Functions.....j [8x+2]/5= 1/4[[8x+2]/5]-3]?

Question

how to get rid of the three???

plz explain

Answer 1

I'm not sure what the j is doing there but without it the equation is

[8x+2]/5= 1/4[[8x+2]/5]-3]

(8/5)x + (2/5) = 1/4[[8x+2]/5]-3]

(8/5)x + (2/5) = (1/4)[8x+2]/5] - (3/4)

(8/5)x + (2/5) = (1/4)[(8/5)x + (2/5)] - (3/4)

(8/5)x + (2/5) = (8/20)x + (2/20) - (3/4)

(8/5)x + (2/5) = (8/20)x + (2/20) - (15/20)

(8/5)x + (2/5) = (8/20)x - (13/20)

(8/5)x = (8/20)x - (13/20) - (2/5)

(8/5)x = (8/20)x - (13/20) - (8/20)

(8/5)x = (8/20)x - (21/20)

(8/5)x - (8/20)x = -(21/20)

(8/5 - 8/20)x = -(21/20)

(32/20 - 8/20)x = -(21/20)

(24/20)x = -(21/20)

24x = -21

x = -21/24

x = -7/8

Hope this helps you!