How do you solve #(2t)/(t+1)+4(t-1)=2#?

1 Answer
Tony B
Mar 31, 2017

#t=+-sqrt(6)/2~~1.2247...#

Explanation:

#color(green)([(2t)/(t+1) ]+[color(white)(.)4(t-1)color(red)(xx1)color(white)(.)]" "=" "[color(white)(.)2color(red)(xx1)color(white)(.)])#

#color(green)([(2t)/(t+1) ]+[color(white)(.)4(t-1)color(red)(xx(t+1)/(t+1))color(white)(.)]" "=" "[color(white)(.)2color(red)(xx(t+1)/(t+1))color(white)(.)])#

#(2t)/(t+1)+ (4(t^2-1))/(t+1)" "=" "(2(t+1))/(t+1)#

Multiply both sides by #(t+1)#

#2t+4t^2-4=2t+2#

Divide both sides by 2

#cancel(color(white)(.)tcolor(white)(.))+2t^2-2=cancel(color(white)(.)tcolor(white)(.))+1#

#2t^2=3#

#t^2=3/2#

#t=+-sqrt(3)/sqrt(2) ~~1.2247...#

#t=+-sqrt(6)/2~~1.2247...#

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