`yltx+k_(1)`
`ygt2x+k_(2)`
Suppose that on a coordinate plane (0, 0) is a solution to the system of inequalities given above. Which of the following conclusions about `k_(1) and k_(2)` must be true?

Options

A. `k_(1)ltk_(2)`
B. `k_(2)ltk_(1)`
C. `|k_(1)|lt|k_(2)|`
D. `k_(1)=-k_(2)`

Explanation

To solve the problem step by step, we need to analyze the given inequalities and determine the conditions for \( k_1 \) and \( k_2 \) based on the fact that the point (0, 0) is a solution to the system of inequalities.

### Step 1: Write down the inequalities
The inequalities given are:
1. \( y < x + k_1 \)
2. \( y > 2x + k_2 \)

### Step 2: Substitute the point (0, 0) into the inequalities
Since (0, 0) is a solution, we substitute \( x = 0 \) and \( y = 0 \) into both inequalities.

For the first inequality:
\[ 0 < 0 + k_1 \]
This simplifies to:
\[ 0 < k_1 \]
This means that \( k_1 \) must be greater than 0.

For the second inequality:
\[ 0 > 2(0) + k_2 \]
This simplifies to:
\[ 0 > k_2 \]
This means that \( k_2 \) must be less than 0.

### Step 3: Analyze the relationship between \( k_1 \) and \( k_2 \)
From the results of the substitutions:
- We found that \( k_1 > 0 \)
- We found that \( k_2 < 0 \)

### Step 4: Conclude the relationship between \( k_1 \) and \( k_2 \)
Since \( k_1 \) is positive and \( k_2 \) is negative, we can conclude that:
\[ k_1 > k_2 \]

### Final Conclusion
The conclusion we can draw from the inequalities is that \( k_1 \) must be greater than \( k_2 \).

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