# y=x + 2 y=-3x

You can solve the system by using the Addition/Elimination Method,

Move all terms containing variables to the left-hand side of the equation.

`y-x=2 `
`y=-3x`

Move all terms containing variables to the left-hand side of the equation.

`y-x=2 `
`y+3x=0`

Multiply each equation by the value that makes the coefficients of `y` equal. This value is found by dividing the least common multiple of the coefficients of `y` by the current coefficient. In this case, the least common multiple is `1` .

`-x+y=2 `
`3x+y=0`

Multiply the first equation by `-1` to make the coefficients of `y` have opposite signs.

`-(-x+y)=-(2)`
`3x+y=0 `

Multiply -1 by each term inside the parentheses.

-(-x+y)=-2
3x+y=0

Reorder the polynomial `-x+y` alphabetically from left to right, starting with the highest order term.

`-(y-x)=-2`
`3x+y=0`

Multiply `-1` by each term inside the parentheses.

`x-y=-2 `
`3x+y=0`

Add the two equations together to eliminate `y` from the system.

`3x + y =0`

`x - y = - 2`

`4x = - 2`

Simplify the equation and solve for `y` .

`x = - 1/2`

Substitute the value found for `x` into the original equation to solve for `y` .

`(-1/2) - y = - 2`

Simplify the equation and solve for `y` .

`y = 3/2`

This is the final solution to the independent system of equations

Independent system
`x = - 1/2`

`y = 3/2`

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Solve the system by substitution.

`y = x+2`

`y = -3x`

In order to solve by substitution, since the second equation is solved for y, we will substitute -3x for y into the first equation.  THis looks like:

`-3x = x + 2` Now solve for x.

`-3x-x = x - x + 2`

`-4x = 2`

`(-4x)/(-4) = 2/(-4)`

`x = -1/2`

Now that x = -1/2, substitute this in for one of the original equations to find y.

`y = -3x`

`y = -3(-1/2)`

` ` `y = 3/2`

The solution to the system is `(-1/2, 3/2).`

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