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m_a_m_a [10]
2 years ago
15

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

Mathematics
1 answer:
mart [117]2 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}:

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)

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There are a few ways we can go about finding r, but the easiest way would be to consider only the constant term in the expansion of the right hand side. We don't have to actually compute the expansion, because we know by properties of multiplication that the constant term will be (-3)(1)(1)(-r)=3r.

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Let the average profit or loss in year 2006 be x millions, 


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