Question

I need the answers thanks :)

Answer 1

1.) y=3
2.) y=13
     2y + y = 3y , 22+17= 39
     3y = 39 , divide 3 on each side (which cancels out the 3 with the y which leaves you with y)
3.) y=10
     3y - y = 2y , 35-15= 20
     2y=20 , do the same as # 2.
4.) y=4
     9y - 5y = 4y , 48-32= 16
     4y=16, do the same as #2 & #3
5.) (4,5)
     x + 3y = 19
   2x - 3y = -7
    solve for x first ; x + 2x= 3x , 19-7 = 12; 3x = 12 ; divide 3 ; x= 4. You know have to make the x's cancel out. so I will multiply -2 on the first equation. -2(x+3y=19)
     -2x - 6y = -38
      2x - 3y = -7
    Now solve for y ; -6y + -3y = -9y , -38 + -7 = -45; -9y=-45 ; divide by -9 (since they are both negative your answer will be positive) ; y= 5
6.) (6,8)
   x + 4y = 38
  -x - 3y = -30
solve for y first ; 4y - 3y = y , 38-30 = 8; y=8 (that simple!) Now you need to find a common multiple for 4 & 3 which is 12. So you will have to multiply each equation by either 4 or 3.
  (x + 4y = 38)*3      =     3x + 12y = 114         
 (-x - 3y = -30)*4     =     -4x - 12y = -120
solve for x ; 3x - 4x = -x , 114-120= -6 ; -x=-6. Since the x has a negative that means there's still a 1 there so -1x=-6 ; you will need to divide this which makes the 6 a positive; x=6