Question

a system of equations has no solution. if $y = 8x + 7$ is one of the equations, which could be the other equation?
$2y = 16x + 14$
$y = 8x - 7$
$y = -8x + 7$
$2y = -16x - 14$

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 \(y = 8x+7\), so its slope \(m_1 = 8\) and y - intercept \(b_1 = 7\).

Step2: Analyze each option

• Option 1: \(2y=16x + 14\). Divide both sides by 2: \(y = 8x+7\). Here, slope \(m = 8\) and y - intercept \(b = 7\). Since \(m = 8\) (same as given equation) and \(b = 7\) (same as given equation), the lines are coincident, and the system has infinitely many solutions.
• Option 2: \(y = 8x-7\). The slope \(m = 8\) (same as \(y = 8x + 7\)) and the y - intercept \(b=-7\) (different from \(7\)). So, the lines are parallel (same slope, different y - intercepts), and the system has no solution.
• Option 3: \(y=-8x + 7\). The slope \(m=-8\) (different from \(8\)), so the lines will intersect at a point, and the system has one solution.
• Option 4: \(2y=-16x-14\). Divide both sides by 2: \(y=-8x - 7\). The slope \(m=-8\) (different from \(8\)), so the lines will intersect at a point, and the system has one solution.

Answer:

\(y = 8x-7\) (the second option)