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

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Answer 1 · 2016-10-01T08:28:32

$x=3$ or $x=15$

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 · 2016-10-01T09:58:06

$x= 15 " or " x = 3$

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,
$x= 15 " or " x = 3$

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