Question 927537
In order to find the small number and large number, we must express it as an algebraic expression.

One positive number is 6 more than twice another of a number can be expressed as:
{{{ y = 6 + 2x }}}
If their product is 308:
{{{ x*y = 308 }}}

Through the expression shown above we can substitute y from the first equation to the second equation.

{{{ x(6 + 2x) = 308 }}}
{{{ 6x + 2x^2) = 308 }}} Distributive Property of Multiplication
{{{ 2x^2 + 6x = 308 }}} Reorder terms in order.
{{{ (2x^2 + 6x)/2 = 308/2 }}} Division Property of Equality
{{{ x^2 + 3x = 154 }}} Simplify
{{{ x^2 + 3x - 154 = 154 - 154 }}} Subtraction Property of Equality
{{{ (x+14)(x-11) }}} Factor
{{{ x = -14, x = 11 }}} Zero Property
So, x = 11 or -14.
In order to find y, we substitute the x-value in the first equation for ALL solutions.
{{{ y = 6 + 2x }}}
{{{ y = 6 + 2(11) }}}
{{{ y = 6 + 22 }}}
{{{ y = 28 }}}
AND
{{{ y = 6 + 2(-14) }}}
{{{ y = 6 - 28 }}}
{{{ y = -22 }}}
So, y = 28 or -22.
So, a solution of {11,28} or {-14,-22} can be plotted on a graph.
HOWEVER, the problem tells you that y is a positive number.
Since {11, 28} is positive. The small number assuming that it is x will be 11 and the large number assuming that it is y will be 28.