How do you solve #6( y + 5) = 54#?

Start with outside of the Parenthesis, using the distributive property. Multiply everything inside the parenthesis by 6. #6*y+6*5=54#. This will give you #6y+30=54# Then subtract 30 from both sides. #6y+30-30=54-30#. This will give you #6y=24#. The last step is to divide both sides by 6. This will give you the answer of #y=4#

Always when working with algebra and brackets, the first thing we need to do is expand the bracket to get the actual equation...

#6 xx y=6y#
#6 xx 5=30#

By removing the brackets and previous values, and plugging these new values in we get...

#6y+30=54#

As we want to solve for #y#, we want #y# on its own, therefore we have to take away the additional constant (in this instance 30). Remember what we want to do to one side, we have to do to another, so take away 30 from both sides leaving us with...

#6y=24#

As we do not want 6y, we want #y# or #1y# on its own, we divide the value of what y is worth by the certain number of #y's, x's, z's#, or similar that we have to find #y#...

#y=24/6#

As this actually equals an integer (as #6# can be divided by #24#), we simplify, if it does not it is fine to leave it as a fraction, or you can turn it to a decimal.

#24/6=4#

So:

#therefore# #y=4#