Question

olving a rational equation using the lcd
solve: \\(\frac{x^2 - x - 6}{x^2} = \frac{x - 6}{2x} + \frac{2x + 12}{x}\\)
what are the solutions to the equation?
\\(x = -6\\) and \\(x = -\frac{2}{3}\\) \\(x = 6\\) and \\(x = \frac{2}{3}\\) \\(x = -6\\) and \\(x = \frac{2}{3}\\)

Explanation:

Step1: Find the LCD

The denominators are \(x^2\), \(2x\), and \(x\). The least common denominator (LCD) is \(2x^2\) (since we take the highest power of each factor: for \(x\) it's \(x^2\), for 2 it's 2).

Step2: Multiply each term by LCD

Multiply each term in the equation \(\frac{x^2 - x - 6}{x^2}=\frac{x - 6}{2x}+\frac{2x + 12}{x}\) by \(2x^2\):

• For the left term: \(2x^2\cdot\frac{x^2 - x - 6}{x^2}=2(x^2 - x - 6)\)
• For the middle term: \(2x^2\cdot\frac{x - 6}{2x}=x(x - 6)\)
• For the right term: \(2x^2\cdot\frac{2x + 12}{x}=2x(2x + 12)\)

So the equation becomes: \(2(x^2 - x - 6)=x(x - 6)+2x(2x + 12)\)

Step3: Expand all terms

Expand each part:

• Left: \(2x^2-2x - 12\)
• Middle: \(x^2-6x\)
• Right: \(4x^2 + 24x\)

Now the equation is \(2x^2-2x - 12=x^2-6x + 4x^2 + 24x\)

Step4: Combine like terms

Combine the right - hand side: \(x^2+4x^2=5x^2\), \(-6x + 24x = 18x\). So the equation is \(2x^2-2x - 12=5x^2+18x\)

Subtract \(2x^2-2x - 12\) from both sides to set the equation to zero:
\(0 = 5x^2+18x-(2x^2-2x - 12)\)
\(0 = 5x^2+18x - 2x^2 + 2x+12\)
Combine like terms: \(5x^2-2x^2=3x^2\), \(18x + 2x=20x\). So \(3x^2+20x + 12 = 0\)

Step5: Solve the quadratic equation

We can factor the quadratic \(3x^2+20x + 12\). We need two numbers that multiply to \(3\times12 = 36\) and add up to 20. The numbers are 18 and 2.
So we rewrite the middle term: \(3x^2+18x+2x + 12 = 0\)
Factor by grouping:
\(3x(x + 6)+2(x + 6)=0\)
\((3x + 2)(x + 6)=0\)

Set each factor equal to zero:

• \(3x+2 = 0\Rightarrow3x=-2\Rightarrow x=-\frac{2}{3}\)
• \(x + 6=0\Rightarrow x=-6\)

We also need to check for extraneous solutions because the original equation has denominators \(x^2\), \(2x\), and \(x\). When \(x = 0\), the denominators are zero, but our solutions \(x=-6\) and \(x =-\frac{2}{3}\) do not make the denominators zero.

Answer:

\(x=-6\) and \(x =-\frac{2}{3}\) (corresponding to the first option: \(x=-6\) and \(x =-\frac{2}{3}\))