If x>y,x+y=36 and 1/x + 1/y = 1/8,then find out the value of x and y
Answers
Answer:
x=24, y=12
Step-by-step explanation:
x>y so let x+y=36 be equation 1
and 1/x + 1/y = 1/8 be equation 2
from equation 2 we get,
x+y/xy = 1/8
Reciprocalise and solve
xy/x+y=8
xy= 8(x+y)
but from equation 1, x+y = 36
hence xy= 8(36) = 288
from equation 1 x= 36-y
y(36-y)= 288
36y-y^2 = 288
multiplying through out negative one we get,
y^2 -36y+288=0
we know sum of 24 and 12 is 36
and product of the same is 288.
we know a quadratic equation can be expressed as in the form of x^2 - (sum of roots)x + product of roots=0
therefore (y-24)(y-12) =0
hence y=24 or 12.
but since x>y
y must be the smaller number hence it is 12
and x is 24.
x=24 , y=12.
Given: x>y,x+y=36 and 1/x + 1/y = 1/8
To Find : Value of x and y
Solution:
x + y = 36
1/x + 1/y = 1/8
=> x + y = xy/8
=> 36 = xy/8
=> xy = 288
x + y = 36
xy = 288
y = 288/x
=> x + 288/x = 36
=> x² - 36x + 288 = 0
=> (x - 24)(x - 12) = 0
=> x = 24 , x = 12
=> y = 12 , y = 24
as x> y
=> x = 24 and y = 12
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