How do you simplify #2/5 (5k + 35) - 8#?

2 Answers
Mar 27, 2018

Answer:

The simplified expression is #2k+6#.

Explanation:

Use the distributive property:

#color(magenta)a(color(red)x+color(blue)y)=color(magenta)a*color(red)x+color(magenta)a*color(blue)y#

Here's this property applied to our expression:

#color(white)=color(magenta)(2/5)(color(red)(5k)+color(blue)35)-8#

#=color(magenta)(2/5)*color(red)(5k)+color(magenta)(2/5)*color(blue)35-8#

#=color(magenta)(2/color(red)cancelcolor(black)5)*color(red)(color(black)cancelcolor(red)5k)+color(magenta)(2/5)*color(blue)35-8#

#=color(magenta)2color(red)k+color(magenta)(2/5)*color(blue)35-8#

#=color(magenta)2color(red)k+color(magenta)((2color(black)*color(blue)35)/5)-8#

#=color(magenta)2color(red)k+color(magenta)(color(purple)70/5)-8#

#=color(magenta)2color(red)k+color(purple)14-8#

#=color(magenta)2color(red)k+6#

That's the expanded expression. Hope this helped!

Mar 27, 2018

Answer:

#2/5(5k+35)-8=color(blue)(2k+6#

Explanation:

Simplify:

#2/5(5k+35)-8#

Expand.

#(10k)/5+70/5-8#

Multiply #8# by #5/5# to get the least common denominator #5#. Multiplying by #5/5# is the same as multiplying by #1#, so the numbers will change, but the value of the fraction will stay the same.

#(10k)/5+70/5-8xx5/5#

Simplify.

#(color(red)cancel(color(black)(10))^2k)/color(red)cancel(color(black)(5))^1+color(red)cancel(color(black)(70))^14/color(red)cancel(color(black)(5))^1-color(red)cancel(color(black)(40))^8/color(red)cancel(color(black)(5))^1#

Simplify.

#2k+14-8#

#2k+6#