Question

solve for all values of x.
$x - \dfrac{x + 5}{x + 7} = \dfrac{2}{x + 7}$

Explanation:

Step1: Identify the common denominator

The denominators are \(x + 7\) and \(x + 7\) (wait, the original equation is \(x-\frac{x + 5}{x + 7}=\frac{2}{x + 7}\)? Wait, maybe I misread. Let's re - express the equation. The equation is \(x-\frac{x + 5}{x + 7}=\frac{2}{x + 7}\). The common denominator for the fractions is \(x + 7\), and we also note that \(x+7
eq0\) (i.e., \(x
eq - 7\)) to avoid division by zero.
Multiply each term in the equation by \(x + 7\) to eliminate the denominators:
\(x(x + 7)-(x + 5)=2\)

Step2: Expand the left - hand side

Expand \(x(x + 7)\) using the distributive property \(a(b + c)=ab+ac\), where \(a = x\), \(b=x\), and \(c = 7\). So \(x(x + 7)=x^{2}+7x\).
The equation becomes \(x^{2}+7x-(x + 5)=2\).
Remove the parentheses: \(x^{2}+7x - x-5 = 2\).

Step3: Simplify the left - hand side

Combine like terms: \(x^{2}+(7x - x)-5=2\), which simplifies to \(x^{2}+6x-5 = 2\).

Step4: Rearrange the equation to standard quadratic form

Subtract 2 from both sides of the equation: \(x^{2}+6x-5 - 2=0\), so \(x^{2}+6x - 7=0\).

Step5: Factor the quadratic equation

We need to find two numbers that multiply to \(-7\) and add up to \(6\). The numbers are \(7\) and \(-1\) because \(7\times(-1)=-7\) and \(7+( - 1)=6\).
So, \(x^{2}+6x - 7=(x + 7)(x - 1)=0\).

Step6: Solve for x

Set each factor equal to zero:

• If \(x+7 = 0\), then \(x=-7\). But we know from Step 1 that \(x

eq - 7\) (because it makes the original denominator zero), so we discard this solution.

• If \(x - 1=0\), then \(x = 1\).

Answer:

\(x = 1\)