Question

1. solve the rational equation \\(\frac{4x}{x + 1} - \frac{3x}{x - 5} = \frac{-132}{x^2 - 4x - 5}\\). show all work. verify the solution(s).

Explanation:

Step1: Factor the denominator

First, factor the quadratic denominator \(x^2 - 4x - 5\). We need two numbers that multiply to \(-5\) and add to \(-4\). Those numbers are \(-5\) and \(1\), so \(x^2 - 4x - 5=(x + 1)(x - 5)\). The equation becomes:
\(\frac{4x}{x + 1}-\frac{3x}{x - 5}=\frac{-132}{(x + 1)(x - 5)}\)

Step2: Find the least common denominator (LCD)

The denominators are \(x + 1\), \(x - 5\), and \((x + 1)(x - 5)\). So the LCD is \((x + 1)(x - 5)\). Multiply each term by the LCD to eliminate the fractions:
\((x + 1)(x - 5)\cdot\frac{4x}{x + 1}-(x + 1)(x - 5)\cdot\frac{3x}{x - 5}=(x + 1)(x - 5)\cdot\frac{-132}{(x + 1)(x - 5)}\)

Step3: Simplify each term

Simplify each term after multiplying by the LCD:

• For the first term: \((x + 1)(x - 5)\cdot\frac{4x}{x + 1}=4x(x - 5)\)
• For the second term: \((x + 1)(x - 5)\cdot\frac{3x}{x - 5}=3x(x + 1)\)
• For the third term: \((x + 1)(x - 5)\cdot\frac{-132}{(x + 1)(x - 5)}=-132\)

So the equation simplifies to:
\(4x(x - 5)-3x(x + 1)=-132\)

Step4: Expand the brackets

Expand \(4x(x - 5)\) and \(3x(x + 1)\):

• \(4x(x - 5)=4x^2-20x\)
• \(3x(x + 1)=3x^2+3x\)

Substitute back into the equation:
\(4x^2-20x-(3x^2+3x)=-132\)
\(4x^2-20x - 3x^2-3x=-132\)

Step5: Combine like terms

Combine the \(x^2\) terms and the \(x\) terms:
\((4x^2-3x^2)+(-20x-3x)=-132\)
\(x^2-23x=-132\)

Step6: Rearrange into standard quadratic form

Add \(132\) to both sides to get the quadratic equation in standard form \(ax^2+bx + c = 0\):
\(x^2-23x + 132 = 0\)

Step7: Solve the quadratic equation

Factor the quadratic \(x^2-23x + 132\). We need two numbers that multiply to \(132\) and add to \(-23\). Those numbers are \(-11\) and \(-12\), so:
\(x^2-23x + 132=(x - 11)(x - 12)=0\)

Set each factor equal to zero:

• \(x - 11 = 0\) gives \(x = 11\)
• \(x - 12 = 0\) gives \(x = 12\)

Step8: Verify the solutions

We need to check if these solutions make the original denominators zero. The denominators are \(x + 1\), \(x - 5\), and \((x + 1)(x - 5)\).

• For \(x = 11\): \(x + 1 = 12

eq0\), \(x - 5 = 6
eq0\), so it's valid.

• For \(x = 12\): \(x + 1 = 13

eq0\), \(x - 5 = 7
eq0\), so it's valid.

Answer:

The solutions are \(x = 11\) and \(x = 12\)