Question

6a. (1 pt.) how many solutions does the following system have? $y = -6x + 2$ and $y + 6x = 5$

• one solution
• infinitely many solutions
• no solution

6b. (1 pt.) how many solutions does the following system have? $y + 3x = 6$ and $y = 3x - 4$

• infinitely many solutions
• no solution
• one solution

6c. (1 pt.) how many solutions does the following system have? $y + 2x = 8$ and $2y = -4x + 16$

• infinitely many solutions
• no solution
• one solution

Explanation:

6a

Step1: Rewrite the second equation

Rewrite \( y + 6x = 5 \) in slope - intercept form (\( y=mx + b \), where \( m \) is the slope and \( b \) is the y - intercept). We get \( y=-6x + 5 \). The first equation is \( y=-6x + 2 \).

Step2: Compare slopes and y - intercepts

For a linear equation \( y = mx + b \), two lines \( y=m_1x + b_1 \) and \( y=m_2x + b_2 \): if \( m_1=m_2 \) and \( b_1
eq b_2 \), the lines are parallel and have no solution. Here, \( m_1=-6 \), \( b_1 = 2 \); \( m_2=-6 \), \( b_2 = 5 \). Since \( m_1=m_2 \) and \( b_1
eq b_2 \), the two lines are parallel and do not intersect.

Step1: Rewrite the first equation

Rewrite \( y + 3x=6 \) in slope - intercept form. We get \( y=-3x + 6 \). The second equation is \( y = 3x-4 \).

Step2: Compare slopes

For the two lines \( y=-3x + 6 \) (slope \( m_1=-3 \)) and \( y = 3x-4 \) (slope \( m_2 = 3 \)). Since \( m_1
eq m_2 \), the two lines are not parallel and will intersect at exactly one point.

Step1: Rewrite the equations in slope - intercept form

Rewrite the first equation \( y + 2x=8 \) as \( y=-2x + 8 \).
Rewrite the second equation \( 2y=-4x + 16 \). Divide both sides by 2: \( y=-2x + 8 \).

Step2: Compare the two equations

Now we have two equations: \( y=-2x + 8 \) and \( y=-2x + 8 \). This means that the two equations represent the same line. So, there are infinitely many solutions (all the points on the line are solutions).

Answer:

No solution

6b