Question

1. solve for x and categorize the solution(s) of

\\(\frac{4}{3(x+1)} = \frac{12}{x^2 - 1}\\)
\\(x=-10\\)
\\(x=1\\)
\\(x=-1\\)
\\(x=0\\)
\\(x=10\\)
solution(s)
answer by dragging
extraneous solution(s)
answer by dragging
answer by dragging

Explanation:

Step1: Factor the denominator

Notice that \(x^2 - 1\) is a difference of squares, so \(x^2 - 1=(x + 1)(x - 1)\). The equation becomes \(\frac{4}{3(x + 1)}=\frac{12}{(x + 1)(x - 1)}\).

Step2: Exclude undefined values

The original equation is undefined when \(x=-1\) (because it makes the denominator \(3(x + 1)\) and \((x + 1)(x - 1)\) zero) and when \(x = 1\) (because it makes \((x + 1)(x - 1)\) zero). So \(x
eq\pm1\).

Step3: Cross - multiply to solve for \(x\)

Cross - multiply: \(4(x^2 - 1)=12\times3(x + 1)\).
Simplify: \(4(x^2 - 1)=36(x + 1)\).
Divide both sides by 4: \(x^2 - 1 = 9(x + 1)\).
Expand the right - hand side: \(x^2 - 1=9x + 9\).
Rearrange to form a quadratic equation: \(x^2-9x - 10 = 0\).
Factor the quadratic equation: \((x - 10)(x+1)=0\).
Set each factor equal to zero: \(x - 10 = 0\) or \(x+1 = 0\).
We get \(x = 10\) or \(x=-1\). But we know from step 2 that \(x=-1\) is an extraneous solution (it makes the original equation undefined), and \(x = 10\) is a valid solution. Wait, there is a mistake in the previous calculation. Let's redo the cross - multiplication:

Starting from \(\frac{4}{3(x + 1)}=\frac{12}{(x + 1)(x - 1)}\), since \(x
eq - 1\), we can multiply both sides by \(3(x + 1)(x - 1)\) (the least common denominator) to get:

\(4(x - 1)=12\times3\)

\(4x-4 = 36\)

\(4x=40\)

\(x = 10\)

Now, check the other values:

• For \(x=-1\): The original equation has denominators \(3(-1 + 1)=0\) and \((-1)^2-1 = 0\), so \(x=-1\) is extraneous.
• For \(x = 1\): The denominator \((1)^2-1 = 0\), so \(x = 1\) is extraneous.
• For \(x = 0\): Substitute into the original equation: \(\frac{4}{3(0 + 1)}=\frac{4}{3}\), \(\frac{12}{0^2-1}=-12\), \(\frac{4}{3}

eq - 12\), so \(x = 0\) is not a solution.

• For \(x=-10\): Substitute into the original equation: \(\frac{4}{3(-10 + 1)}=\frac{4}{3\times(-9)}=-\frac{4}{27}\), \(\frac{12}{(-10)^2-1}=\frac{12}{99}=\frac{4}{33}\), \(-\frac{4}{27}

eq\frac{4}{33}\), so \(x=-10\) is not a solution.

Wait, there is a mistake in the initial cross - multiplication. Let's do it correctly:

\(\frac{4}{3(x + 1)}=\frac{12}{x^2 - 1}\), \(x^2-1=(x + 1)(x - 1)\)

Multiply both sides by \(3(x + 1)(x - 1)\) ( \(x
eq\pm1\)):

\(4(x - 1)=12\times3\)

\(4x-4 = 36\)

\(4x=40\)

\(x = 10\)

Now, let's check the given options:

• \(x = 10\) is a solution.
• \(x=-1\) is extraneous (as it makes the denominator zero).
• \(x = 1\) is extraneous (as it makes the denominator zero).
• \(x = 0\) is not a solution.
• \(x=-10\) is not a solution.

Answer:

Solution: \(x = 10\)
Extraneous Solution: \(x=-1\), \(x = 1\)

(If we consider the options given in the problem, there is a mistake in the options. But following the correct solution process, the valid solution is \(x = 10\) and the extraneous solutions are \(x=-1\) and \(x = 1\))