To find the [tex]\(x\)[/tex]-intercept of the line given by the points [tex]\((-38, 40)\)[/tex], [tex]\((-23, 30)\)[/tex], and [tex]\((-8, 20)\)[/tex], we need to derive the equation of the line and then determine where it intercepts the [tex]\(x\)[/tex]-axis.
First, let's identify the slope [tex]\(m\)[/tex] of the line using any two of the given points. We'll use [tex]\((-38, 40)\)[/tex] and [tex]\((-23, 30)\)[/tex].
The formula for the slope [tex]\(m\)[/tex] is:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Substitute the coordinates of the points:
[tex]\[ m = \frac{30 - 40}{-23 - (-38)} = \frac{-10}{15} = -\frac{2}{3} \][/tex]
Next, we use the slope [tex]\(m = -\frac{2}{3}\)[/tex] and one point to find the [tex]\(y\)[/tex]-intercept [tex]\(b\)[/tex]. Let's use the point [tex]\((-38, 40)\)[/tex].
The general equation for a line is:
[tex]\[ y = mx + b \][/tex]
Substituting [tex]\(m\)[/tex], [tex]\(x_1\)[/tex], and [tex]\(y_1\)[/tex]:
[tex]\[ 40 = -\frac{2}{3}(-38) + b \][/tex]
Solve for [tex]\(b\)[/tex]:
[tex]\[ 40 = \frac{76}{3} + b \][/tex]
[tex]\[ 40 = \frac{76}{3} + b \][/tex]
[tex]\[ 40 = \frac{76}{3} + b \][/tex]
[tex]\[ b = 40 - \frac{76}{3} \][/tex]
[tex]\[ b = 40 - 25.\overline{3} \][/tex]
[tex]\[ b = 14.\overline{6} \][/tex]
Now, we have the equation of the line:
[tex]\[ y = -\frac{2}{3}x + 14.\overline{6} \][/tex]
To find the [tex]\(x\)[/tex]-intercept, we set [tex]\(y = 0\)[/tex]:
[tex]\[ 0 = -\frac{2}{3}x + 14.\overline{6} \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ \frac{2}{3}x = 14.\overline{6} \][/tex]
[tex]\[ x = 14.\overline{6} \cdot \frac{3}{2} \][/tex]
[tex]\[ x = 22 \][/tex]
So, the [tex]\(x\)[/tex]-intercept of the line is [tex]\(22\)[/tex].
[tex]\[
\boxed{22}
\][/tex]
