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

2 Answers
Oct 1, 2016

Answer:

#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#

Oct 1, 2016

Answer:

#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#