Question

svlc algebra 1a - standard (15260)
solving systems of linear equations: graphing
how many solutions does this linear system have?
y = -\frac{1}{2}x + 4
x + 2y = -8
one solution: (0, 8)
infinite number of solutions
one solution: (8, 0)
no solution
pretest
mark and return
6 of 10
save & exit

Explanation:

Step1: Rewrite the second equation

We have the second equation \(x + 2y=-8\). Let's solve for \(y\) to get it in slope - intercept form (\(y = mx + b\), where \(m\) is the slope and \(b\) is the y - intercept).
Subtract \(x\) from both sides: \(2y=-x - 8\).
Divide both sides by 2: \(y=-\frac{1}{2}x-4\).

Step2: Compare the slopes and y - intercepts

The first equation is \(y = -\frac{1}{2}x + 4\), so its slope \(m_1=-\frac{1}{2}\) and y - intercept \(b_1 = 4\).
The second equation (after rewriting) is \(y=-\frac{1}{2}x - 4\), so its slope \(m_2=-\frac{1}{2}\) and y - intercept \(b_2=-4\).

Since the slopes of the two lines (\(m_1 = m_2=-\frac{1}{2}\)) are equal and the y - intercepts (\(b_1 = 4\) and \(b_2=-4\)) are different, the two lines are parallel. Parallel lines never intersect, so the system of linear equations has no solution.

Answer:

no solution