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Sagot :
let 'x' be the negative integer
let 'y' be the positive integer
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since we are talking about the distance of the two numbers in the number line then we take the absolute values of the two
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since they are 18 units apart then,
|x|+|y| = 18 -------equation 1
the absolute value of the positive integer is 5 times the absolute value of the negative integer
|y| = 5|x| ------equation 2
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since we've already assigned signs to the variables, then we may take equations 1&2 in its positive sense
x+y =18 -----equation 1'
y =5x ------equation 2'
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substitute equation 2' to equation 1'
x+y =18
x+(5x) = 18
6x = 18
divide the whole equation with 6
x = 3
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substitute x=3 to equation 2'
y = 5x
y = 5 (3)
y = 15
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Since we assign 'x' as negative and 'y' as positive then
x=-3
y=15
the two numbers are -3 & 15
let 'y' be the positive integer
--------
since we are talking about the distance of the two numbers in the number line then we take the absolute values of the two
-------------
since they are 18 units apart then,
|x|+|y| = 18 -------equation 1
the absolute value of the positive integer is 5 times the absolute value of the negative integer
|y| = 5|x| ------equation 2
----------
since we've already assigned signs to the variables, then we may take equations 1&2 in its positive sense
x+y =18 -----equation 1'
y =5x ------equation 2'
---------
substitute equation 2' to equation 1'
x+y =18
x+(5x) = 18
6x = 18
divide the whole equation with 6
x = 3
----------
substitute x=3 to equation 2'
y = 5x
y = 5 (3)
y = 15
----------
Since we assign 'x' as negative and 'y' as positive then
x=-3
y=15
the two numbers are -3 & 15
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