Question

if p=2x² + 3x²y +3xy - 5y² Q=-5x²- 4x²y + 2xy+3xy& R= -3x²-x²y+5xy-2y² show that p+q-r=0

Answer 1

p=2x² + 3x²y +3xy - 5y² Q=-5x²- 4x²y + 2xy+3y^2 R= -3x²-x²y+5xy-2y² ==========================show that p+q-r=0==========================================

VERSION 1:
2x² + 3x²y +3xy - 5y² +(-5x²- 4x²y + 2xy+3y^2)-(-3x²-x²y+5xy-2y² )=2x² + 3x²y +3xy - 5y²-5x²- 4x²y + 2xy+3y^2+3x²+x²y-5xy+2y²=
2x^2-5x^2+3x^2+3x^2y-4x^2y+x^2y+3xy+2xy-5xy-5y^2+3y^2+2Y2=
0+0+0=0 TRUE

VERSION 2:

p=2x² + 3x²y +3xy - 5y² Q=-5x²- 4x²y + 2xy+3y^2R= -3x²-x²y+5xy-2y² ==========================p+q-r=0for x^2: 2x^2-5x^2+3x^2=0
for x^2y : 3x^2y-4x^2y+x^2y=0
for xy : 3xy+2xy-5xy=0
for y^2=-5y^2+2y^2-2y^2=0

Answer 2

P  =  2 x² + 3 x² y + 3 x y - 5 y²
Q = -5 x²  - 4 x² y + 2 x y + 3 y²
=======================  add P +Q
      -3 x²  - x² y    + 5 x y - 2 y²
========================
Compare P+Q with R .  So P+Q = R  or  P+Q - R = 0