what is x over x minus 2 plus x plus 7 over x plus 2 equals 5? find x

We are given the equation:

\[
\frac{x}{x - 2} + \frac{x + 7}{x + 2} = 5
\]

### Step 1: Find a common denominator.
The denominators are \(x - 2\) and \(x + 2\). The least common denominator (LCD) is the product of these two, i.e., \((x - 2)(x + 2)\).

### Step 2: Rewrite each term with the common denominator.
\[
\frac{x}{x - 2} = \frac{x(x + 2)}{(x - 2)(x + 2)} \quad \text{and} \quad \frac{x + 7}{x + 2} = \frac{(x + 7)(x - 2)}{(x - 2)(x + 2)}
\]

So, the equation becomes:
\[
\frac{x(x + 2)}{(x - 2)(x + 2)} + \frac{(x + 7)(x - 2)}{(x - 2)(x + 2)} = 5
\]

### Step 3: Combine the numerators.
Since the denominators are now the same, we can combine the numerators:
\[
\frac{x(x + 2) + (x + 7)(x - 2)}{(x - 2)(x + 2)} = 5
\]

Expand both numerators:
\[
x(x + 2) = x^2 + 2x
\]
\[
(x + 7)(x - 2) = x^2 - 2x + 7x - 14 = x^2 + 5x - 14
\]

Now, combine the terms in the numerator:
\[
x^2 + 2x + x^2 + 5x - 14 = 2x^2 + 7x - 14
\]

So, the equation becomes:
\[
\frac{2x^2 + 7x - 14}{(x - 2)(x + 2)} = 5
\]

### Step 4: Eliminate the denominator.
Multiply both sides of the equation by \((x - 2)(x + 2)\) to eliminate the denominator:
\[
2x^2 + 7x - 14 = 5(x - 2)(x + 2)
\]

### Step 5: Expand the right-hand side.
Recall that \((x - 2)(x + 2) = x^2 - 4\), so:
\[
5(x^2 - 4) = 5x^2 - 20
\]

Now the equation is:
\[
2x^2 + 7x - 14 = 5x^2 - 20
\]

### Step 6: Rearrange the equation.
Move all terms to one side:
\[
2x^2 + 7x - 14 - 5x^2 + 20 = 0
\]
Simplify:
\[
-3x^2 + 7x + 6 = 0
\]

### Step 7: Solve the quadratic equation.
We now solve \(-3x^2 + 7x + 6 = 0\) using the quadratic formula. The general form of the quadratic formula is:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For our equation, \(a = -3\), \(b = 7\), and \(c = 6\). Plug these values into the formula:
\[
x = \frac{-7 \pm \sqrt{7^2 - 4(-3)(6)}}{2(-3)}
\]
\[
x = \frac{-7 \pm \sqrt{49 + 72}}{-6}
\]
\[
x = \frac{-7 \pm \sqrt{121}}{-6}
\]
\[
x = \frac{-7 \pm 11}{-6}
\]

### Step 8: Find the two possible solutions.
1. \(x = \frac{-7 + 11}{-6} = \frac{4}{-6} = -\frac{2}{3}\)
2. \(x = \frac{-7 - 11}{-6} = \frac{-18}{-6} = 3\)

### Step 9: Check for restrictions.
The original equation has restrictions that \(x \neq 2\) and \(x \neq -2\) because these values would make the denominators zero. Neither of our solutions (\(x = -\frac{2}{3}\) and \(x = 3\)) violate these restrictions.

### Final Answer:
The solutions are:
\[
x = -\frac{2}{3} \quad \text{or} \quad x = 3
\]

Would you like any further explanation or details on the steps?

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Here are some related questions to explore:

1. How does factoring help in solving quadratic equations?
2. What are other methods to solve a quadratic equation besides the quadratic formula?
3. How do you find the least common denominator of algebraic fractions?
4. Can you explain how to check for extraneous solutions?
5. What happens when a quadratic equation has no real solutions?

**Tip:** Always check for values that make the denominator zero when solving rational equations to avoid invalid solutions.

Explore how to solve the rational equation x/(x-2) + (x+7)/(x+2) = 5 with a step-by-step solution. This guide covers key algebraic concepts like finding a common denominator, simplifying rational expressions, and solving quadratic equations using the quadratic formula. Suitable for students in Grades 9-11.

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