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Sagot :
In this problem, we have the following system of inequalities:
[tex]\begin{gathered} y>x^2-4x-5, \\ y<-x^2-5x+6. \end{gathered}[/tex]We must check which of the given pair of values (x, y) is not a solution of the system. To do that, we replace the values (x, y) in the system, and then we check if the inequality holds.
A) Replacing (x, y) = (1, -5), we have:
[tex]\begin{gathered} -5>1^2-4\cdot1-5=-8\Rightarrow-5>-8\text{ }✓ \\ -5<-1^2-5\cdot1+6=0\Rightarrow-5<0\text{ }✓ \end{gathered}[/tex]B) Replacing (x, y) = (-1, 6), we have:
[tex]\begin{gathered} 6>(-1)^2-4\cdot(-1)-5=0\Rightarrow6>0\text{ }✓ \\ 6<-(-1)^2-5\cdot(-1)+6=10\Rightarrow6<10\text{ }✓ \end{gathered}[/tex]C) Replacing (x, y) = (1, 7), we have:
[tex]\begin{gathered} 7>1^2-4\cdot1-5=-8\Rightarrow7>-8\text{ }✓ \\ 7<-1^2-5\cdot1+6=0\Rightarrow7<0\text{ }✖ \end{gathered}[/tex]We see that the second inequality doesn't hold. So the ordered pair (1, 7) is not a solution of the system.
D) Replacing (x, y) = (1, -7), we have:
[tex]\begin{gathered} -7>1^2-4\cdot1-5=-8\text{ }✓ \\ -7<-1^2-5\cdot1+6=0\text{ }✓ \end{gathered}[/tex]Answer
C) (1, 7)
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