How do you subtract #(2x ^ { 2} + y ^ { 2} - 8) - ( 6x ^ { 2} + y ^ { 2} - 5)#?

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Answer 1 · 2017-06-18T20:12:40

See a solution process below:

First, remove all of the terms from parenthesis. Be careful to handle the signs of each individual term correctly:

#2x^2 + y^2 - 8 - 6x^2 - y^2 + 5#

Next, group like terms:

#2x^2 - 6x^2 + y^2 - y^2 - 8 + 5#

Now, combine like terms:

#(2 - 6)x^2 + (y^2 - y^2) + (-8 + 5)#

#-4x^2 + 0 + (-3)#

#-4x^2 - 3#

Answer 2 · 2017-06-18T20:19:01

Just another way of writing the same thing!

#-4x^2-3#

#+2x^2+y^2-8#
#ul(+6x^2+y^2-5)larr" Subtract"#
#-4x^2+0-3#

Answer 3 · 2017-06-18T20:32:53

#4x^2-3#

Subtracting is the same as ADDING the INVERSE:

#3# subtracted from #6# can be written as:

#(6)color(red)(-(+3)) = +3#

Or, as an addition of the inverse of #3#

#(6) color(red)(+ (-3)) = +3#

To make the INVERSE, just change the signs:

Subtract:
#+2x^2+y^2-8#
#ul(color(red)(+6x^2+y^2-5))#

can be changed to:

Add
#+2x^2+y^2-8#
#ul(color(red)(-6x^2-y^2+5))" "larr# the signs have changed.
#-4x^2" "-3#