Question 113579


Can someone help me solve the system:
{{{y=3x+9}}}…………..(1)
{{{y=x^2-2x-5}}}.………(2)


{{{y=x^2-2x-5}}}.………substitute {{{y}}} from (1) in (2)

{{{3x+9 = x^2-2x-5}}}

{{{0 = x^2-2x – 3x -5 - 9}}}

{{{ x^2 - 2x – 3x -5 – 9 = 0}}}

{{{ x^2 – 5x  – 14 = 0}}}….use square root formula


{{{x[1,2]=(-b +- sqrt (b^2 -4*a*c )) / (2*a)}}}

{{{x[1,2]=(-(-5) +- sqrt ((-5)^2 - 4*1*(-14) )) / (2*1)}}}

{{{x[1,2]=(5 +- sqrt (25 + 56 )) / 2}}}

{{{x[1,2]=(5 +- sqrt (81 )) / 2}}}

{{{x[1,2]=(5 +- 9 ) / 2}}}


{{{x[1]=(5 + 9 ) / 2}}}


{{{x[1]= 14 / 2}}}

{{{x[1]= 7}}}

{{{x[2]=(5 - 9 ) / 2}}}

{{{x[2]= - 4  / 2}}}

{{{x[2]= - 2}}}