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To determine whether a point belongs to the solution region of the given system of inequalities, we need to verify that it satisfies both inequalities:
[tex]\[ \begin{array}{l} y > 1.5^x + 4 \\ y < \frac{2}{3}x + 6 \end{array} \][/tex]
Let's choose a point [tex]\((x, y)\)[/tex] and check whether it satisfies both inequalities.
Let's test the point [tex]\((2, 6)\)[/tex]:
Step 1: Check the first inequality [tex]\( y > 1.5^x + 4 \)[/tex]
For [tex]\( x = 2 \)[/tex]:
[tex]\[ y > 1.5^2 + 4 \][/tex]
Simplifying the right-hand side:
[tex]\[ y > 1.5 \times 1.5 + 4 \][/tex]
[tex]\[ y > 2.25 + 4 \][/tex]
[tex]\[ y > 6.25 \][/tex]
Substitute [tex]\( y = 6 \)[/tex]:
[tex]\[ 6 > 6.25 \][/tex]
This inequality is False.
Step 2: Check the second inequality [tex]\( y < \frac{2}{3}x + 6 \)[/tex]
For [tex]\( x = 2 \)[/tex]:
[tex]\[ y < \frac{2}{3} \times 2 + 6 \][/tex]
Simplifying the right-hand side:
[tex]\[ y < \frac{4}{3} + 6 \][/tex]
[tex]\[ y < \frac{4}{3} + \frac{18}{3} \][/tex]
[tex]\[ y < \frac{22}{3} \][/tex]
[tex]\[ y < 7.\overline{3} \][/tex]
Substitute [tex]\( y = 6 \)[/tex]:
[tex]\[ 6 < 7.\overline{3} \][/tex]
This inequality is True.
Conclusion:
For the point [tex]\((2, 6)\)[/tex], we checked both inequalities:
- The first inequality is False.
- The second inequality is True.
Since a point must satisfy both inequalities to belong to the solution region, the point [tex]\((2, 6)\)[/tex] does not satisfy the first inequality.
Thus, the point [tex]\( (2, 6) \)[/tex] does not belong to the solution region of this system of inequalities.
[tex]\[ \begin{array}{l} y > 1.5^x + 4 \\ y < \frac{2}{3}x + 6 \end{array} \][/tex]
Let's choose a point [tex]\((x, y)\)[/tex] and check whether it satisfies both inequalities.
Let's test the point [tex]\((2, 6)\)[/tex]:
Step 1: Check the first inequality [tex]\( y > 1.5^x + 4 \)[/tex]
For [tex]\( x = 2 \)[/tex]:
[tex]\[ y > 1.5^2 + 4 \][/tex]
Simplifying the right-hand side:
[tex]\[ y > 1.5 \times 1.5 + 4 \][/tex]
[tex]\[ y > 2.25 + 4 \][/tex]
[tex]\[ y > 6.25 \][/tex]
Substitute [tex]\( y = 6 \)[/tex]:
[tex]\[ 6 > 6.25 \][/tex]
This inequality is False.
Step 2: Check the second inequality [tex]\( y < \frac{2}{3}x + 6 \)[/tex]
For [tex]\( x = 2 \)[/tex]:
[tex]\[ y < \frac{2}{3} \times 2 + 6 \][/tex]
Simplifying the right-hand side:
[tex]\[ y < \frac{4}{3} + 6 \][/tex]
[tex]\[ y < \frac{4}{3} + \frac{18}{3} \][/tex]
[tex]\[ y < \frac{22}{3} \][/tex]
[tex]\[ y < 7.\overline{3} \][/tex]
Substitute [tex]\( y = 6 \)[/tex]:
[tex]\[ 6 < 7.\overline{3} \][/tex]
This inequality is True.
Conclusion:
For the point [tex]\((2, 6)\)[/tex], we checked both inequalities:
- The first inequality is False.
- The second inequality is True.
Since a point must satisfy both inequalities to belong to the solution region, the point [tex]\((2, 6)\)[/tex] does not satisfy the first inequality.
Thus, the point [tex]\( (2, 6) \)[/tex] does not belong to the solution region of this system of inequalities.
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