Question 696476
{{{3x^2-4x-5 < y}}}.......1.... and 
{{{4x+3y< 2}}}..............2
______________________

{{{3x^2-4x-5 < y}}}.......1.set {{{y=0}}}

{{{3x^2-4x-5 < 0}}}..solve as equation to find {{{x}}}

{{{3x^2-4x-5 =0}}}

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

{{{x = (-(-4) +- sqrt( (-4)^2-4*3*(-5) ))/(2*3) }}}


{{{x = (4 +- sqrt(16+60 ))/6 }}}

{{{x = (4 +- sqrt(76 ))/6 }}}

{{{x = (4 +- 8.7177978870813471044739639677192)/6 }}}

solutions:

{{{x = (4 + 8.7177978870813471044739639677192)/6 }}}

{{{x = (12.718)/6 }}}

{{{x=2.12}}}


{{{x = (4 - 8.718)/6 }}}

{{{x = (-4.718)/6 }}}

{{{x=-0.79}}}

so, {{{-0.79<x<2.12}}}


{{{3y>-4x+2}}}..............2..set {{{y=0}}}

{{{0>-4x+2}}}

{{{4x>2}}}

{{{x>2/4}}}

{{{x>1/2}}}

system solution is

C. {{{-0.79<x<2.12}}} 

{{{ graph( 600, 600, -6, 10, -10, 10, (4/3)x- 2/3, 3x^2-4x-5) }}}