Sure, let's find the slope and [tex]\( y \)[/tex]-intercept for each of the given equations.
### First Equation: [tex]\( 2x + 3y = 15 \)[/tex]
1. Express in slope-intercept form ( [tex]\( y = mx + b \)[/tex] ):
[tex]\[
2x + 3y = 15
\][/tex]
2. Solve for [tex]\( y \)[/tex]:
[tex]\[
3y = -2x + 15
\][/tex]
[tex]\[
y = -\frac{2}{3}x + 5
\][/tex]
3. Identify the slope ([tex]\( m \)[/tex]) and [tex]\( y \)[/tex]-intercept ([tex]\( b \)[/tex]):
- The slope [tex]\( m = -\frac{2}{3} \)[/tex]
- The [tex]\( y \)[/tex]-intercept [tex]\( b = 5 \)[/tex]
### Second Equation: [tex]\( x - \sqrt{3}y - \sqrt{3} = 0 \)[/tex]
1. Express in slope-intercept form ( [tex]\( y = mx + b \)[/tex] ):
[tex]\[
x - \sqrt{3}y - \sqrt{3} = 0
\][/tex]
2. Solve for [tex]\( y \)[/tex]:
[tex]\[
\sqrt{3}y = x - \sqrt{3}
\][/tex]
[tex]\[
y = \frac{1}{\sqrt{3}}x - 1
\][/tex]
3. Identify the slope ([tex]\( m \)[/tex]) and [tex]\( y \)[/tex]-intercept ([tex]\( b \)[/tex]):
- Since [tex]\( \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \)[/tex], the slope [tex]\( m = \frac{\sqrt{3}}{3} \)[/tex]
- The [tex]\( y \)[/tex]-intercept [tex]\( b = -1 \)[/tex]
### Summary:
- For the equation [tex]\( 2x + 3y = 15 \)[/tex]:
- Slope [tex]\( m = -\frac{2}{3} \)[/tex]
- [tex]\( y \)[/tex]-intercept [tex]\( b = 5 \)[/tex]
- For the equation [tex]\( x - \sqrt{3}y - \sqrt{3} = 0 \)[/tex]:
- Slope [tex]\( m = \frac{\sqrt{3}}{3} \)[/tex]
- [tex]\( y \)[/tex]-intercept [tex]\( b = -1 \)[/tex]
So, the slopes and [tex]\( y \)[/tex]-intercepts of the given equations are:
- [tex]\( 2x + 3y = 15 \)[/tex]: Slope = [tex]\( -\frac{2}{3} \)[/tex], [tex]\( y \)[/tex]-intercept = 5
- [tex]\( x - \sqrt{3}y - \sqrt{3} = 0 \)[/tex]: Slope = [tex]\( \frac{\sqrt{3}}{3} \)[/tex], [tex]\( y \)[/tex]-intercept = -1
