Question

1. -3x + 4y = 8

x + 2y + 6 = 0

Explanation:

Step1: Rewrite equations in slope - intercept form

For the first equation \(-3x + 4y=8\), we solve for \(y\):
\(4y = 3x + 8\), then \(y=\frac{3}{4}x + 2\). The slope \(m_1=\frac{3}{4}\) and the \(y\) - intercept \(b_1 = 2\).
For the second equation \(x + 2y+6 = 0\), we solve for \(y\):
\(2y=-x - 6\), then \(y=-\frac{1}{2}x-3\). The slope \(m_2 =-\frac{1}{2}\) and the \(y\) - intercept \(b_2=-3\).

Step2: Find intersection point (solve the system)

We can use the substitution or elimination method. Let's use elimination. Multiply the second equation \(x + 2y=-6\) by \(3\) to get \(3x+6y=-18\).
Add this to the first equation \(-3x + 4y = 8\):
\((-3x+4y)+(3x + 6y)=8+(-18)\)
\(10y=-10\), so \(y=-1\).
Substitute \(y = - 1\) into the second equation \(x+2(-1)+6 = 0\), \(x-2 + 6=0\), \(x=-4\). So the intersection point is \((-4,-1)\).

Step3: Graph the lines

• For \(y=\frac{3}{4}x + 2\): Plot the \(y\) - intercept \((0,2)\). Then use the slope \(\frac{3}{4}\) (rise 3, run 4) to find another point. From \((0,2)\), moving up 3 units and right 4 units gives \((4,5)\).
• For \(y=-\frac{1}{2}x-3\): Plot the \(y\) - intercept \((0,-3)\). Then use the slope \(-\frac{1}{2}\) (rise - 1, run 2) to find another point. From \((0,-3)\), moving down 1 unit and right 2 units gives \((2,-4)\). Also, we know the intersection point is \((-4,-1)\), which should lie on both lines.

Answer:

The system of equations \(

$$\begin{cases}-3x + 4y = 8\\x + 2y+6 = 0\end{cases}$$

\) has a solution \(x=-4,y = - 1\) (the intersection point is \((-4,-1)\)), and the two lines can be graphed using their slopes and \(y\) - intercepts.