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)`

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

Solving for the positive square root:

`x -(3)/(2) = \sqrt{(81)/(4)} \\ x -(3)/(2) = (9)/(2) \\ x = (9)/(2) +(3)/(2) \\ x = (12)/(2) \\ x = 6`

Solving for the negative square root:

`x -(3)/(2) = -\sqrt{(81)/(4)} \\ x -(3)/(2) = -(9)/(2) \\ x = -(9)/(2) +(3)/(2) \\ x = -(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)`