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Sagot :
Present age of Byron = b
Present age of Doug = d
Byron is 3 years older than Doug
b=d+3.........(1)
The product of their age is 40
bd=40...........(2)
Substitute the value of b from (1) to (2)
(d+3)d=40
d^2+3d=40
d^2+3d-40=0
Now solve for d
d^2+8d-5d-40=0
d(d+8)-5(d+8)=0
(d-5)(d+8)=0
d-5=0 or d+8=0
d=5 or d=-8
d=-8 (inadmissible)
So d=5
Put the value of d in (1)
b=d+3.........(1)
b=5+3
b=8
Present age of Byron = b = 8 years old
Present age of Doug = d = 5 years old
Present age of Doug = d
Byron is 3 years older than Doug
b=d+3.........(1)
The product of their age is 40
bd=40...........(2)
Substitute the value of b from (1) to (2)
(d+3)d=40
d^2+3d=40
d^2+3d-40=0
Now solve for d
d^2+8d-5d-40=0
d(d+8)-5(d+8)=0
(d-5)(d+8)=0
d-5=0 or d+8=0
d=5 or d=-8
d=-8 (inadmissible)
So d=5
Put the value of d in (1)
b=d+3.........(1)
b=5+3
b=8
Present age of Byron = b = 8 years old
Present age of Doug = d = 5 years old
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