How do you solve #9y-4(y+5)=40#?

1 Answer
Aug 1, 2015

This equation has one variable #(y)#; also called an unknown.

Solving this equation means to find all the real values that #y# can assume.

Explanation:

Here are the steps that you can follow to get the value(s) of #y#:

#"Step 1 :"# Open the brackets.

This means the you multiply the number outside the brakets#(color(red)(-4))# by every terms that's in the brackets : #(color(red)y# and #color(red)5)#

So, #-4(y+5)# becomes #-4*y+(-4)*5color(orange)=-4y-20#

Now we can write #9y-4(y+5)=40" "# as #" "9ycolor(orange)(-4y-20)=40#
#=>5y-20=40#

  • Why is #9y-4y=5y#?
  • Answer : Compare #9y-4y# to "nine yens minus four yens". The result is five yens. It's the exact same thing.

#"Step 2 :"# Group similar terms together.

Add #20# on both sides.
Result : #5y-20+20=40+20#
#=>5y=60#

#"Step 3 :"# Nos that you have grouped the like the terms. Make #y# to be alone on one side.

Divide both sides by #5#

Result : #(5y)/5=60/5#

#=>color(blue)(y=12)#