Answer:
(-2, -1) and (3, 14)
Step-by-step explanation:
[tex]\left\{\begin{array}{ccc}y=x^2+2x-1\\y-3x=5&\text{add}\ 3x\ \text{to both sides}\end{array}\right\\\left\{\begin{array}{ccc}y=x^2+2x-1&(1)\\y=3x+5&(2)\end{array}\right[/tex]
Substitute (1) to (2):
[tex]x^2+2x-1=3x+5\qquad|\text{subtract}\ 3x\ \text{and 5 from both sides}\\\\x^2-x-6=0\\\\x^2+2x-3x-6=\\\\x(x+2)-3(x+2)=0\\\\(x+2)(x-3)=0\iff x+2=0\ \vee\ x-3=0\\\\x=-2\ \vee\ x=3[/tex]
Substitute the values of x to the (2):
[tex]\text{for}\ x=-2\\\\y=3(-2)+5=-6+5=-1\\\\\text{for}\ x=3\\\\y=3(3)+5=9+5=14[/tex]
