https://file.aiquickdraw.com/mathbot/file/17304978445575316244u.jpg

The problem in the image presents a system of linear equations. Let's break down and solve it step-by-step.

### System of Equations
The system given in the image appears as:
\[
5x - 2y = 4
\]
\[
\frac{x}{5} - \frac{y}{6} = -\frac{1}{6}
\]

We will solve this system for \( x \) and \( y \).

### Step 1: Solve the First Equation for \( x \)
From the first equation:
\[
5x - 2y = 4
\]
We can express \( x \) in terms of \( y \) (or vice versa) or use it directly in substitution.

### Step 2: Rewrite the Second Equation in a More Usable Form
For the second equation:
\[
\frac{x}{5} - \frac{y}{6} = -\frac{1}{6}
\]
Let's eliminate the fractions by multiplying every term by the least common multiple of 5 and 6, which is 30:
\[
30 \cdot \frac{x}{5} - 30 \cdot \frac{y}{6} = 30 \cdot -\frac{1}{6}
\]
This simplifies to:
\[
6x - 5y = -5
\]

### Revised System
Now, our system is:
\[
5x - 2y = 4
\]
\[
6x - 5y = -5
\]

### Step 3: Solve by Elimination or Substitution
Let's use the elimination method to solve for \( x \) and \( y \).

1. Multiply the first equation by 6 and the second equation by 5 to align the coefficients of \( x \):
   \[
   6(5x - 2y) = 6 \cdot 4 \Rightarrow 30x - 12y = 24
   \]
   \[
   5(6x - 5y) = 5 \cdot -5 \Rightarrow 30x - 25y = -25
   \]

2. Subtract the first modified equation from the second:
   \[
   (30x - 25y) - (30x - 12y) = -25 - 24
   \]
   This simplifies to:
   \[
   -13y = -49
   \]
   \[
   y = \frac{49}{13}
   \]

3. Substitute \( y = \frac{49}{13} \) back into the first equation to solve for \( x \):
   \[
   5x - 2\left(\frac{49}{13}\right) = 4
   \]
   \[
   5x - \frac{98}{13} = 4
   \]
   Multiply every term by 13 to clear the fraction:
   \[
   65x - 98 = 52
   \]
   \[
   65x = 150
   \]
   \[
   x = \frac{150}{65} = \frac{30}{13}
   \]

### Solution
The solution to the system is:
\[
x = \frac{30}{13}, \quad y = \frac{49}{13}
\]

Would you like a more detailed explanation of any specific step, or have any questions?

### Related Questions
1. How would the solution change if the signs in one equation were altered?
2. What methods can be used to check if the solution to a system of equations is correct?
3. How does the elimination method differ from the substitution method?
4. What are some applications of solving systems of equations in real life?
5. How would you solve this system if only one equation were linear?

### Tip
When dealing with fractions in systems of equations, try multiplying through by the least common multiple to simplify calculations.

Решите систему уравнений: 5x - 2y = 4 и (x / 5) - (y / 6) = -1 / 6

Learn how to solve a system of linear equations with fractions in algebra: 5x - 2y = 4 and (x / 5) - (y / 6) = -1 / 6. This solution involves the elimination method and fraction elimination techniques, suitable for students in Grades 9-11.

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