Question

Solve each system using elimination PLEASE HELP!!!

Answer 1

1.
$8x+5y=6 \ \ \ |\times 2 \\ 3x-2y=10 \ \ \ |\times 5 \\ \\ 16x+10y=12 \\ \underline{15x-10y=50} \\ 16x+15x=12+50 \\ 31x=62 \ \ \ |\div 31 \\ x=2 \\ \\ 3x-2y=10 \\ 3 \times 2-2y=10 \\ 6-2y=10 \ \ \ |-6 \\ -2y=4 \ \ \ |\div (-2) \\ y=-2 \\ \\ (x,y)=(2,-2)$

2.
$-3x+8y=10 \ \ \ |\times 5 \\ 5x+6y=80 \ \ \ |\times 3 \\ \\ -15x+40y=50 \\ \underline{15x+18y=240 \ \ } \\ 40y+18y=50+240 \\ 58y=290 \ \ \ |\div 58 \\ y=5 \\ \\ 5x+6y=80 \\ 5x+6 \times 5=80 \\ 5x+30=80 \ \ \ |-30 \\ 5x=50 \ \ \ |\div 5 \\ x=10 \\ \\ (x,y)=(10,5)$

3.
$4x+12y=-24 \\ 3x-6y=-18 \ \ \ |\times 2 \\ \\ 4x+12y=-24 \\ \underline{6x-12y=-36} \\ 4x+6x=-24-36 \\ 10x=-60 \ \ \ |\div 10 \\ x=-6 \\ \\ 3x-6y=-18 \\ 3 \times (-6)-6y=-18 \\ -18-6y=-18 \ \ \ |+18 \\ -6y=0 \ \ \ |\div (-6) \\ y=0 \\ \\ (x,y)=(-6,0)$

4.
$5x-6y=48 \ \ \ |\times 2 \\ 2x+5y=-3 \ \ \ |\times (-5) \\ \\ 10x-12y=96 \\ \underline{-10x-25y=15} \\ -12y-25y=96+15 \\ -37y=111 \ \ \ |\div (-37) \\ y=-3 \\ \\ 2x+5y=-3 \\ 2x+5 \times (-3)=-3 \\ 2x-15=-3 \ \ \ |+15 \\ 2x=12 \ \ \ |\div 2 \\ x=6 \\ \\ (x,y)=(6,-3)$