I need help with this problem please? #14-1/5(j-10)=2/5(25+j)#

2 Answers
Apr 6, 2018

#color(magenta)(j=10#

Explanation:

#14-1/5(j-10)=2/5(25+j)#

[Distributive Property]

#=>14-1/5xxj+1/5xx10=2/5xx25+2/5xxj#

#=>14-j/5+2=10+(2j)/5#

#=>16-10=(2j)/5+j/5#

#=>6=(3j)/5#

#=>3j=6xx5#

#=>3j=30#

#=>j=30/3#

#color(magenta)(=>j=10#

~Hope this helps! :)

Apr 6, 2018

In this equation #j=10#

Explanation:

  1. In this equation, we want to get #j# onto one side of the equation by itself.

  2. In order to do this we are going to distribute the #-1/5# throughout the parentheses on the left side of equation giving us #14-1/5j+2# (pay close attention to the negative sign).
    On the other side we're going to distribute the #2/5# through the parentheses on the right side of the equation giving us #10+2/5#j.

  3. So at the moment we have #14-1/5j+2=10+2/5j#.

  4. Now in order to get #j# by itself on one side were are going to add #1/5j# to both sides of the equation giving us #14+2=10+3/5j#

  5. Then we are going to subtract 10 from both sides of the equation giving us #6=3/5j#

  6. Now here comes the tricky part (please pay close attention), since #j# is a fraction, we are going to multiply both side of the equation by its reciprocal fraction to give us #1 x j# (any number multiplied by its reciprocal gives you #1#) giving us:
    #(5/3)6=(5/3)(3/5)j#.

  7. Simplifying this give the answer #10=j#.