Find the coordinates of the points where the straight line y=2x-3 intersects the curve x^2+y^2+xy+x=30. Help please?

1 Answer
Dec 2, 2017

(3,3)

(-1,-5)

Explanation:

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We have to set the two functions equal to each other and solve for x and y. It is simpler to find the value of terms from one and substitute them into the other.

From:

y=2x-3

We get:

y^2=(2x-3)^2=4x^2-12x+9

xy=2x^2-3x

Now, we substitute:

x^2+4x^2-12x+9+2x^2-3x+x=30

7x^2-14x-21=0

7(x^2-2x-3)=0

x^2-2x-3=0

(x-3)(x+1)=0

x-3=0 results in x=3 and plugging it into one of the equations we find y=3. So one intersection point is (3,3).

x+1=0 results in x=-1. Plugging it in gives us y=-5.

The second intersection point is (-1,-5)