# How do you solve this equation?: 4/5x + 4/5 = 16/x+12

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

Uff.! there was a mistake in putting the value of 14^^^^
`14^2 = 196`
so, the roots are: `x = (14+-sqrt(196+80))/2`

`rArrx = (14+sqrt(276))/(2) and (14-sqrt(276))/(2)`
` `` ``rArr`` ``x = (14+16.613247)/2 and (14-16.613247)/2`` `

Hence x=15.3066, and -1.3066 are the two possible solutions of the given problem.

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

The problem, as presented, is ambiguous.

If it is`(4/(5x)+4/5= 16/x+12)`

Then,`(4)/(5x) - 16/x = 12-4/5`

`(4-5*16)/(5x) = (12*5-4)/5`

or, `-(76)/(5x) = 56/5`

removing 5 from the denominator of both sides,

`-(76)/x = 56`

By cross multiplying,

`56x = -76`

or, `x = -(76)/(56) = -(19)/(14)= -1 5/14`

` ` Again if it is `(4)/(5)x+ 4/5 = 16/x + 12`

Rearranging, `(4)/(5)x-16/x = 12-4/5`

or, `(4x^2-16*5)/(5x) = (12*5-4)/(5)`

or, `(4x^2-80)/(5x) = 56/5`

Removing 5 from the denominator of both sides,

we get, `(4x^2-80)/(x) = 56`

Cross multiplying,

`4x^2-80 = 56*x`

or, `4x^2-56x-80 = 0`

dividing by 4 on both sides,

`x^2-14x-20 = 0`

This is a quadratic equation of the type, `ax^2+bx+c` whose roots are obtainable by the Acharya formula, `x = (-b+-sqrt(b^2-4*a*c))/(2a)` Therefore, `x = (-(-14)+-sqrt((-14)^2-4*1*(-20)))/(2*1)`

`rArr x = (14+-sqrt(256+80))/2`

`= (14+sqrt(336))/2 and (14-sqrt(336))/2`

`= (14+18.33)/2 and (14-18.33)/2`

`= 16.165 and -2.165`

Hence x=16.165, and -2.165 are the two possible solutions of the given problem.