The sum of two number is 18 the sum of these resiprocal is 1/4 find the number
Answers
Answered by
0
Answer:
Step-by-step explanation:
Plzz check the attachment
Attachments:
Answered by
0
Let the two numbers be x & y,
from first condition,
x+y=18 ...(1)
from second condition,
1/x+1/y=1/4 ...(2)
x+y=18
x=18-y ...(3)
substituting eqn (3) in eqn (2)
1/(18-y)+1/y=1/4
(y+18-y)/(18-y)y=1/4
18/(18y-y^2)=1/4
18×4=18y-y^2
y^2-18y+72=0
y^2-12y-6y+72=0
y(y-12)-6(y-12)=0
y-12=0 or y-6=0
y=12 or y=6
if y=12 then x = 18-12 ...(from (3))
x=6
Ans: the required numbers are 12 or 6
from first condition,
x+y=18 ...(1)
from second condition,
1/x+1/y=1/4 ...(2)
x+y=18
x=18-y ...(3)
substituting eqn (3) in eqn (2)
1/(18-y)+1/y=1/4
(y+18-y)/(18-y)y=1/4
18/(18y-y^2)=1/4
18×4=18y-y^2
y^2-18y+72=0
y^2-12y-6y+72=0
y(y-12)-6(y-12)=0
y-12=0 or y-6=0
y=12 or y=6
if y=12 then x = 18-12 ...(from (3))
x=6
Ans: the required numbers are 12 or 6
Similar questions