Question

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

Answer 1

1.
$x-2y=-1 \\ 2x+y=4 \ \ \ |\times 2 \\ \\ x-2y=-1 \\ \underline{4x+2y=8 \ } \\ x+4x=8-1 \\ 5x=7 \\ x=\frac{7}{5} \\ x=1 \frac{2}{5} \\ \\ x-2y=-1 \\ \frac{7}{5}-2y=-1 \\ -2y=-1-\frac{7}{5} \\ -2y=-\frac{5}{5}-\frac{7}{5} \\ -2y=-\frac{12}{5} \\ y=\frac{6}{5} \\ y=1 \frac{1}{5} \\ \\ \boxed{(x,y)=(1\frac{2}{5}, 1 \frac{1}{5}) \Leftarrow \hbox{one solution}}$

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

3.
$y=-\frac{1}{2}x-3 \ \ \ |\times (-2) \\ x+2y=-6 \\ \\ -2y=x+6 \\ \underline{x+2y=-6 \ } \\ x=x+6-6 \\ x-x=6-6 \\ 0=0 \\ \\ \boxed{\hbox{infinitely many solutions}}$

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