Question 856324
Solve the linear equation for x in terms of y  (or the other way around if you want); substitute this into the conic section equation, and simplify.  You are looking to solve a quadratic equation in ONE variable which may have two solutions, or possibly one solution.  Can you do this without more help?


"No."


{{{2x-3y-10=0}}}
{{{2x=3y+10}}}
{{{highlight_green(x=(3/2)y+5)}}}---Use this in the quadratic equation and again later after finding values for y.
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The substitution:
{{{(3y/2+5)^2+(3y/2+5)y+y^2+2(3y/2+5)+y-6=0}}}
{{{(9/4)y^2+15y+25+(3/2)y^2+5y+y^2+3y+10+y-6=0}}}
...combine like terms and simplify ...
{{{(29/4)y^2+(33/2)y+29=0}}}
{{{highlight_green(29y^2+66y+116=0)}}}   ---- you could try to factor this, but much easier to use the general solution to a quadratic equation.
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Discriminant is {{{66^2-4*29*116=4356-113456=-109100}}}
The solution will contain an Imaginary component.  The discriminant is less than zero.
!  you would expect the two graphs to share not any points; no intersections.
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Another try done on paper but solving first for x instead of y in the linear equation and then substituting, gave me after simplifications, {{{highlight_green(19x^2-46x+732=0)}}};
and the discriminant, "b^2-4*a*c", is {{{highlight_green(-53516)}}}, a value less than zero.  This again means that the solution for x will contain an Imaginary component.  The given conic section and the given line DO NOT INTERSECT!