Question 157926
Let {{{x}}} -----> 1st number
{{{y}}} ----------> 2nd number
1st condition:
{{{2x=3y+1}}} --------> eqn 1: twice the 1st exceeds 3 times the second by 1
{{{3x=2y+14}}} -------> eqn 2: 3 times the 1st exceeds twice the 2nd by 14
In eqn 1 we get,
{{{x=(3y+1)/2}}} ------> eqn 3
Put eqn 3 to eqn 2:
{{{(3)*((3y+1)/2)=2y+14}}}, cross multiply
{{{(2)(2y+14)=(3)(3y+1)}}}
{{{4y+28=9y+3}}}
{{{28-3=9y-4y}}}
{{{25=5y}}} -------> {{{cross(25)5/cross(5)=cross(5)y/cross(5)}}}
{{{y=5}}}, ANSWER
In eqn 3:
{{{x=((3*5)+1)/2=(15+1)/2=16/2}}}
{{{x=8}}}, ANSWER
In doubt? go back eqn 1& 2:
via eqn 1: {{{(2)(8)=(3)(5)+1}}}
{{{16=15+1}}}
{{{16=16}}}
via eqn 2: {{{(3)(8)=(2*5)+14}}}
{{{24=10+14}}}
{{{14=14}}}
Thank you,
Jojo