Question

Which of the following systems of linear equations has no solution?(1 point)%0D%0AResponses%0D%0A%0D%0A{4x−y=−174x+y=3%0D%0A{%0D%0A4%0D%0A𝑥%0D%0A−%0D%0A𝑦%0D%0A=%0D%0A−%0D%0A17%0D%0A4%0D%0A𝑥%0D%0A+%0D%0A𝑦%0D%0A=%0D%0A3%0D%0A%0D%0A{4x−y=174x−y=3%0D%0A{%0D%0A4%0D%0A𝑥%0D%0A−%0D%0A𝑦%0D%0A=%0D%0A17%0D%0A4%0D%0A𝑥%0D%0A−%0D%0A𝑦%0D%0A=%0D%0A3%0D%0A%0D%0A{4x−y=174x+y=3%0D%0A{%0D%0A4%0D%0A𝑥%0D%0A−%0D%0A𝑦%0D%0A=%0D%0A17%0D%0A4%0D%0A𝑥%0D%0A+%0D%0A𝑦%0D%0A=%0D%0A3%0D%0A%0D%0A{4x+y=174x−y=3

Answer 1

To determine which system of linear equations has no solution, we need to analyze each option in terms of their slopes and intercepts, as no-solution scenarios typically arise from parallel lines (same slope, different intercepts).

Let's evaluate the given systems:

1. **System 1**:
\[
\begin{align*}
4x - y &= -17 \quad (1)\\
4x + y &= 3 \quad (2)
\end{align*}
\]

For (1), rearranging gives \(y = 4x + 17\). For (2), rearranging gives \(y = -4x + 3\).
The slopes are 4 and -4, which are different. Therefore, this system has a unique solution.

2. **System 2**:
\[
\begin{align*}
4x - y &= 17 \quad (1)\\
4x - y &= 3 \quad (2)
\end{align*}
\]

Here both equations (1) and (2) are identical in term structure, but they cannot both be true simultaneously since 17 does not equal 3. This system has no solution due to being parallel lines that do not intersect.

3. **System 3**:
\[
\begin{align*}
4x - y &= 17 \quad (1)\\
4x + y &= 3 \quad (2)
\end{align*}
\]

From (1), rearranging gives \(y = 4x - 17\). From (2), rearranging gives \(y = -4x + 3\).
The slopes remain different (4 and -4), hence this system has a unique solution.

4. **System 4**:
\[
\begin{align*}
4x + y &= 17 \quad (1)\\
4x - y &= 3 \quad (2)
\end{align*}
\]

From (1), rearranging gives \(y = -4x + 17\). From (2), rearranging gives \(y = 4x - 3\).
The slopes are again different (4 and -4), resulting in a unique solution.

**Conclusion**:

The system with no solution is **System 2**:
\[
4x - y = 17 \quad \text{and} \quad 4x - y = 3
\]