[tex](6,-3)[/tex] and [tex](-3,6)[/tex]
The [tex]xy[/tex]-coordinates of intersection between the graphs of two equations are just the [tex]xy[/tex]-values that satisfies both equation.
Let's find the coordinates with the given system of equations:
[tex]x +y =3[/tex]
[tex]x^2 +3y = 27[/tex]
First let's rewrite one of the equations to make some eliminations. The easiest way that I can think of is to multiply both sides of [tex]x +y =3[/tex] by [tex]3[/tex] so when we subtract it from [tex]x^2 +3y = 27[/tex] the [tex]y[/tex] t terms are eliminated.
[tex]x +y =3 \\ 3(x +y) = 3(3) \\ 3x +3y = 9 [/tex]
Subtracting [tex]\bold{3x +3y = 9}[/tex] from [tex]\bold{x^2 +3y = 27}[/tex]:
[tex]3x +3y -(x^2 +3y) = 9 -27 \\ 3x +3y -x^2 -3y = 9 -27 \\ -x^2 +3x (3 -3)y = 9 -27 \\ -x^2 +3x + 0y = -18[/tex]
We can disregard the term with [tex]0[/tex] as its coefficient so the result is [tex]-x^2 +3x = -18[/tex].
Now we can solve the value of [tex]x[/tex] with the resulting equation.
[tex]-x^2 +3x = -18 \\ x^2 -3x = 18 \\ x^2 -3x +(\frac{3}{2})^2 = 18 +(\frac{3}{2})^2 \\ ( -\frac{3}{2})^2 = 18 +\frac{9}{4} \\ ( -\frac{3}{2})^2 = \frac{72}{4} +\frac{9}{4} \\ ( -\frac{3}{2})^2 = \frac{81}{4} \\ x -\frac{3}{2} = ±\sqrt{\frac{81}{4}} [/tex]
Solving for the positive square root:
[tex]x -\frac{3}{2} = \sqrt{\frac{81}{4}} \\ x -\frac{3}{2} = \frac{9}{2} \\ x = \frac{9}{2} +\frac{3}{2} \\ x = \frac{12}{2} \\ x = 6[/tex]
Solving for the negative square root:
[tex]x -\frac{3}{2} = -\sqrt{\frac{81}{4}} \\ x -\frac{3}{2} = -\frac{9}{2} \\ x = -\frac{9}{2} +\frac{3}{2} \\ x = -\frac{6}{2} \\ x = -3[/tex]
We have two [tex]x[/tex]-values that satisfy both of the equation. We also have two respective [tex]y[/tex]-values that satisfy both of the equation. This all means that both equations intersect twice.
Let's solve for the corresponding [tex]y[/tex]-values of each of the [tex]x[/tex]-values with [tex]x +y =3[/tex].
Solving [tex]\bold{y}[/tex] with [tex]\bold{x = 6}[/tex]:
[tex]6 +y = 3 \\ y = 3 -6 \\ y = -3[/tex]
Now we know that both equations intersect at [tex](6,-3)[/tex].
Solving [tex]\bold{y}[/tex] with [tex]\bold{x = -3}[/tex]:
[tex]-3 +y =3 \\ y = 3 +3 \\ y = 6[/tex]
Now we know that both equations also intersect at [tex](-3,6)[/tex]
