How do you solve #2f(5f-2)-10(f^2-3f+6)=-8f(f+4)+4(2f^2-7f)#?

1 Answer
Mar 12, 2018

Answer:

#46/27#

Explanation:

expand all brackets:

#2f(5f-2) = 10f^2-4f#

#-10(f^2-3f+6) = -10f^2+30f-60#

#-8f(f+4) = -8f^2-32#

#4(2f^2-7f) = 8f^2-28f#

input expanded brackets into equation:

#10f^2-4f-10f^2+30f-60 = -8f^2-32+8f^2-28f#

collect like terms on either side:

#10f^2 - 10f^2 - 4f + 30f - 60 = -8f^2+8f^2-28f+32#

cancel out terms that add to #0#:

#cancel(10f^2) cancel(- 10f^2) - 4f + 30f - 60 = cancel(-8f^2)cancel(+8f^2)-28f+32#

add like terms:

#26f - 60 = -28f + 32#

add #28f:#

#54f - 60 = 32#

add #60:#

#54f = 92#

divide by #54:#

#f = 92/54 = 46/27#