Find the coordinates of the mid-point of the straight line joining the points of intersection of the curve $\displaystyle x^2+2y^2+5x=68$ and the line $\displaystyle 2y+3x=9$
$\displaystyle x^2 + 2y^2 + 5x = 68$ and $\displaystyle y = \frac{9-3x}{2}$.
Substituting the second into the first gives
$\displaystyle x^2 + \left(\frac{9-3x}{2}\right)^2 + 5x = 68$
$\displaystyle \frac{4x^2}{4} + \frac{81 -54x + 9x^2}{4} + \frac{20x}{4} = \frac{272}{4}$
$\displaystyle \frac{13x^2 - 34x - 191}{4} = 0$
$\displaystyle 13x^2 - 34x - 191 = 0$
Now solve for $\displaystyle x$ and substitute back to find the corresponding $\displaystyle y$ values. This will give you the two points, so you can use the midpoint rule to find the midpoint.
The kind of obvious thing to do is to find the two point of intersection. From 2y+ 3x= 9, y= 9/2- (3/2)x. Replace y in $\displaystyle x^2+ 2y^2+ 5x= 68$ and you get a quadratic equation for x which should have two solutions. Once you have found those two points of intersection, find the point midway between them.
Do you know that the point midway between $\displaystyle (x_0, y_0)$ and $\displaystyle (x_1, y_1)$ is $\displaystyle \left(\frac{x_0+ x_1}{2}, \frac{y_0+ y_1}{2}\right)$?