Answer :
[tex]y=3x+1 \\
-x+2y=-3 \\ \\
\hbox{substitute 3x+1 for y in the 2nd equation:} \\
-x+2(3x+1)=-3 \\
-x+6x+2=-3 \\
5x+2=-3 \\
5x=-3-2 \\
5x=-5 \\
x=\frac{-5}{5} \\
x=-1 \\ \\
y=3x+1=3 \times (-1) +1=-3+1=-2 \\ \\
\boxed{(x,y)=(-1,-2)}
[/tex]
Substitute 3x + 1 for 'y' in the 2nd equation:
-x + 2(3x + 1) = -3
Distribute 2:
-x + 6x + 2 = -3
Combine like terms:
5x + 2 = -3
Subtract 2 to both sides:
5x = -5
Divide 5 to both sides:
x = -1
Plug this into any of the two equations to find 'y':
y = 3x + 1
y = 3(-1) + 1
y = -3 + 1
y = -2
So our solution is (-1, -2).
-x + 2(3x + 1) = -3
Distribute 2:
-x + 6x + 2 = -3
Combine like terms:
5x + 2 = -3
Subtract 2 to both sides:
5x = -5
Divide 5 to both sides:
x = -1
Plug this into any of the two equations to find 'y':
y = 3x + 1
y = 3(-1) + 1
y = -3 + 1
y = -2
So our solution is (-1, -2).
