Question 481057
{{{2x^2 - y^2 = 14}}}.....1

{{{x-y = 1}}}.........2
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{{{x-y = 1}}}.........2...solve for {{{y}}}

{{{x-1= y}}}....substitute it in 1


{{{2x^2 - (x-1)^2 = 14}}}.....1...solve for {{{x}}}

{{{2x^2 - (x^2-2x+1) = 14}}}

{{{2x^2 -x^2+2x-1 = 14}}}

{{{x^2 +2x-1 = 14}}}

{{{x^2 +2x-1 -14=0}}}

{{{x^2 +2x-15=0}}}.....use quadratic formula



{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}...note {{{a=1}}}, {{{b=2}}} and {{{c=-15}}}

{{{x = (-2+- sqrt( 2^2-4*1*(-15) ))/(2*1) }}}

{{{x = (-2+- sqrt( 4+60) )/2 }}}

{{{x = (-2+- sqrt( 64 ))/2 }}}

{{{x = (-2+- 8)/2 }}}

solutions:

{{{x = (-2+8)/2 }}}

{{{x = 6/2 }}}

{{{x = 3 }}}

or


{{{x = (-2-8)/2 }}}

{{{x = -10/2 }}}

{{{x = -5 }}}


now find {{{y}}}


{{{y=x-1}}}

{{{y=3-1}}}

{{{y=2}}}

or

{{{y=-5-1}}}

{{{y=-6}}}

so, you have two sets of solutions and they are:


{{{highlight(x = 3 )}}} and {{{highlight(y=2)}}}

or

{{{highlight(x = -5 )}}} and {{{highlight(y=-6)}}}