^2 + 33y - 7 = 0\).

Notice how more manageable this new equation is. Using substitution can make solving equations less intimidating and easier to understand.

The quadratic formula is instrumental in solving quadratic equations, like the one we obtained through substitution: \(10y^2 + 33y - 7 = 0\). The formula to solve such an equation is:

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \(a\), \(b\), and \(c\) are coefficients from the quadratic equation \(ax^2 + bx + c = 0\). For our equation: \(a = 10\), \(b = 33\), and \(c = -7\).

We first calculate the discriminant, \(b^2 - 4ac\):
\[ 33^2 - 4(10)(-7) = 1089 + 280 = 1369 \]

With the discriminant calculated, we can now find \(y\):
\[ y = \frac{-33 \pm \sqrt{1369}}{20} \]
This solves to:
\[ y = \frac{-33 + 37}{20} = \frac{4}{20} = \frac{1}{5} \]
\[ y = \frac{-33 - 37}{20} = \frac{-70}{20} = -3.5 \]

Hence, the solutions for \(y\) are \(\frac{1}{5}\) and \(-3.5\). Using the quadratic formula consistently helps find roots of quadratic equations efficiently.

After solving for \(y\), we must revert to the original variable \(x\) using back-substitution. Remember, our substitution was \(y = x^{-1}\), i.e., \(y = \frac{1}{x}\).

For \(y = \frac{1}{5}\):
We know \(\frac{1}{x} = \frac{1}{5}\), so \(x = 5\).

For \(y = -3.5\):
We know \(\frac{1}{x} = -3.5\), so \(x = \frac{1}{-3.5} = -\frac{2}{7}\).

Thus, the final solutions for \(x\) are \(x = 5\) and \(x = -\frac{2}{7}\).

The back-substitution process transforms back the solutions we obtained for the substituted variable into solutions for the original variable. This ensures our final answers are relevant to the initial problem.