Answer :

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[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]
iGreen
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).

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