How do you solve #4(x-2)=8(4-x)+12#?

Question

1561 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-14T17:13:32

#x = 13/3#

#4(x-2) = 8 (4-x)+12#

#4(x) + 4 (-2) = 8 (4) + 8 (-x)+12#

#4x -8 = 32 -8x+12#

#4x +8x = 32 +8+12#

#12x = 52#

#x = 52/12#

#x = 13/3#