Question 975958


{{{x = 4a+7}}} and {{{y = 16a^2}}}, 

{{{y}}} in terms of {{{x}}}:

{{{x = 4a+7}}}..........solve for {{{a}}}

{{{x-7 = 4a}}}

{{{(x-7)/4 = a}}}.......substitute in {{{y = 16a^2}}}


{{{y = 16((x-7)/4)^2}}}

{{{y = 16((x-7)^2/4^2)}}}

{{{y = cross(16)((x-7)^2/cross(16))}}}

{{{y = (x-7)^2}}}

{{{y = x^2-14x+49}}}