Question

given $y = 2x + 8$, which equation would make a system with one solution? *
$\bigcirc$ $y = 2x - 7$
$\bigcirc$ $y = -4x + 3$
$\bigcirc$ $y = 2x - 2$
$\bigcirc$ $y = 8 + 2x$

Explanation:

Step1: Recall system solution conditions

A system of linear equations \(y = m_1x + b_1\) and \(y = m_2x + b_2\) has one solution when \(m_1
eq m_2\) (different slopes, 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 \(b_1 = b_2\) (same line). The given equation is \(y = 2x + 8\) (slope \(m_1 = 2\), y - intercept \(b_1 = 8\)).

Step2: Analyze each option

• Option 1: \(y = 2x - 7\). Slope \(m = 2\) (same as given), \(b=-7

eq8\). Parallel, no solution.

• Option 2: \(y = - 4x + 3\). Slope \(m=-4

eq2\). Different slopes, so the system will have one solution.

• Option 3: \(y = 2x - 2\). Slope \(m = 2\) (same as given), \(b=-2

eq8\). Parallel, no solution.

• Option 4: \(y = 8 + 2x\). Slope \(m = 2\), \(b = 8\) (same as given). Same line, infinitely many solutions.

Answer:

B. \(y = -4x + 3\)