Question

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

Explanation:

Step1: Recall no - solution system condition

A system of linear equations \(y = m_1x + b_1\) and \(y=m_2x + b_2\) has no solution when the slopes are equal (\(m_1=m_2\)) and the y - intercepts are different (\(b_1
eq b_2\)). The given equation is \(y=-9x + 8\), where the slope \(m_1=-9\) and the y - intercept \(b_1 = 8\).

Step2: Analyze each option

• Option 1: For the equation \(y=-9x - 7\), the slope \(m_2=-9\) (which is equal to \(m_1=-9\)) and the y - intercept \(b_2=-7\) (which is not equal to \(b_1 = 8\)).
• Option 2: For the equation \(y=-4x + 3\), the slope \(m_2=-4

eq - 9\) (so this equation does not satisfy the no - solution condition).

• Option 3: For the equation \(y = 2x-2\), the slope \(m_2 = 2

eq - 9\) (so this equation does not satisfy the no - solution condition).

• Option 4: For the equation \(y=-9x + 8\), the slope \(m_2=-9\) and the y - intercept \(b_2 = 8\) (since \(b_1=b_2\), this system has infinitely many solutions, not no solution).

Answer:

A. \(y = - 9x-7\)