how many solutions does this linear system have?\ny = 2x - 5\n-8x - 4y = -20\none solution: (-2.5, 0)\none…

Explanation:

Step1: Rewrite the second - equation

Rewrite $-8x - 4y=-20$ in slope - intercept form $y = mx + b$. First, solve for $y$: [ \begin{align*} -8x-4y&=-20\ -4y&=8x - 20\ y&=-2x + 5 \end{align*} ]

Step2: Compare the two equations

The first equation is $y = 2x-5$ with slope $m_1 = 2$ and $y$-intercept $b_1=-5$. The second equation $y=-2x + 5$ has slope $m_2=-2$ and $y$-intercept $b_2 = 5$. Since the slopes $m_1\neq m_2$, the two lines intersect at a single point. We can also solve the system by substitution. Substitute $y = 2x-5$ into $-8x-4y=-20$: [ \begin{align*} -8x-4(2x - 5)&=-20\ -8x-8x + 20&=-20\ -16x&=-40\ x&=2.5 \end{align*} ] Substitute $x = 2.5$ into $y = 2x-5$: $y=2\times2.5-5=0$.

Answer:

one solution: $(2.5,0)$