How do you solve $\frac{a + 3}{a} - \frac{6}{5 a} = \frac{1}{a}$ $(a+3)/a-6/(5a)=1/a$ ?

Question

1788 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-12T14:21:09

$a = - \frac{4}{5}$ $color(blue)(a=-4/5$

$\frac{a + 3}{a} - \frac{6}{5 a} = \frac{1}{a}$ $(a+3)/a-6/(5a)=1/a$

$:.(a+3)/a-1/a=6/(5a)$

$:.(a+3-1)/a=6/(5a)$

multiply L.H.S and R.H.S by $5a$

$:.(5cancela)^color(green)1/1 xx (a+3-1)/cancela^color(green)1=6/cancel(5a)^color(green)1 xx cancel (5a)^color(green)1/1$

$:.5(a+3-1)=6$

$:.5(a+2)=6$

$5a+10=6$

$:.5a=6-10$

$:.5a=-4$

$:.color(blue)(a=-4/5$

substitute $color(blue)(a=-4/5$

$:.(-(4/5)+3)/(-4/5)-6/(5(-4/5))=1/(-4/5)$

$:.(-4/5+15/5)/(-4/5)-6/(-20/5)=1 xx -5/4$

$:.(11/5)/(-4/5)-6/-4=-5/4$

$:.11/cancel5^1 xx -cancel5^1/4=-5/4$

$:.11/-4-6/-4=-5/4$

$:.color(blue)(5/-4=5/-4$

How do you solve (a+3)/a-6/(5a)=1/aa+3a−65a=1a(a+3)/a-6/(5a)=1/a?