Answer:
Hello,
Answer B (-5/3,-4/3)
Step-by-step explanation:
I am going to use the substitution 's method.
[tex]\left\{\begin{array}{ccc}x+y&=&-3\\y&=&2x+2\\\end {array} \right.\\\\\\\left\{\begin{array}{ccc}y&=&2x+2\\x+2x+2&=&-3\\\end {array} \right.\\\\\\\left\{\begin{array}{ccc}y&=&2x+2\\3x&=&-5\\\end {array} \right.\\\\\\\left\{\begin{array}{ccc}x&=&-\dfrac{5}{3} \\\\y&=&2*(-\dfrac{5}{3})+2\\\end {array} \right.\\\\\\\boxed{\left\{\begin{array}{ccc}x&=&-\dfrac{5}{3} \\\\y&=&-\dfrac{4}{3}\\\end {array} \right.\\}[/tex]