Question 765198
One number is 1 more than 2 times another. Their product is 7 more than 3 times their sum. Find the numbers.

let one number be {{{x}}} and another number{{{y}}}

if one number is 1 more than 2 times another, then we have

{{{x=2y}}}......eq.1

if their product is 7 more than 3 times their sum, then we have

{{{xy=3(x+y)+7}}}......eq.2

solve the system:

{{{x=2y}}}......eq.1
{{{xy=3(x+y)+7}}}......eq.2
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{{{xy=3(x+y)+7}}}......eq.2....substitute {{{x}}} from eq.1

{{{2y*y=3(2y+y)+7}}}.......solve for {{{y}}}

{{{2y^2=6y+3y+7}}}

{{{2y^2=9y+7}}}

{{{2y^2-9y-7=0}}}


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

{{{y = (-(-9) +- sqrt( (-9)^2-4*2*(-7) ))/(2*2) }}}

{{{y = (9 +- sqrt( 81+56 ))/4 }}}

{{{y = (9 +- sqrt( 137))/4 }}}

{{{y = (9 +- 11.7)/4 }}}

solutions:

{{{y = (9 + 11.7)/4 }}}

{{{y =20.7/4 }}}

{{{y =5.175 }}}


{{{y = (9 -11.7)/4 }}}

{{{y =-2.7/4 }}}

{{{y =-0.675 }}}

the {{{x}}} is

{{{x=2*5.175}}}=>{{{x=10.35}}}

{{{x=2*(-0.675)}}}=>{{{x=-1.35}}}

so, numbers are:

{{{x=10.35}}} and {{{y =5.175 }}}

or

{{{x=-1.35}}} and {{{y =-0.675 }}}

check:

{{{x=2y}}}......eq.1
{{{10.35=2*5.175}}}
{{{10.35=10.35}}}

{{{xy=3(x+y)+7}}}......eq.2

{{{10.35*5.175=3(10.35+5.175)+7}}}

{{{53.56125=3(15.525)+7}}}

{{{53.56125=46.575+7}}}

{{{53.56125=53.575}}}...both rounded to one decimal are

{{{53.6=53.6}}}