Question

Find the coordinates where the line x+y=3 and the curve x^2+3y=27 intersect

Answer 1

Solving System of Equations by Eliminations

Answer:

$(6,-3)$ and $(-3,6)$

Step-by-step explanation:

The $xy$-coordinates of intersection between the graphs of two equations are just the $xy$-values that satisfies both equation.

Let's find the coordinates with the given system of equations:

$x +y =3$

$x^2 +3y = 27$

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 $x +y =3$ by $3$ so when we subtract it from $x^2 +3y = 27$ the $y$ t terms are eliminated.

$x +y =3 \\ 3(x +y) = 3(3) \\ 3x +3y = 9$

Subtracting $\bold{3x +3y = 9}$ from $\bold{x^2 +3y = 27}$:

$3x +3y -(x^2 +3y) = 9 -27 \\ 3x +3y -x^2 -3y = 9 -27 \\ -x^2 +3x (3 -3)y = 9 -27 \\ -x^2 +3x + 0y = -18$

We can disregard the term with $0$ as its coefficient so the result is $-x^2 +3x = -18$.

Now we can solve the value of $x$ with the resulting equation.

$-x^2 +3x = -18 \\ x^2 -3x = 18 \\ x^2 -3x +(\frac{3}{2})^2 = 18 +(\frac{3}{2})^2 \\ ( -\frac{3}{2})^2 = 18 +\frac{9}{4} \\ ( -\frac{3}{2})^2 = \frac{72}{4} +\frac{9}{4} \\ ( -\frac{3}{2})^2 = \frac{81}{4} \\ x -\frac{3}{2} = ±\sqrt{\frac{81}{4}}$

Solving for the positive square root:

$x -\frac{3}{2} = \sqrt{\frac{81}{4}} \\ x -\frac{3}{2} = \frac{9}{2} \\ x = \frac{9}{2} +\frac{3}{2} \\ x = \frac{12}{2} \\ x = 6$

Solving for the negative square root:

$x -\frac{3}{2} = -\sqrt{\frac{81}{4}} \\ x -\frac{3}{2} = -\frac{9}{2} \\ x = -\frac{9}{2} +\frac{3}{2} \\ x = -\frac{6}{2} \\ x = -3$

We have two $x$-values that satisfy both of the equation. We also have two respective $y$-values that satisfy both of the equation. This all means that both equations intersect twice.

Let's solve for the corresponding $y$-values of each of the $x$-values with $x +y =3$.

Solving $\bold{y}$ with $\bold{x = 6}$:

$6 +y = 3 \\ y = 3 -6 \\ y = -3$

Now we know that both equations intersect at $(6,-3)$.

Solving $\bold{y}$ with $\bold{x = -3}$:

$-3 +y =3 \\ y = 3 +3 \\ y = 6$

Now we know that both equations also intersect at $(-3,6)$