Question #1a1c4

Question

1192 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-04T14:56:13

#x^2+y^2=26#
#xy=5#

#x+y=6# called @
#x-y=4# called @@
then @-@@
get# 2y=2# so #y=1#
then #x=5#
then #x^2+y^2=26#
#xy=5#

Answer 2 · 2017-05-04T15:15:57

#x^2+y^2=26#
#xy=5#

Method 1
Since #x+y=6#, #(x+y)^2=36#. Expand the left-hand side to get #x^2+2xy+y^2=36#.

Since #x-y=4#, #(x-y)^2=16#. Expand the left-hand side to get #x^2-2xy+y^2=16#.

Now, add the first equation to the second equation: #(x^2+2xy+y^2)+(x^2-2xy+y^2)=36+16#. This gives #2x^2+2y^2=52#, or #x^2+y^2=26#.

Now, subtract the second equation from the first equation: #(x^2+2xy+y^2)-(x^2-2xy+y^2)=36-16#. This gives #4xy=20#, or #xy=5#.

Method 2
Add #x+y=6# to #x-y=4# to get #(x+y)+(x-y)=6+4#, or #x=5#. Subtract #x-y=4# from #x+y=6# to get #(x+y)-(x-y)=6-4#, or #y=1#.

Thus, #x^2+y^2=5^2+1^2=26# and #xy=5*1=5#.