Question 255158
{{{x^2 - x + 6 = y }}} Start with the first equation.



{{{y = x^2 - x + 6}}} Rearrange the equation.



{{{2x + 2y = 36}}} Move onto the second equation



{{{2x + 2(x^2 - x + 6) = 36}}} Plug in {{{y = x^2 - x + 6}}} 



{{{2x + 2x^2 - 2x + 12 = 36}}} Distribute



{{{2x+2x^2-2x+12-36=0}}} Get every term to the left side.



{{{2x^2+0x-24=0}}} Combine like terms.



Notice that the quadratic {{{2x^2+0x-24}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=2}}}, {{{B=0}}}, and {{{C=-24}}}



Let's use the quadratic formula to solve for "x":



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = ((0) +- sqrt( (0)^2-4(2)(-24) ))/(2(2))}}} Plug in  {{{A=2}}}, {{{B=0}}}, and {{{C=-24}}}



{{{x = (0 +- sqrt( 0-4(2)(-24) ))/(2(2))}}} Square {{{0}}} to get {{{0}}}. 



{{{x = (0 +- sqrt( 0--192 ))/(2(2))}}} Multiply {{{4(2)(-24)}}} to get {{{-192}}}



{{{x = (0 +- sqrt( 0+192 ))/(2(2))}}} Rewrite {{{sqrt(0--192)}}} as {{{sqrt(0+192)}}}



{{{x = (0 +- sqrt( 192 ))/(2(2))}}} Add {{{0}}} to {{{192}}} to get {{{192}}}



{{{x = (0 +- sqrt( 192 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (0 +- 8*sqrt(3))/(4)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (0)/(4) +- (8*sqrt(3))/(4)}}} Break up the fraction.  



{{{x = "" +- 2*sqrt(3)}}} Reduce.  



{{{x = 2*sqrt(3)}}} or {{{x = -2*sqrt(3)}}} Break up the expression.  



So the solutions in terms of 'x' are {{{x = 2*sqrt(3)}}} or {{{x = -2*sqrt(3)}}} 



which approximate to {{{x=3.464}}} or {{{x=-3.464}}} 



Now let's find the corresponding solution for y when {{{x=2*sqrt(3)}}}



{{{y = x^2 - x + 6}}} Start with the given equation.



{{{y = (2*sqrt(3))^2 - 2*sqrt(3) + 6}}} Plug in {{{x=2*sqrt(3)}}}



{{{y = 12 - 2*sqrt(3) + 6}}} Square {{{2*sqrt(3)}}} to get {{{(2*sqrt(3))^2=2^2*(sqrt(3))^2=4*3=12}}}



{{{y = 18 - 2*sqrt(3)}}} Combine like terms.



So when {{{x=2*sqrt(3)}}}, {{{y = 18 - 2*sqrt(3)}}} which approximates to {{{y=14.536}}}




Now let's find the corresponding solution for y when {{{x=-2*sqrt(3)}}}



{{{y = x^2 - x + 6}}} Start with the given equation.



{{{y = (-2*sqrt(3))^2 + 2*sqrt(3) + 6}}} Plug in {{{x=-2*sqrt(3)}}}



{{{y = 12 + 2*sqrt(3) + 6}}} Square {{{-2*sqrt(3)}}} to get {{{(-2*sqrt(3))^2=(-2)^2*(sqrt(3))^2=4*3=12}}}



{{{y = 18 + 2*sqrt(3)}}} Combine like terms.



So when {{{x=-2*sqrt(3)}}}, {{{y = 18 - 2*sqrt(3)}}} which approximates to {{{y=21.464}}}