Question

How do you solve for x in `x=18/(x+3)`?

Answer 1

`x=-6, x=3.`

Explanation:

`x=18/(x+3)`
`rArr x(x+3)=18.`
`rArr x^2+3x-18=0.`
`rArr x^2+6x-3x-18=0,.........[6xx3=18, 6-3=3.]`
`rArr x(x+6)-3(x+6)=0.`
`rArr (x+6)(x-3)=0.`
`rArr x=-6, x=3.`

Clearly, these satisfy the given eqn. hence, the roots are confirmed.

Answer 2

So solutions are `x=3" and "x=-6`

Explanation:

Multiply both sides by `(x+3)`

`x(x+3)=18xx(x+3)/(x+3)`

But `(x+3)/(x+3)=1` giving;

`x(x+3)=18`

Subtract `18` from both sides

`x(x+3)-18=18-18`

`x(x+3)-18=0`

Expand the brackets

`x^2+3x-18=0 larr" solve as a quadratic"`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Note that `(-3)xx6=-18` and that `-3+6=+3`

`=>(x-3)(x+6)=0`

So solutions are `x=3" and "x=-6`