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
Sean
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)#

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