Answer :

LammettHash

Solve the first equation for [tex]x[/tex].

[tex]x+3y = -2 \implies x = -3y - 2[/tex]

Substitute this into the second equation and solve for [tex]y[/tex].

[tex]-3(-3y-2) + y = 6 \implies 9y + 6 + y = 6 \implies 10y = 0 \implies y=0[/tex]

Now solve for the value of [tex]x[/tex].

[tex]x = -3\cdot0 - 2 = -2[/tex]

So the solution to the system is the point [tex](x,y) = \boxed{(-2,0)}[/tex]

semsee45

Answer:

[tex]\left( \: \boxed{-2}\:, \: \boxed{0} \: \right)[/tex]

Step-by-step explanation:

Given system of equations:

[tex]\begin{cases}x+3y=-2\\-3x+y=6 \end{cases}[/tex]

To solve using the method of substitution, solve one equation for one of the variables.

Solving equation 2 for y:

[tex]\implies -3x+y=6[/tex]

[tex]\implies -3x+y+3x=3x+6[/tex]

[tex]\implies y=3x+6[/tex]

Substitute this expression into the other equation and solve for x:

[tex]\implies x+3y=-2[/tex]

[tex]\implies x+3(3x+6)=-2[/tex]

[tex]\implies x+9x+18=-2[/tex]

[tex]\implies 10x+18=-2[/tex]

[tex]\implies 10x+18-18=-2-18[/tex]

[tex]\implies 10x=-20[/tex]

[tex]\implies 10x \div 10=-20 \div 10[/tex]

[tex]\implies x=-2[/tex]

Substitute the found value of x into the expression for y, and solve for y:

[tex]\implies y=3x+6[/tex]

[tex]\implies y=3(-2)+6[/tex]

[tex]\implies y=-6+6[/tex]

[tex]\implies y=0[/tex]

Therefore, the solution to the given system of equations is:

[tex]\left( \: \boxed{-2}\:, \: \boxed{0} \: \right)[/tex]

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