What is #y# in #5y + 2x = 5 # when #x = 5#?

1 Answer
Dec 6, 2017

#y=-1#

Explanation:

Substitute #5# for #x# in the equation #5y+2x=5#

So...

#5y+2(color(red)5)=5#

#5y+10=5#

Now we can solve for the variable #y#

Subtract 10 from both sides:

#5y+cancel(10color(red)(-10))=5color(red)(-10)#

#5y=-5#

Divide #5# from both sides:

#cancel5/cancelcolor(red)5y=-5/color(red)5#

#y=-1#