Question 935077


you did this:
{{{(x-3)}}}  identified the answer as {{{3p+27+q}}}.... (Equation 1)
 {{{(x+1)}}}  identified the answer as {{{-p-1+q}}} .....(Equation 2) 

just solve this system

{{{3p+27+q}}}.... (Equation 1)
{{{-p-1+q}}} .....(Equation 2) 
__________________________subtract eq.2 from eq.1

{{{3p+27+q -(-p-1+q )=0 }}}  

{{{3p+27+cross(q) +p+1-cross(q) =0}}}  

{{{3p+27 +p+1 =0}}}
  
{{{4p+28 =0 }}} 

{{{4p =-28}}}

{{{highlight(p =-7)}}}

go to eq.2 plug in value for {{{p}}}

{{{-p-1+q}}} .....(Equation 2)

{{{-(-7)-1+q =0 }}} 

{{{7-1+q =0 }}}
 
{{{6+q =0  }}}

{{{highlight(q=-6)}}}