Question

equations: linear combinations
which statement is true about the equations -3x + 4y = 12 and \frac{1}{4}x - \frac{1}{3}y = 1?
the system of the equations has no solution; the two lines are parallel.
the system of the equations has exactly one solution at (-8, 3).
the system of the equations has exactly one solution at (-4, 3).
the system of the equations has an infinite number of solutions represented by either equation.

Explanation:

Step1: Rewrite equations in slope - intercept form

For the first equation \(-3x + 4y=12\), solve for \(y\):
Add \(3x\) to both sides: \(4y = 3x+12\)
Divide by \(4\): \(y=\frac{3}{4}x + 3\)

For the second equation \(\frac{1}{4}x-\frac{1}{3}y = 1\), solve for \(y\):
Subtract \(\frac{1}{4}x\) from both sides: \(-\frac{1}{3}y=-\frac{1}{4}x + 1\)
Multiply both sides by \(- 3\): \(y=\frac{3}{4}x-3\)

Step2: Analyze slopes and y - intercepts

The slope - intercept form of a line is \(y = mx + b\), where \(m\) is the slope and \(b\) is the y - intercept.
For the first line \(y=\frac{3}{4}x + 3\), the slope \(m_1=\frac{3}{4}\) and \(y\) - intercept \(b_1 = 3\).
For the second line \(y=\frac{3}{4}x-3\), the slope \(m_2=\frac{3}{4}\) and \(y\) - intercept \(b_2=-3\).

Since the slopes \(m_1 = m_2=\frac{3}{4}\) (so the lines are parallel) and the y - intercepts \(b_1
eq b_2\) (so the lines are not coincident), the two lines are parallel and the system of equations has no solution.

Answer:

The system of the equations has no solution; the two lines are parallel.