Question

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?

Answer 1

`(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)`