Question

cphs : advanced algebra: concepts and connections - block (27.0831030)
rational equations
what are the solutions to the equation $x - \frac{7}{x} = 6$?
$x = -6$ and $x = -1$
$x = -1$ and $x = 7$
$x = 1$ and $x = 6$
$x = -7$ and $x = 1$

Explanation:

Step1: Multiply both sides by \( x \) ( \( x

eq 0 \) )
To eliminate the denominator, we multiply each term in the equation \( x - \frac{7}{x} = 6 \) by \( x \). This gives us \( x \cdot x - x \cdot \frac{7}{x} = 6 \cdot x \). Simplifying each term, we have \( x^2 - 7 = 6x \).

Step2: Rearrange into standard quadratic form

We rearrange the equation \( x^2 - 7 = 6x \) to the standard quadratic form \( ax^2 + bx + c = 0 \). Subtracting \( 6x \) from both sides, we get \( x^2 - 6x - 7 = 0 \).

Step3: Factor the quadratic equation

We factor the quadratic \( x^2 - 6x - 7 \). We need two numbers that multiply to \( -7 \) and add to \( -6 \). These numbers are \( -7 \) and \( 1 \). So, the factored form is \( (x - 7)(x + 1) = 0 \).

Step4: Solve for \( x \)

Using the zero - product property, if \( (x - 7)(x + 1) = 0 \), then either \( x - 7 = 0 \) or \( x + 1 = 0 \). Solving \( x - 7 = 0 \) gives \( x = 7 \), and solving \( x + 1 = 0 \) gives \( x = - 1 \). We also need to check that these solutions do not make the original denominator zero. For \( x = 7 \) and \( x=-1 \), the denominator \( x
eq0 \), so both are valid.

Answer:

\( x = - 1 \) and \( x = 7 \) (the option with \( x=-1 \) and \( x = 7 \))