Question

given $y = 2x + 8$, which equation would make a system with no solution? *
$\bigcirc$ $y = -9x - 7$
$\bigcirc$ $y = -4x + 3$
$\bigcirc$ $y = 2x - 2$
$\bigcirc$ $y = 2x + 8$

Explanation:

Step1: Recall condition for no solution

A system of linear equations \( y = m_1x + b_1 \) and \( y = m_2x + b_2 \) has no solution when \( m_1 = m_2 \) (same slope) and \( b_1
eq b_2 \) (different y - intercepts). The given equation is \( y = 2x+8 \), so its slope \( m_1 = 2 \) and y - intercept \( b_1 = 8 \).

Step2: Analyze each option

• Option 1: For \( y=-9x - 7 \), slope \( m=-9

eq2 \). So, the system will have a solution (intersecting lines).

• Option 2: For \( y = - 4x+3 \), slope \( m=-4

eq2 \). So, the system will have a solution (intersecting lines).

• Option 3: For \( y = 2x-2 \), slope \( m = 2\) (same as the given equation's slope) and y - intercept \( b=-2

eq8 \) (different y - intercepts). This satisfies the condition for no solution.

• Option 4: For \( y = 2x + 8 \), slope \( m = 2\) and y - intercept \( b = 8\) (same as the given equation). The system will have infinitely many solutions (coinciding lines).

Answer:

\( y = 2x-2 \) (the third option: \( y = 2x - 2 \))