Question

a system of equations has 1 solution. if $4x - y = 5$ is one of the equations, which could be the othe\
$\circ$ $y = -4x + 5$\
$\circ$ $y = 4x - 5$\
$\circ$ $2y = 8x - 10$\
$\circ$ $-2y = -8x - 10$

Explanation:

Step1: Analyze the given equation

First, rewrite the given equation \(4x - y = 5\) in slope - intercept form (\(y=mx + b\), where \(m\) is the slope and \(b\) is the y - intercept).
We get \(y = 4x-5\). The slope of this line is \(m_1 = 4\) and the y - intercept is \(b_1=-5\).

Step2: Analyze the slopes of the options

• For the equation \(y=-4x + 5\), the slope \(m_2=-4\). Since \(m_1

eq m_2\), the two lines are not parallel and will intersect at exactly one point.

• For the equation \(y = 4x-5\), the slope \(m_3 = 4\) and the y - intercept \(b_3=-5\). This is the same line as the given equation \(4x - y = 5\) (infinitely many solutions).
• For the equation \(2y=8x - 10\), divide both sides by 2 to get \(y = 4x-5\). This is also the same line as the given equation (infinitely many solutions).
• For the equation \(-2y=-8x - 10\), divide both sides by - 2 to get \(y = 4x + 5\). The slope \(m_4 = 4\) and \(b_4 = 5\). Since the slope is the same as the given line and the y - intercept is different, the lines are parallel (no solution).

Answer:

\(y=-4x + 5\) (the first option)