Question

lesson 25 review - solving rational equations
3
\\(\frac{x}{x - 2} - \frac{2}{x + 4} = \frac{12}{x^2 + 2x - 8}\\)
lcd = \\((x + 4)(x - 2)\\)
① factor denominators
② find lcd
③ multiply to create lcd
④ solve using numerator
⑤ check for extraneous solutions!
multiple - choice question
your turn now. this step should be worked out on paper first.
we should now have \\(x(x + 4) - 2(x - 2) = 12\\) on our papers.
if you perform distributive property where needed, combine like terms on the left, and then bring over the 12 to the left, which equivalent simplified equation will result?
\\(\square\\) \\(x^2 + 2x - 4 = 12\\)
\\(\square\\) \\(x^2 - 2x + 8 = 0\\)
\\(\square\\) \\(x^2 - 2x - 8 = 0\\)
\\(\square\\) \\(x^2 + 2x - 8 = 0\\)

Explanation:

Step1: Distribute terms

Distribute \( x \) in \( x(x + 4) \) to get \( x^2 + 4x \), and distribute \( -2 \) in \( -2(x - 2) \) to get \( -2x + 4 \). So the left - hand side becomes \( x^2+4x - 2x + 4 \).

Step2: Combine like terms

Combine the like terms \( 4x-2x = 2x \) on the left - hand side. So the left - hand side is \( x^2+2x + 4 \), and the equation is \( x^2+2x + 4=12 \).

Step3: Subtract 12 from both sides

Subtract 12 from both sides of the equation \( x^2 + 2x+4 = 12 \). We get \( x^2+2x + 4-12=0 \), which simplifies to \( x^2+2x - 8 = 0 \)? Wait, no, wait. Wait, original equation after multiplying by LCD is \( x(x + 4)-2(x - 2)=12 \). Let's re - do the distribution:

\( x(x + 4)=x^{2}+4x \), \( -2(x - 2)=-2x + 4 \). Then \( x^{2}+4x-2x + 4 = 12 \). Combine like terms: \( 4x-2x = 2x \), so \( x^{2}+2x + 4=12 \). Now, subtract 12 from both sides: \( x^{2}+2x+4 - 12=0\Rightarrow x^{2}+2x - 8 = 0 \)? Wait, no, 4 - 12=-8. Wait, but let's check the options again. Wait, maybe I made a mistake. Wait, let's start over.

Wait, the equation is \( x(x + 4)-2(x - 2)=12 \)

Expand \( x(x + 4)=x^{2}+4x \)

Expand \( -2(x - 2)=-2x + 4 \)

Now, combine the left - hand side: \( x^{2}+4x-2x + 4=x^{2}+2x + 4 \)

So the equation is \( x^{2}+2x + 4 = 12 \)

Now, subtract 12 from both sides: \( x^{2}+2x+4 - 12=x^{2}+2x - 8 = 0 \)

Wait, but the option is \( x^{2}+2x - 8 = 0 \)? Wait, the options are:

1. \( x^{2}+2x - 4 = 12 \)
2. \( x^{2}-2x + 8 = 0 \)
3. \( x^{2}-2x - 8 = 0 \)
4. \( x^{2}+2x - 8 = 0 \)

Ah, I see. When we have \( x^{2}+2x + 4 = 12 \), we can also move 12 to the left: \( x^{2}+2x+4 - 12 = 0\Rightarrow x^{2}+2x - 8 = 0 \), which is the fourth option.

Wait, let's check the expansion again. \( x(x + 4)=x^{2}+4x \), \( -2(x - 2)=-2x + 4 \). So \( x^{2}+4x-2x + 4 = 12 \). \( 4x-2x = 2x \), so \( x^{2}+2x + 4 = 12 \). Then \( x^{2}+2x+4 - 12 = 0\Rightarrow x^{2}+2x - 8 = 0 \). So the correct option is the fourth one: \( x^{2}+2x - 8 = 0 \)

Answer:

\( x^{2}+2x - 8 = 0 \) (the fourth option)