https://file.aiquickdraw.com/mathbot/file/1725439212167r3582xpw.jpg

Let's solve the equations provided in the image one by one and verify the solutions.

### 1. \(\frac{1}{5} = \frac{x - 3}{2x + 2}\)

First, cross-multiply to eliminate the fractions:
\[
1 \cdot (2x + 2) = 5 \cdot (x - 3)
\]
\[
2x + 2 = 5x - 15
\]

Now, isolate \(x\):
\[
2x - 5x = -15 - 2
\]
\[
-3x = -17
\]
\[
x = \frac{17}{3}
\]

**Solution:** \(x = \frac{17}{3}\)

**Check:** Substitute \(x = \frac{17}{3}\) into the original equation:
\[
\frac{1}{5} = \frac{\frac{17}{3} - 3}{2 \cdot \frac{17}{3} + 2}
\]

Simplify:
\[
\frac{1}{5} = \frac{\frac{8}{3}}{\frac{40}{3}} = \frac{8}{40} = \frac{1}{5}
\]
The solution is correct.

### 2. \(\frac{2x + 1}{x - 3} = \frac{5}{3}\)

Cross-multiply:
\[
3(2x + 1) = 5(x - 3)
\]
\[
6x + 3 = 5x - 15
\]

Isolate \(x\):
\[
6x - 5x = -15 - 3
\]
\[
x = -18
\]

**Solution:** \(x = -18\)

**Check:** Substitute \(x = -18\) into the original equation:
\[
\frac{2(-18) + 1}{-18 - 3} = \frac{-36 + 1}{-21} = \frac{-35}{-21} = \frac{5}{3}
\]
The solution is correct.

### 3. \(\frac{x - 3}{x - 2} = \frac{x + 1}{x}\)

Cross-multiply:
\[
x(x - 3) = (x - 2)(x + 1)
\]
\[
x^2 - 3x = x^2 - x - 2x - 2
\]
\[
x^2 - 3x = x^2 - 3x - 2
\]

Simplify and isolate \(x\):
\[
0 = -2 \quad \text{(this statement is false)}
\]
Thus, **no solution** exists for this equation.

### 4. \(\frac{x + 2}{4} = \frac{x - 3}{3}\)

Cross-multiply:
\[
3(x + 2) = 4(x - 3)
\]
\[
3x + 6 = 4x - 12
\]

Isolate \(x\):
\[
3x - 4x = -12 - 6
\]
\[
-x = -18
\]
\[
x = 18
\]

**Solution:** \(x = 18\)

**Check:** Substitute \(x = 18\) into the original equation:
\[
\frac{18 + 2}{4} = \frac{20}{4} = 5, \quad \frac{18 - 3}{3} = \frac{15}{3} = 5
\]
The solution is correct.

### 5. \(\frac{6}{5} = \frac{8x + 5}{2x + 5}\)

Cross-multiply:
\[
6(2x + 5) = 5(8x + 5)
\]
\[
12x + 30 = 40x + 25
\]

Isolate \(x\):
\[
12x - 40x = 25 - 30
\]
\[
-28x = -5
\]
\[
x = \frac{5}{28}
\]

**Solution:** \(x = \frac{5}{28}\)

**Check:** Substitute \(x = \frac{5}{28}\) into the original equation:
\[
\frac{6}{5} = \frac{8 \cdot \frac{5}{28} + 5}{2 \cdot \frac{5}{28} + 5}
\]

Simplify:
\[
\frac{6}{5} = \frac{\frac{40}{28} + \frac{140}{28}}{\frac{10}{28} + \frac{140}{28}} = \frac{\frac{180}{28}}{\frac{150}{28}} = \frac{180}{150} = \frac{6}{5}
\]
The solution is correct.

---

Would you like further details or have any questions?

Here are some related questions you might find interesting:
1. How can we solve systems of linear equations using cross-multiplication?
2. What does it mean when an equation has no solution?
3. How do you solve equations involving absolute values?
4. What are the methods to solve quadratic equations?
5. How does multiplying both sides of an equation by the denominator help in solving the equation?

**Tip:** Always check your solutions by substituting them back into the original equation to ensure they satisfy the equation.

Solve each equation and check the solution.

This page provides step-by-step solutions for solving various linear equations using cross-multiplication. The solutions involve isolating variables, simplifying expressions, and verifying the results. Suitable for students in Grades 7-9.

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