Question 565975
Two numbers differ by 2. Twice the smaller number is added to the larger number and the result is 6. Find the numbers.
I have tried solving this equation and i only got this:
Let the smaller number be 'x', and the larger number be 'y'.
x = y - 2 (since they differ by 2)
Equation 1) 2x + y = 6
2(y - 2) + y = 6
2y - 4 + y = 6
3y - 4 = 6
3y = 10
Continue
y = {{{10/3}}}
then
x = y - 2
x = {{{10/3}}} - 2
x = {{{10/3}}} - {{{6/3}}}
x = {{{4/3}}}
:
You know what you were doing, but because the solutions were not integers, you doubted. 
:
Check our solution in the 2nd equation: 2x + y = 6
2({{{4/3}}}) + {{{10/3}}} = 6
({{{8/3}}} + {{{10/3}}} = 6
{{{18/3}}} = 6
6 = 6