Question

given $y = -9x + 8$, which equation would make a system with one solution? *
$\bigcirc$ $y = -9x - 7$
$\bigcirc$ $y = -9x + 3$
$\bigcirc$ $y = 2x - 2$
$\bigcirc$ $y = 8 - 9x$

Explanation:

Step1: Recall linear system solution rules

A system of linear equations \( y = m_1x + b_1 \) and \( y = m_2x + b_2 \) has one solution when \( m_1
eq m_2 \) (intersecting lines), no solution when \( m_1 = m_2 \) and \( b_1
eq b_2 \) (parallel lines), and infinitely many solutions when \( m_1 = m_2 \) and \( b_1 = b_2 \) (coinciding lines). The given equation is \( y=-9x + 8 \), so \( m_1=-9 \), \( b_1 = 8 \).

Step2: Analyze each option

• Option 1: \( y=-9x - 7 \). Here \( m_2=-9 \), \( b_2=-7 \). Since \( m_1 = m_2 \) and \( b_1

eq b_2 \), the lines are parallel, no solution.

• Option 2: \( y=-9x + 3 \). Here \( m_2=-9 \), \( b_2 = 3 \). Since \( m_1 = m_2 \) and \( b_1

eq b_2 \), the lines are parallel, no solution.

• Option 3: \( y = 2x-2 \). Here \( m_2 = 2 \), \( b_2=-2 \). Since \( m_1=-9

eq m_2 = 2 \), the lines will intersect, so one solution.

• Option 4: \( y=8 - 9x \) can be rewritten as \( y=-9x + 8 \). Here \( m_2=-9 \), \( b_2 = 8 \). Since \( m_1 = m_2 \) and \( b_1 = b_2 \), the lines coincide, infinitely many solutions.

Answer:

C. \( y = 2x - 2 \) (assuming the options are labeled as A: \( y=-9x - 7 \), B: \( y=-9x + 3 \), C: \( y = 2x - 2 \), D: \( y=8 - 9x \))