https://file.aiquickdraw.com/mathbot/file/1727206790758hbja2lpu.jpg

Let's solve the equations provided in the image.

### 2. Solve:
\[
\frac{2x - 7}{2x + 4} = \frac{2}{3}
\]

To solve this, let's cross-multiply:
\[
3(2x - 7) = 2(2x + 4)
\]

Expanding both sides:
\[
6x - 21 = 4x + 8
\]

Now, move all terms involving \(x\) to one side and constants to the other:
\[
6x - 4x = 8 + 21
\]
\[
2x = 29
\]

Solving for \(x\):
\[
x = \frac{29}{2}
\]

So, the solution for the second equation is:
\[
x = \frac{29}{2}
\]

---

### 3. Solve:
\[
\frac{1}{x + 3} + \frac{5}{x} = \frac{2}{x - 3}
\]

To solve this, let's first find a common denominator for the fractions. The common denominator will be \((x + 3)(x)(x - 3)\).

Rewrite each term with this denominator:
\[
\frac{(x)(x - 3)}{x(x + 3)(x - 3)} + \frac{5(x + 3)(x - 3)}{x(x + 3)(x - 3)} = \frac{2(x + 3)(x)}{(x - 3)(x)(x + 3)}
\]

Now that the denominators are the same, we can equate the numerators:
\[
x(x - 3) + 5(x + 3)(x - 3) = 2(x + 3)(x)
\]

Expanding each term:
\[
x^2 - 3x + 5((x^2 - 9)) = 2(x^2 + 3x)
\]

Expanding further:
\[
x^2 - 3x + 5x^2 - 45 = 2x^2 + 6x
\]

Now, combine like terms:
\[
6x^2 - 3x - 45 = 2x^2 + 6x
\]

Move all terms to one side:
\[
6x^2 - 3x - 45 - 2x^2 - 6x = 0
\]
\[
4x^2 - 9x - 45 = 0
\]

This is a quadratic equation. We can solve it using the quadratic formula:
\[
x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(4)(-45)}}{2(4)}
\]
\[
x = \frac{9 \pm \sqrt{81 + 720}}{8}
\]
\[
x = \frac{9 \pm \sqrt{801}}{8}
\]

Thus, the solutions for \(x\) are:
\[
x = \frac{9 \pm \sqrt{801}}{8}
\]

---

### 4. Solve:
\[
\frac{2x - 4}{2x - 4} - 2 = \frac{1}{x - 2}
\]

Since \(\frac{2x - 4}{2x - 4} = 1\) (assuming \(x \neq 2\)), the equation simplifies to:
\[
1 - 2 = \frac{1}{x - 2}
\]
\[
-1 = \frac{1}{x - 2}
\]

Now, multiply both sides by \(x - 2\):
\[
-1(x - 2) = 1
\]
\[
- x + 2 = 1
\]
\[
- x = -1
\]
\[
x = 1
\]

So, the solution for the fourth equation is:
\[
x = 1
\]

---

Let me know if you need more details or explanations on these solutions!

Here are some related questions for practice:
1. Can you cross-multiply both sides of any equation to solve it?
2. How do you find the common denominator when working with fractions?
3. How does the quadratic formula work for different kinds of equations?
4. What are the restrictions for \(x\) in rational equations?
5. What are the steps to simplify complex rational equations?

**Tip:** Always check for restrictions in rational equations (e.g., values that make the denominator zero) before proceeding with the solution!

Solve the equations provided in the image: 2x - 7 / 2x + 4 = 2 / 3, (1 / x + 3) + 5 / x^2 - 9 = 2 / x - 3, 2x - 4 / 2x - 4 - 2 = 1 / x - 2

Step-by-step solutions for solving rational and quadratic equations involving cross-multiplication, common denominators, and quadratic formula applications. Problems include algebraic manipulation and finding restrictions on variables to avoid undefined expressions. Suitable for high school students (Grades 9-12).

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