Question 1022479
{3x+2y = 1, 3x^2+4 = y^2}
<pre>
Solve 3x+2y = 1 for y

{{{3x+2y = 1}}}
{{{2y = 1-3x}}}
{{{y = (1-3x)/2}}}

Substitute in

{{{3x^2 - y^2 + 4 = 0}}}

{{{3x^2 - ((1-3x)/2)^2 + 4 = 0}}}

{{{3x^2 - (1-3x)^2/4 + 4 = 0}}}

{{{12x^2 - (1-3x)^2 + 16 = 0}}}

{{{12x^2 - (1-6x+9x^2) + 16 = 0}}}

{{{12x^2 - 1+6x-9x^2 + 16 = 0}}}

{{{3x^2+6x + 15 = 0}}}

{{{x^2+2x + 5 = 0}}}

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{x = (-(2) +- sqrt( (2)^2-4*(1)*(5) ))/(2*(1)) }}}

{{{x = (-2 +- sqrt(4-20))/2 }}}

{{{x = (-2 +- sqrt(-16))/2 }}}

{{{x = (-2 +- 4i)/2 }}}

{{{x = (2(-1 +- 2i))/2 }}}

{{{x = (cross(2)(-1 +- 2i))/cross(2) }}}

{{{x = -1 +- 2i }}}

Substitute (-1+2i) in

{{{y = (1-3x)/2}}}

{{{y = (1-3(-1 + 2i))/2}}}

{{{y = (1+3 - 6i)/2}}}

{{{y = (4 - 6i)/2}}}

{{{y = (4 - 6i)/2}}}

{{{y = (2(2 - 3i))/2}}}

{{{y = (cross(2)(2 - 3i))/cross(2)}}}

{{{y = 2 - 3i}}}

One solution is (-1+2i, 2-3i)

Substituting (-1-2i) the same way gives  2+3i

So the other solution is (-1-2i, 2+3i)

These solutions are complex imaginary, so the graphs
should not cross:

{{{drawing(400,400,-10,10,-10,10,

graph(400,400,-10,10,-10,10,sqrt(3x^2+4)),

graph(400,400,-10,10,-10,10,-sqrt(3x^2+4)),

line(-11,17,11,-16) )}}}

And, as we see the black line, which is the graph of
{{{3x+2y = 1}}}, does not cross the red hyperbola, which is
the graph of {{{3x^2+4 = y^2}}}  

Edwin</pre>