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m_a_m_a [10]
3 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]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}:

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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tester [92]
Let s = the smaller one of the two.
Since they are consecutive, the second one is (s + 1)

Since they are natural numbers, they cannot be negative.

We know that s(s + 1) = s + (s + 1) + 155

So:
s(s + 1) = 2s + 156
s^2 + s = 2s + 156
s^2 - s - 156 = 0
(s - 13)(s + 12) = 0
From this, it looks like s can be either 13 or -12. But since it's a natural number, it cannot be negative. So, it is 13.

s was the smaller one. The larger one is s + 1 = 14
Check: 13*14=182=13+14+155
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