Question

solve the equation.\\(\dfrac{6}{x^2 - 6x - 7} - \dfrac{3}{x^2 - 49} = \dfrac{4}{x^2 + 8x + 7}\\)

Explanation:

Step 1: Factor the denominators

First, factor each quadratic denominator:

• For \(x^{2}-6x - 7\), we find two numbers that multiply to \(-7\) and add to \(-6\). The numbers are \(-7\) and \(1\), so \(x^{2}-6x - 7=(x - 7)(x+1)\).
• For \(x^{2}-49\), this is a difference of squares, so \(x^{2}-49=(x - 7)(x + 7)\).
• For \(x^{2}+8x + 7\), we find two numbers that multiply to \(7\) and add to \(8\). The numbers are \(7\) and \(1\), so \(x^{2}+8x + 7=(x + 7)(x+1)\).

The equation becomes:
\(\frac{6}{(x - 7)(x + 1)}-\frac{3}{(x - 7)(x + 7)}=\frac{4}{(x + 7)(x + 1)}\)

Step 2: Find the least common denominator (LCD)

The LCD of \((x - 7)(x + 1)\), \((x - 7)(x + 7)\), and \((x + 7)(x + 1)\) is \((x - 7)(x + 7)(x + 1)\).

Step 3: Multiply each term by the LCD

Multiply each term by \((x - 7)(x + 7)(x + 1)\) to eliminate the denominators:

• For the first term: \(\frac{6}{(x - 7)(x + 1)}\times(x - 7)(x + 7)(x + 1)=6(x + 7)\)
• For the second term: \(-\frac{3}{(x - 7)(x + 7)}\times(x - 7)(x + 7)(x + 1)=-3(x + 1)\)
• For the third term: \(\frac{4}{(x + 7)(x + 1)}\times(x - 7)(x + 7)(x + 1)=4(x - 7)\)

So the equation becomes:
\(6(x + 7)-3(x + 1)=4(x - 7)\)

Step 4: Expand and simplify

Expand each term:

• \(6(x + 7)=6x+42\)
• \(-3(x + 1)=-3x - 3\)
• \(4(x - 7)=4x-28\)

Now combine like terms on the left side:
\(6x + 42-3x - 3=(6x-3x)+(42 - 3)=3x+39\)

So the equation is:
\(3x+39 = 4x-28\)

Step 5: Solve for \(x\)

Subtract \(3x\) from both sides:
\(39=x - 28\)

Add \(28\) to both sides:
\(x=39 + 28=67\)

Step 6: Check for extraneous solutions

We need to check if \(x = 67\) makes any of the original denominators zero.

• For \(x^{2}-6x - 7=(67)^{2}-6\times67 - 7=4489-402 - 7=4080

eq0\)

• For \(x^{2}-49=(67)^{2}-49=4489 - 49=4440

eq0\)

• For \(x^{2}+8x + 7=(67)^{2}+8\times67+7=4489+536 + 7=5032

eq0\)

So \(x = 67\) is a valid solution.

Answer:

\(x = 67\)