Question 4321
The x-intercept is where y=0.  So substitute y=0 into the equation:
{{{8x/3 + 5 - 2y = 0}}}
{{{8x/3 + 5 - 2*0 = 0}}}


Add -5 to each side of the equation:
{{{8x/3 + 5 - 5 = 0 - 5}}}
{{{8x/3 = -5}}}


Multiply both sides of the equation by the reciprocal of {{{8/3}}} which is {{{3/8}}}
{{{(3/8)* (8x/3) = (3/8) * (-5)}}}
{{{x=-15/8}}}


R^2 at SCC