How do you solve #a-5/a=4#?

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Answer 1 · 2016-12-22T18:39:05

#a = 5#

#a-5/a=4#

multiply by #a#:

#a^2 - 5 = 4a#

add #5#:

#a^2 = 4a + 5#

#a^2>4a#

#a>4#

trial:
#a = 5#

#a^2 = 4a+5#

#= 20+5 = 25#

#a^2 = 25#

#a = 5#

Answer 2 · 2016-12-26T14:57:47

#a= 5 " or "a =-1#

#a -5/a = 4" "larr# we can see #a !=0#

Multiply by #a# to get rid of the fraction.

#a^2 -5 =4a" "larr# a quadratic, make it equal to 0

#a^2 -4a -5 =0" "larr# factorise

#(a-5)(a+1)= 0" "larr# either factor can be =0

If #a-5 = 0, " then " a = 5#

If #a +1 = 0, " then " a =-1#