Question

Tentukan nilai r pada persamaan bentuk aljabar (2×+3y)(p×+qy) = rײ+23×y+12y²

Answer 1

( 2x + 3y ) ( px + qy )          = rx² + 23xy + 12y²
2px² + 2qxy + 3ypx + 3qy² = rx² + 23xy + 12y²

Dari persamaan 3qy² = 12y² didapat :
3q = 12
  q =   4

Masukan ke 8xy + 3ypx = 23xy
( 8 + 3p )xy = 23xy
  8 + 3p       = 23
        3p       = 15
          p       =   5

Masukan lagi :
2px² = rx²
10x² = rx²
      r = 10

Answer 2

(2×+3y)(p×+qy) = rײ+23×y+12y²
2pײ + 2qxy + 3qxy + 3qy2 = rײ+23×y+12y²
2pײ + (2q + 3p)xy + 3qy2  = rײ+23×y+12y²
Jadi , 3qy²  = 12y²                       (2q + 3p)xy = 23×y         
          3q  = 12                               (2q + 3p)= 23         
            q = 12/3                               2(4)+ 3p=23            
            q = 4                                     8 + 3p  = 23                                                          3p   = 23-8                                                             3p = 15                                                               p =   15/3                                                               p=5  
Sehingga.  2pײ      = rײ              
                       2p   = r             
                          r  = 2p               
                           r = 2(5)             
                             r = 10