To determine which points are solutions to the inequality [tex]\( y < \frac{2}{3}x + 2 \)[/tex], we need to check each point by substituting the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] coordinates into the inequality. Here is the step-by-step evaluation for each point:
### Point [tex]\( (0, 3) \)[/tex]
1. Substitute [tex]\( x = 0 \)[/tex] and [tex]\( y = 3 \)[/tex] into the inequality:
[tex]\[
y < \frac{2}{3} \cdot 0 + 2
\][/tex]
2. Simplify the right side:
[tex]\[
3 < 2
\][/tex]
This statement is false.
### Point [tex]\( (-3, 1) \)[/tex]
1. Substitute [tex]\( x = -3 \)[/tex] and [tex]\( y = 1 \)[/tex] into the inequality:
[tex]\[
y < \frac{2}{3} \cdot (-3) + 2
\][/tex]
2. Simplify the right side:
[tex]\[
1 < \frac{2}{3} \cdot (-3) + 2
\][/tex]
[tex]\[
1 < -2 + 2
\][/tex]
[tex]\[
1 < 0
\][/tex]
This statement is false.
### Point [tex]\( (3, 5) \)[/tex]
1. Substitute [tex]\( x = 3 \)[/tex] and [tex]\( y = 5 \)[/tex] into the inequality:
[tex]\[
y < \frac{2}{3} \cdot 3 + 2
\][/tex]
2. Simplify the right side:
[tex]\[
5 < \frac{2}{3} \cdot 3 + 2
\][/tex]
[tex]\[
5 < 2 + 2
\][/tex]
[tex]\[
5 < 4
\][/tex]
This statement is false.
### Point [tex]\( (1, 2) \)[/tex]
1. Substitute [tex]\( x = 1 \)[/tex] and [tex]\( y = 2 \)[/tex] into the inequality:
[tex]\[
y < \frac{2}{3} \cdot 1 + 2
\][/tex]
2. Simplify the right side:
[tex]\[
2 < \frac{2}{3} \cdot 1 + 2
\][/tex]
[tex]\[
2 < \frac{2}{3} + 2
\][/tex]
[tex]\[
2 < \frac{2}{3} + \frac{6}{3}
\][/tex]
[tex]\[
2 < \frac{8}{3}
\][/tex]
[tex]\[
2 < 2.67
\][/tex]
This statement is true.
### Conclusion
The point that is a solution to the inequality [tex]\( y < \frac{2}{3}x + 2 \)[/tex] is [tex]\( (1, 2) \)[/tex].
