Find pair of integers whose product is (‐36) and whose difference is 15.
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Answered by
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it is said that x+y=-36
=y=-36-x
so x+y=-36 and it is also given that x-y=15
so x-(36-x)=15
=x-36+x=15
=2x-36=15
=2x=15+36
=2x=51
=x=51\2
x=25.5
1st number is 25.5
2nd number is 36-25.5=10.5
regards
=y=-36-x
so x+y=-36 and it is also given that x-y=15
so x-(36-x)=15
=x-36+x=15
=2x-36=15
=2x=15+36
=2x=51
=x=51\2
x=25.5
1st number is 25.5
2nd number is 36-25.5=10.5
regards
Answered by
0
According to the question
xy=(-36)
& x-y=15
⇒x=15+y
As xy=(-36) & x=15+y
⇒(15+y)y= -36
⇒y^2 + 15y = -36
⇒y^2 + 15y +36=0
Solving the equation you will get
y= -3 or y= -12
So if y is (–3) then x is 12 & if y is (–12) then x is 3.
Hope it helps..... :-)
xy=(-36)
& x-y=15
⇒x=15+y
As xy=(-36) & x=15+y
⇒(15+y)y= -36
⇒y^2 + 15y = -36
⇒y^2 + 15y +36=0
Solving the equation you will get
y= -3 or y= -12
So if y is (–3) then x is 12 & if y is (–12) then x is 3.
Hope it helps..... :-)
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