How do you solve #3-2y=2(3y-2)-5y#?

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Answer 1 · 2018-07-27T04:33:24

#y = 3/2# or #1.5#

#3 - 2y = 2(3y-2) - 5y#

Simplify the right side using the distributive property:
#3 - 2y = 6y - 4 - 5#

Simplify the right side:
#3 - 2y = 6y - 9#

Subtract #color(blue)(6y)# from both sides:
#3 - 2y quadcolor(blue)(-quad6y) = 6y - 9 quadcolor(blue)(-quad6y)#

#3 - 8y = -9#

Subtract #color(blue)3# from both sides:
#3 - 8y quadcolor(blue)(-quad3) = -9 quadcolor(blue)(-quad3)#

#-8y = -12#

Divide both sides by #color(blue)(-8)#:
#(-8y)/color(blue)(-8) = (-12)/color(blue)(-8)#

#y = 12/8#

#y = 3/2# or #1.5#

Hope this helps!