Question

solve the system by graphing. if the system does not have one unique solution, also state the number of solutions and whether the equations are dependent.
2x - 4y = -6
-x - 3y = -2
part: 0 / 2
part 1 of 2
graph the lines.

Explanation:

Step1: Rewrite equations in slope - intercept form

For $2x - 4y=-6$, solve for $y$:
$-4y=-2x - 6$, so $y=\frac{1}{2}x+\frac{3}{2}$.
For $-x - 3y=-2$, solve for $y$:
$-3y=x - 2$, so $y=-\frac{1}{3}x+\frac{2}{3}$.

Step2: Identify slope and y - intercept

For $y=\frac{1}{2}x+\frac{3}{2}$, slope $m_1=\frac{1}{2}$ and y - intercept $b_1 = \frac{3}{2}$.
For $y=-\frac{1}{3}x+\frac{2}{3}$, slope $m_2=-\frac{1}{3}$ and y - intercept $b_2=\frac{2}{3}$.

Step3: Graph the lines

Plot the y - intercepts for each line. For $y=\frac{1}{2}x+\frac{3}{2}$, plot the point $(0,\frac{3}{2})$. Then use the slope $\frac{1}{2}$ (rise 1, run 2) to find another point. For $y=-\frac{1}{3}x+\frac{2}{3}$, plot the point $(0,\frac{2}{3})$, then use the slope $-\frac{1}{3}$ (rise - 1, run 3) to find another point. Draw the lines through the points.

Step4: Find the intersection point

By graphing or solving the system algebraically (using substitution or elimination), we find the intersection point.
Multiply the second equation $-x - 3y=-2$ by 2 to get $-2x-6y = - 4$.
Add it to the first equation $2x - 4y=-6$:
\[

$$\begin{align*} (2x-4y)+(-2x - 6y)&=-6+( - 4)\\ -10y&=-10\\ y& = 1 \end{align*}$$

\]
Substitute $y = 1$ into $2x-4y=-6$:
$2x-4\times1=-6$, $2x=-2$, $x=-1$.

Answer:

The solution of the system is the point $(-1,1)$. The system has one unique solution and the equations are independent.