To find the equation of a line that is parallel to a given line and has an [tex]\( x \)[/tex]-intercept of [tex]\(-3\)[/tex], we need to start by identifying the slope of the given lines.
1. We are given the following lines:
[tex]\[
y = \frac{2}{3}x + 3,
\][/tex]
[tex]\[
y = \frac{2}{3}x + 2,
\][/tex]
[tex]\[
y = -\frac{3}{2}x + 3,
\][/tex]
[tex]\[
y = \frac{3}{2}x + 2.
\][/tex]
2. Observing these equations, we see that the first and second lines have a slope of [tex]\(\frac{2}{3}\)[/tex]. The third line has a slope of [tex]\(-\frac{3}{2}\)[/tex], and the fourth line has a slope of [tex]\(\frac{3}{2}\)[/tex].
3. Since the new line must be parallel to one of the given lines, it must have the same slope as one of them. The valid choices are the lines with a slope of [tex]\(\frac{2}{3}\)[/tex].
4. The slope of the new line will be [tex]\(\frac{2}{3}\)[/tex].
5. The next step is to use the [tex]\( x \)[/tex]-intercept of [tex]\(-3\)[/tex]. In the Slope-Intercept form of a line, [tex]\( y = mx + b \)[/tex], the coordinates of the [tex]\( x \)[/tex]-intercept are [tex]\((-3, 0)\)[/tex].
6. Plugging the point [tex]\((-3, 0)\)[/tex] into the equation to find the intercept [tex]\( b \)[/tex]:
[tex]\[
0 = \frac{2}{3}(-3) + b.
\][/tex]
7. Solving for [tex]\( b \)[/tex]:
[tex]\[
0 = -2 + b \implies b = 2.
\][/tex]
Thus, the equation of the new line that is parallel to the given line and has an [tex]\( x \)[/tex]-intercept of [tex]\(-3\)[/tex] is:
[tex]\[
y = \frac{2}{3}x + 2.
\][/tex]
