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The image shows a mathematical problem with two given equations and a question asking for the sum of \(a + b + c + d\), where \(a\), \(b\), \(c\), and \(d\) are positive integers.

The equations provided are:
\[
ab + cd = 34
\]
\[
ac - bd = 19
\]

The goal is to find \(a + b + c + d\), with the provided answer options: 16, 32, 24, 12, 20.

### Solution:

We have two equations:
1. \( ab + cd = 34 \)
2. \( ac - bd = 19 \)

Let's try to solve these using trial and error, based on the fact that \(a\), \(b\), \(c\), and \(d\) are positive integers.

We will start by testing small integer values for \(a\), \(b\), \(c\), and \(d\).

### Trial: \(a = 2\)

- Plugging \(a = 2\) into the equations:
  
For the first equation:
\[
2b + 2d = 34 \quad \Rightarrow \quad b + d = 17
\]

For the second equation:
\[
2c - 2d = 19 \quad \Rightarrow \quad c - d = \frac{19}{2}
\]
This is not valid since \(c\) and \(d\) must be integers.

### Trial: \(a = 3\)

For \(a = 3\), let's solve:

1. \( 3b + 3d = 34 \Rightarrow b + d = \frac{34}{3} \) → This doesn’t work because the sum is not an integer.

Continue with this process for \(a = 2, b

ab + cd = 34, ac - bd = 19 where a, b, c, d are positive integers. Find a + b + c + d.

Learn how to solve the system of linear equations ab + cd = 34 and ac - bd = 19 using trial and error methods. This problem is suitable for students in Grades 8-10 and requires understanding of algebraic manipulation.

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