Problem 15 - Precalculus Student Solutions Manual 5th

Question

Solve each equation. $$\frac{2 x+1}{x-2}+\frac{3}{x}=\frac{-6}{x^{2}-2 x}$$

Step-by-Step Solution

Answer

The solutions are \(x = 0\) and \(x = -2\).

Step 1: Identify a common denominator

The equation involves fractions with different denominators. The left side has denominators of \(x-2\) and \(x\), and the right side has \(x^2 - 2x = x(x-2)\). Identify that \(x(x-2)\) can be used as a common denominator.

Step 2: Rewrite each term with the common denominator

Rewrite the equation so that each term has the common denominator \(x(x-2)\): ewline \[\frac{(2x+1)x}{x(x-2)} + \frac{3(x-2)}{x(x-2)} = \frac{-6}{x(x-2)}\]

Step 3: Combine the fractions

Combine the fractions on the left side of the equation: ewline \[\frac{(2x+1)x + 3(x-2)}{x(x-2)} = \frac{-6}{x(x-2)}\] ewline Simplify the numerator on the left side: \[2x^2 + x + 3x - 6\] which combines to \[2x^2 + 4x - 6\]

Step 4: Set up the equation

Since the denominators are the same, you can set the numerators equal to each other: ewline \[2x^2 + 4x - 6 = -6\]

Step 5: Solve the quadratic equation

To solve the quadratic equation, first add 6 to both sides: ewline \[2x^2 + 4x - 6 + 6 = 0\] ewline Simplifying to: ewline \[2x^2 + 4x = 0\]

Step 6: Factor out the greatest common factor

Factor out \(2x\): ewline \[2x(x + 2) = 0\]

Step 7: Solve for x

Set each factor equal to zero and solve: ewline \[2x = 0\] so \(x = 0\) ewline \[x + 2 = 0\] so \(x = -2\)

Key Concepts

common denominator

When solving equations involving fractions, a crucial step is to identify a common denominator. This allows you to combine the fractions more easily. The common denominator should be a value that all individual denominators can divide into without leaving a remainder.

In our exercise, the denominators on the left side are \(x-2\) and \(x\), and on the right side, it's \(x^2 - 2x = x(x-2)\). Therefore, the common denominator for this equation is \x(x-2)\. Using this common denominator will make it easier to rewrite the original fractions and eventually solve the equation.

Here is a quick recap to identify a common denominator:

• Find the least common multiple (LCM) of all denominators.
• Rewrite each fraction so each one has the common denominator.
• Combine the fractions and proceed with the solution.

quadratic equation

A quadratic equation is of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents an unknown variable. Solving a quadratic equation involves finding the value(s) of \(x\) that satisfy the equation.

In the given example, once we find a common denominator and combine the terms, we transform the equation into a quadratic form: \[2x^2 + 4x - 6 = -6\]. Simplifying further, we add 6 to both sides to get: \[2x^2 + 4x = 0\].

To solve quadratic equations:

• Bring all terms to one side so that the equation equals zero.
• Factor the quadratic equation if possible.
• Use the quadratic formula \(-b \pm \sqrt{b^2 - 4ac}) / 2a\) if factoring is challenging.
• Set each factor equal to zero and solve for \x\.

factoring

Factoring is the process of breaking down an expression into a product of simpler expressions. It's a pivotal step in solving quadratic equations. In our exercise, we end up with the quadratic expression \[2x^2 + 4x = 0\]

To factor out the greatest common factor, we notice both terms in the quadratic expression share a common factor which is \2x\.

Factoring it out, we get: \[2x(x + 2) = 0\]. Now, we have two separate expressions \2x\ and \(x + 2\) to solve:

• Set \2x = 0\, giving the solution \x=0\.
• Set \x + 2 = 0\, leading to the solution \x=-2\.
• These solutions are the roots of the quadratic equation.

Factoring helps simplify complex equations and is a strategy you’ll use often in algebra and precalculus problems. If the quadratic equation doesn't factor easily, remember you can always use the quadratic formula.