y = 5x - 1\n-15x - 3y = 3\nhow many solutions does this linear system have?\none solution: (0, -1)\none…

Explanation:

Step1: Rewrite the second - equation

Rewrite $-15x - 3y=3$ in slope - intercept form $y = mx + b$. First, solve for $y$: [ \begin{align*} -15x-3y&=3\ -3y&=15x + 3\ y&=- 5x-1 \end{align*} ]

Step2: Compare the slopes and y - intercepts

The first equation is $y = 5x-1$, with slope $m_1 = 5$ and y - intercept $b_1=-1$. The second equation $y=-5x - 1$ has slope $m_2=-5$ and y - intercept $b_2=-1$. Since the slopes $m_1\neq m_2$, the two lines intersect at exactly one point. We can also check by substitution. Substitute $y = 5x-1$ into $-15x-3y = 3$: [ \begin{align*} -15x-3(5x - 1)&=3\ -15x-15x + 3&=3\ -30x&=0\ x&=0 \end{align*} ] Substitute $x = 0$ into $y = 5x-1$, we get $y=-1$.

Answer:

one solution: $(0, - 1)$