Question 188800
{{{y =  sqrt (x) + 4}}} Start with the first equation.



{{{ 2x + 1 =  sqrt (x) + 4}}} Plug in {{{y=2x+1}}}



{{{ 2x + 1 -4=  sqrt (x) }}} Subtract 4 from both sides.



{{{ 2x - 3=  sqrt (x) }}}  Combine like terms.



{{{ (2x - 3)^2=  x }}} Square both sides



{{{ 4x^2-12x+9=  x }}} FOIL



{{{ 4x^2-12x+9-x= 0 }}} Subtract x from both sides.



{{{ 4x^2-13x+9= 0 }}} Combine like terms.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=4}}}, {{{b=-13}}}, and {{{c=9}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(-13) +- sqrt( (-13)^2-4(4)(9) ))/(2(4))}}} Plug in  {{{a=4}}}, {{{b=-13}}}, and {{{c=9}}}



{{{x = (13 +- sqrt( (-13)^2-4(4)(9) ))/(2(4))}}} Negate {{{-13}}} to get {{{13}}}. 



{{{x = (13 +- sqrt( 169-4(4)(9) ))/(2(4))}}} Square {{{-13}}} to get {{{169}}}. 



{{{x = (13 +- sqrt( 169-144 ))/(2(4))}}} Multiply {{{4(4)(9)}}} to get {{{144}}}



{{{x = (13 +- sqrt( 25 ))/(2(4))}}} Subtract {{{144}}} from {{{169}}} to get {{{25}}}



{{{x = (13 +- sqrt( 25 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (13 +- 5)/(8)}}} Take the square root of {{{25}}} to get {{{5}}}. 



{{{x = (13 + 5)/(8)}}} or {{{x = (13 - 5)/(8)}}} Break up the expression. 



{{{x = (18)/(8)}}} or {{{x =  (8)/(8)}}} Combine like terms. 



{{{x = 9/4}}} or {{{x = 1}}} Simplify. 



So the <i>possible</i> answers for "x" are {{{x = 9/4}}} or {{{x = 1}}} 

  


However, if you plug in {{{x = 1}}} into the equation {{{ 2x + 1 =  sqrt (x) + 4}}}, you'll find a contradiction. So this is an extraneous solution (ie we must ignore this value).



So the only solution for "x" is {{{x = 9/4}}}



Now plug in {{{x = 9/4}}} into {{{y=2x+1}}} to find "y"