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

Mathematics

1 answer:

mart [117]3 years ago

6 0

<h2>Solving System of Equations by Eliminations</h2>

<h3>Answer:</h3>

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

<h3>Step-by-step explanation:</h3>

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 $\bold{x = 6}$ :