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Sagot :
Let the first number be x.
The second number is 5 times less than x => x - 5
Therefore, we can write the statement as
[tex]\begin{gathered} x\times(x-5)=-4 \\ x^2-5x=-4 \\ \therefore \\ x^2-5x+4=0 \end{gathered}[/tex]Solving the quadratic equation:
Let us replace -5x in the equation with -4x and -x to be able to factorize.
Hence,
[tex]\begin{gathered} x^2-4x-x+4=0 \\ x(x-4)-1(x-4)=0 \\ (x-4)(x-1)=0 \\ \text{Therefore} \\ x-4=0\text{ } \\ or \\ x-1=0 \end{gathered}[/tex]Hence,
[tex]\begin{gathered} x=1 \\ or \\ x=4 \end{gathered}[/tex]Therefore, the number can be 1 or 4.
The second number can be
[tex]\begin{gathered} 1-5=-4 \\ or \\ 4-5=-1 \end{gathered}[/tex]Therefore, the pair of numbers can be
[tex](1,-4)\text{ or (4, -1)}[/tex]
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