Question

If the same values of x and y satisfy the following equations; find the value of p:
                             3x + 7y + 5 =0
                             4x - 3y - 8 = 0
                             px + y - 1 = 0

Answer 1

         x            y            1    
  7           5            3           7
  -3         -8            4           -3
 x /-56+15 = y/20+24  = 1/-9-28
x/-41        = y/44        =  1/-37
x = -41/-37 , y = 44/-37
x = 41/37   , y = -44/37
Substitute x,y values in px + y -1 = 0
p(41/37) + (-44/37) - 1 = 0
(41p-44-37)/37 = 0
     41p-81       = 0
      41p          = 81
         p           = 81/41

     

Answer 2

First find the values of x and y which satisfy the first two equations.
3x + 7y + 5 =0    ---------(1)
4x - 3y - 8 = 0     ----------(2)

multiply eqn(1) with 4 and eqn(2) with 3
4×(3x + 7y + 5) =4×0  
⇒12x + 28y + 20 = 0   -------------(3)
3×(4x - 3y - 8) = 3×0
⇒ 12x -9y - 24 = 0       -------------(4)

subtract eqn(4) from eqn(3), we get
  12x + 28y + 20 = 0
   12x -  9y  - 24 = 0
  -     +       +         
          37y + 44= 0

          ⇒ y = $\frac{-44}{37}$

putting y=(-44/37) in equation (1), we get

$3x + 7( \frac{-44}{37}) + 5 =0 \\ \\ 3x+ \frac{-308}{37}+5=0 \\ \\ \frac{111x-308+185}{37}=0 \\ \\ 111x-123=0 \\ \\ x= \frac{123}{111}= \frac{41}{37}$

So x=41/37 and y=(-44/37)

$px+y-1=0 \\ \\ p (\frac{41}{37})+( \frac{-44}{37} ) -1=0 \\ \\ \frac{41p-44-37}{37}=0 \\ \\ 41p-81=0 \\ \\ 41p=81 \\ \\ p= \frac{81}{41}$