Question

Solve each system using elimination. tell whether the system has one solution, infinitely many solutions, or no solution.

x-2y=-1
2x+y=4


x+3y=4
2x-6y=8


y=-1/2x-3
x+2y=-6


6x-3y=-18
-2x+4y=18

Answer 1

1.

`x-2y=-1 \\ 2x+y=4 \ \ \ |* 2 \\ \\ x-2y=-1 \\ \underline{4x+2y=8 \ } \\ x+4x=8-1 \\ 5x=7 \\ x=(7)/(5) \\ x=1 (2)/(5) \\ \\ x-2y=-1 \\ (7)/(5)-2y=-1 \\ -2y=-1-(7)/(5) \\ -2y=-(5)/(5)-(7)/(5) \\ -2y=-(12)/(5) \\ y=(6)/(5) \\ y=1 (1)/(5) \\ \\ \boxed{(x,y)=(1(2)/(5), 1 (1)/(5)) \Leftarrow \hbox{one solution}}`

2.

`x+3y=4 \ \ \ |* 2 \\ 2x-6y=8 \\ \\ 2x+6y=8 \\ \underline{2x-6y=8} \\ 2x+2x=8+8 \\ 4x=16 \\ x=4 \\ \\ x+3y=4 \\ 4+3y=4 \\ 3y=4-4 \\ 3y=0 \\ y=0 \\ \\ \boxed{(x,y)=(4,0) \Leftarrow \hbox{one solution}}`

3.

`y=-(1)/(2)x-3 \ \ \ |* (-2) \\ x+2y=-6 \\ \\ -2y=x+6 \\ \underline{x+2y=-6 \ } \\ x=x+6-6 \\ x-x=6-6 \\ 0=0 \\ \\ \boxed{\hbox{infinitely many solutions}}`

4.

`6x-3y=-18 \\ -2x+4y=18 \ \ \ |* 3 \\ \\ 6x-3y=-18 \\ \underline{-6x+12y=54} \\ -3y+12y=54-18 \\ 9y=36 \\ y=4 \\ \\ 6x-3y=-18 \\ 6x-3 * 4=-18 \\ 6x-12=-18 \\ 6x=-18+12 \\ 6x=-6 \\ x=-1 \\ \\ \boxed{(x,y)=(-1,4) \Leftarrow \hbox{one solution}}`