Problem 60 - College Algebra

Question

Solve each equation by making an appropriate substitution. $$\left(y-\frac{10}{y}\right)^{2}+6\left(y-\frac{10}{y}\right)-27-0$$

Step-by-Step Solution

Answer

The solutions for the equation are \( y = 4, -2, 5, -3 \).

Step 1: Introduction of the Substitution

Introduce the substitution variable \( z \), such that \( z = y - \frac{10}{y} \). This will simplify the equation substantially.

Step 2: Substitution into the Equation

Substitute \( z \) into the equation to obtain \( z^{2} + 6z -27 = 0 \). This is a simple quadratic equation.

Step 3: Solving the Quadratic Equation

Solve the quadratic equation \( z^{2} + 6z - 27 = 0 \) by factoring, to find that \( z = -9 \) and \( z = 3 \).

Step 4: Solving for y

Replace \( z \) with its original expression in terms of \( y \). This will generate two equations: \( -9 = y - \frac{10}{y} \) and \( 3 = y - \frac{10}{y} \). Multiply through by \( y \) to clear fractions and solve to find the solutions for \( y \).

Key Concepts

Substitution Method

The substitution method is like swapping a tricky part of our problems with something simpler. In complex mathematical problems, you often come across expressions that repeat in the equations. By identifying these repeating parts, you can introduce a new variable, making things much easier to handle.
For our problem, we introduced a new variable, \( z \), such that \( z = y - \frac{10}{y} \). By doing this, the complex expression gets transformed into something manageable, opening the door to simpler arithmetic or algebraic manipulation.

• Identify a complex or repeating expression.
• Substitute it with a new variable.
• Solve the resulting simpler equation.

After solving the simple equation, remember to return to the original variables. Unsubstituting is crucial since it connects the new solution to your initial problem.

Factoring Quadratic Equations

Factoring quadratic equations involves breaking down a more complex quadratic expression into simpler ones that, when multiplied, give the original equation. For our simplified problem \( z^2 + 6z - 27 = 0 \), we identify factors of \(-27\) that add up to \(6\), which are \(-3\) and \(9\).
By factoring, we write this as \((z + 9)(z - 3) = 0\). This approach lets us find solutions rapidly because, when any product equals zero, at least one of the factors must equal zero.

• Look for two numbers that multiply to the constant term (\(-27\)) and add to the coefficient of the linear term (\(6\)).
• Create factors \((z + 9)\) and \((z - 3)\) from these numbers.
• Solve each factor for zero to find \(z\) values: \(z = -9\) and \(z = 3\).

Factoring offers a simple yet powerful technique in algebra, especially helpful since quadratic equations are common in many math problems.

Solving Rational Equations

Rational equations, those that involve fractions or variable expressions in the denominator, can seem intimidating initially. However, solving them often involves multiplying through by a common denominator to eliminate the fractions.
In our exercise, once substitution was done, we ended up with rational expressions like \(-9 = y - \frac{10}{y}\) and \(3 = y - \frac{10}{y}\). From here:

• Multiply through by \( y \) to eliminate the fractional part.
• Rearrange and solve the resulting quadratic or linear equations.
• Check that solutions do not make the denominator zero, which would invalidate them.

Solving rational equations gives you a comprehensive understanding of both algebraic manipulation and equation solving, ensuring that you can handle not just pure quadratics, but also equations where rational expressions are involved.