Question

which system has no solution? 1. $y = 3x - 1$, $y = -x + 5$ 2. $y = x + 2$, $y = -x + 2$ 3. $y = x$, $y = x + 1$ 4. $y = 2x + 1$, $y = 2x - 3$

Explanation:

To determine which system of linear equations has no solution, we use the fact that a system of linear equations \( y = m_1x + b_1 \) and \( y = m_2x + b_2 \) has no solution if the slopes are equal (\( m_1 = m_2 \)) and the y - intercepts are different (\( b_1
eq b_2 \)).

Step 1: Analyze System 1 (\( y = 3x - 1 \) and \( y=-x + 5 \))

For the first equation \( y = 3x-1 \), the slope \( m_1=3 \) and the y - intercept \( b_1=-1 \).
For the second equation \( y=-x + 5 \), the slope \( m_2=-1 \) and the y - intercept \( b_2 = 5 \).
Since \( m_1
eq m_2 \), the two lines intersect, and the system has a solution.

Step 2: Analyze System 2 (\( y=x + 2 \) and \( y=-x + 2 \))

For the first equation \( y=x + 2 \), the slope \( m_1 = 1 \) and the y - intercept \( b_1=2 \).
For the second equation \( y=-x + 2 \), the slope \( m_2=-1 \) and the y - intercept \( b_2 = 2 \).
Since \( m_1
eq m_2 \), the two lines intersect, and the system has a solution.

Step 3: Analyze System 3 (\( y = x \) and \( y=x + 1 \))

For the first equation \( y=x \), the slope \( m_1=1 \) and the y - intercept \( b_1 = 0 \).
For the second equation \( y=x + 1 \), the slope \( m_2=1 \) and the y - intercept \( b_2=1 \).
Since \( m_1=m_2 \) (both slopes are 1) and \( b_1
eq b_2 \) (0≠1), the two lines are parallel and do not intersect. So the system has no solution.

Step 4: Analyze System 4 (\( y = 2x+1 \) and \( y=2x - 3 \))

Wait, let's correct this. Wait, the fourth system: Let's re - check. Wait, the fourth system is \( y = 2x + 1 \) and \( y=2x-3 \)? Wait, no, in the image, the fourth system is \( y = 2x+1 \) and \( y = 2x-3 \)? Wait, no, the user's image shows the fourth system as \( y = 2x + 1 \) and \( y=2x-3 \)? Wait, no, let's re - check the third system. The third system is \( y=x \) and \( y=x + 1 \).
For the first equation \( y=x \), slope \( m_1 = 1 \), y - intercept \( b_1=0 \).
For the second equation \( y=x + 1 \), slope \( m_2=1 \), y - intercept \( b_2 = 1 \).
Since \( m_1=m_2 \) and \( b_1
eq b_2 \), the lines are parallel.
For the fourth system (let's check it as well for completeness): For \( y = 2x+1 \), slope \( m_1 = 2 \), y - intercept \( b_1=1 \); for \( y=2x-3 \), slope \( m_2=2 \), y - intercept \( b_2=-3 \). But we already found that system 3 has \( m_1=m_2 \) and \( b_1
eq b_2 \).

Answer:

The system with equations \( y = x \) and \( y=x + 1 \) (the third system) has no solution.