How do you solve #2v=8/(v-1/5)#?

1 Answer
Mar 16, 2016

Answer:

#v=(1+-sqrt(401))/10#

Explanation:

#1#. Multiply both sides of the equation by #v-1/5# to get rid of the denominator.

#2v=8/(v-1/5)#

#2v(v-1/5)=(8/(v-1/5))(v-1/5)#

#2v(v-1/5)=(8/(color(red)cancelcolor(black)(v-1/5)))(color(red)cancelcolor(black)(v-1/5))#

#color(orange)(2v)(color(blue)v-color(purple)(1/5))=8#

#2#. Use the distributive property, #color(orange)a(color(blue)b+color(purple)c)=color(orange)acolor(blue)b+color(orange)acolor(purple)c#, to expand the left side of the equation.

#color(orange)(2v)(color(blue)(v))+color(orange)(2v)(color(purple)(-1/5))=8#

#2v^2-(2v)/5=8#

#3#. Multiply the whole equation by #5# to get rid of the denominator.

#5(2v^2-(2v)/5)=5(8)#

#10v^2-2v=40#

#4#. Subtract 40 from both sides.

#10v^2-2v-40=0#

#5#. Factor out #2# from the left side of the equation.

#2(color(teal)5v^2# #color(violet)(-1)v# #color(brown)(-20))=0#

#6#. Use the quadratic formula to factor the trinomial.

#color(teal)(a=5)color(white)(XXXXX)color(violet)(b=-1)color(white)(XXXXX)color(brown)(c=-20)#

#v=(-b+-sqrt(b^2-4ac))/(2a)#

#v=(-(color(violet)(-1))+-sqrt((color(violet)(-1))^2-4(color(teal)5)(color(brown)(-20))))/(2(color(teal)5))#

#color(green)(|bar(ul(color(white)(a/a)v=(1+-sqrt(401))/10color(white)(a/a)|)))#