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The image contains a system of linear equations to solve for \( x \) and \( y \). The equations are:

\[
\frac{3}{4}x - \frac{1}{3}y = 5
\]
\[
\frac{1}{2}x + \frac{2}{3}y = -2
\]

We are asked to solve for \( x = a \) and \( y = b \), and then determine the value of \( a - 3b \).

Let me solve this system of equations step by step.

### 1. Eliminate the fractions
To make the calculations easier, we can multiply both equations by the least common denominator of the fractions involved.

- The first equation: Multiply by 12 (LCM of 4 and 3)
  \[
  12 \times \left( \frac{3}{4}x - \frac{1}{3}y \right) = 12 \times 5
  \]
  This simplifies to:
  \[
  9x - 4y = 60
  \]

- The second equation: Multiply by 6 (LCM of 2 and 3)
  \[
  6 \times \left( \frac{1}{2}x + \frac{2}{3}y \right) = 6 \times (-2)
  \]
  This simplifies to:
  \[
  3x + 4y = -12
  \]

### 2. Add the equations to eliminate \( y \)
Now, we can add both equations:
\[
(9x - 4y) + (3x + 4y) = 60 + (-12)
\]
This simplifies to:
\[
12x = 48
\]
So, we find:
\[
x = 4
\]

### 3. Substitute \( x = 4 \) into one of the equations
Let's substitute \( x = 4 \) into the second equation:
\[
\frac{1}{2}x + \frac{2}{3}y = -2
\]
Substitute \( x = 4 \):
\[
\frac{1}{2}(4) + \frac{2}{3}y = -2
\]
This simplifies to:
\[
2 + \frac{2}{3}y = -2
\]
Subtract 2 from both sides:
\[
\frac{2}{3}y = -4
\]
Multiply both sides by \( \frac{3}{2} \):
\[
y = -6
\]

### 4. Calculate \( a - 3b \)
We now have \( a = 4 \) and \( b = -6 \). We need to find \( a - 3b \):
\[
a - 3b = 4 - 3(-6) = 4 + 18 = 22
\]

### Final Answer
The value of \( a - 3b \) is \( 22 \).

Would you like further details or have any other questions? Here are 5 related problems you might want to try:

1. Solve a similar system of linear equations using substitution method.
2. Solve the system \( 2x - 3y = 7 \) and \( 4x + y = 1 \).
3. What happens if both equations are multiplied by different constants and then subtracted?
4. Can you solve a system of three variables like \( x + y + z = 6 \), \( 2x - y + z = 3 \), and \( x - 2y + 2z = -1 \)?
5. What are the conditions for a system of linear equations to have no solution or infinite solutions?

**Tip**: When solving systems of linear equations, multiplying to eliminate fractions can simplify calculations and reduce the chance of errors.

Solve the system of linear equations: (3/4)x - (1/3)y = 5 and (1/2)x + (2/3)y = -2, then find the value of a - 3b where x = a and y = b.

This problem involves solving a system of linear equations: (3/4)x - (1/3)y = 5 and (1/2)x + (2/3)y = -2. Learn the step-by-step process using the elimination method to find x = a and y = b, and calculate a - 3b. Suitable for students in Grades 9-11.

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