Question

How do you solve `x/9 + 5/x = 2`?

Answer 1

`x=3` or `x=15`

Explanation:

As `x/9+5/x=2`

we have `(x xx x+5xx9)/(9x)=2`

or `(x^2+45)/(9x)=2`

Assuming `x!=0` and multiplying both sides by `9x`

`x^2+45=18x` or `x^2-18x+45=0`

or `x^2-3x-15x+45=0`

or `x(x-3)-15(x-3)=0`

or `(x-15)(x-3)=0`

i.e. either `x-3 =0` i.e. `x=3`

or `x-15=0` i.e. `x=15`

Answer 2

`x= 15 " or " x = 3`

Explanation:

If you have an equation which has fractions, you can get rid of the fractions by multiplying every term by the LCD.

In `x/9+5/x= 2" "larr" "LCD= color(red)(9x)`

`(color(red)(9x)xx x)/9+(color(red)(9x)xx5)/x= color(red)(9x)xx2" "larr` cancel the denominators.

`(color(red)(cancel9x)xx x)/cancel9+(color(red)(9cancelx)xx5)/cancelx= color(red)(9x)xx2`

`x^2 +45 = 18x" "larr` make the quadratic = 0

`x^2 -18x +45 =0`

`(x-15)(x-3) = 0`

Making each factor = 0 gives,
`x= 15 " or " x = 3`