Question

graph the system below and write its solution.\

$$\begin{cases}y = \\dfrac{1}{3}x - 2\\\\-3x + y = 6\\end{cases}$$

\
note that you can also answer
o solution\ or \infinitely many\ solutions.

Explanation:

Step1: Rewrite the second equation

Rewrite \(-3x + y = 6\) in slope - intercept form (\(y=mx + b\), where \(m\) is the slope and \(b\) is the y - intercept). We can solve for \(y\) by adding \(3x\) to both sides of the equation. So, \(y=3x + 6\).

Step2: Analyze the slopes of the two lines

The first equation is \(y=\frac{1}{3}x-2\), so its slope \(m_1=\frac{1}{3}\). The second equation (after rewriting) is \(y = 3x+6\), so its slope \(m_2 = 3\).

Since \(m_1
eq m_2\) (the slopes are not equal), the two lines are not parallel. Also, since \(m_1\times m_2=\frac{1}{3}\times3 = 1
eq - 1\), the lines are not perpendicular. But we can also solve the system of equations algebraically to find the solution.

We have the system:
\(

$$\begin{cases}y=\frac{1}{3}x - 2\\y=3x + 6\end{cases}$$

\)

Set the two expressions for \(y\) equal to each other:
\(\frac{1}{3}x-2=3x + 6\)

Step3: Solve for \(x\)

Subtract \(\frac{1}{3}x\) from both sides:
\(-2=3x-\frac{1}{3}x + 6\)
\(-2=\frac{9x - x}{3}+6\)
\(-2=\frac{8x}{3}+6\)

Subtract 6 from both sides:
\(-2-6=\frac{8x}{3}\)
\(-8=\frac{8x}{3}\)

Multiply both sides by \(\frac{3}{8}\):
\(x=-8\times\frac{3}{8}=-3\)

Step4: Solve for \(y\)

Substitute \(x = - 3\) into the first equation \(y=\frac{1}{3}x-2\):
\(y=\frac{1}{3}\times(-3)-2=-1 - 2=-3\)

Answer:

The solution of the system is \(x=-3,y = - 3\) or the ordered pair \((-3,-3)\)