```Question 188362

{{{x= 1+2y}}} Add 2y to both sides to isolate "x".

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{{{x^2 - 3y^2 = 13}}} Move onto the first equation

{{{(1+2y)^2 - 3y^2 = 13}}} Plug in {{{x= 1+2y}}}

{{{1+4y+4y^2- 3y^2 = 13}}} FOIL

{{{1+4y+4y^2- 3y^2 - 13=0}}} Subtract 13 from both sides.

{{{y^2+4y - 12=0}}} Combine like terms.

Notice we have a quadratic equation in the form of {{{ay^2+by+c}}} where {{{a=1}}}, {{{b=4}}}, and {{{c=-12}}}

Let's use the quadratic formula to solve for y

{{{y = (-(4) +- sqrt( (4)^2-4(1)(-12) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=4}}}, and {{{c=-12}}}

{{{y = (-4 +- sqrt( 16-4(1)(-12) ))/(2(1))}}} Square {{{4}}} to get {{{16}}}.

{{{y = (-4 +- sqrt( 16--48 ))/(2(1))}}} Multiply {{{4(1)(-12)}}} to get {{{-48}}}

{{{y = (-4 +- sqrt( 16+48 ))/(2(1))}}} Rewrite {{{sqrt(16--48)}}} as {{{sqrt(16+48)}}}

{{{y = (-4 +- sqrt( 64 ))/(2(1))}}} Add {{{16}}} to {{{48}}} to get {{{64}}}

{{{y = (-4 +- sqrt( 64 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}.

{{{y = (-4 +- 8)/(2)}}} Take the square root of {{{64}}} to get {{{8}}}.

{{{y = (-4 + 8)/(2)}}} or {{{y = (-4 - 8)/(2)}}} Break up the expression.

{{{y = (4)/(2)}}} or {{{y =  (-12)/(2)}}} Combine like terms.

{{{y = 2}}} or {{{y = -6}}} Simplify.

So the answers for "y" are {{{y = 2}}} or {{{y = -6}}}

Now simply plug each solution of "y" into {{{x=1+2y}}} to find "x"

Plug in {{{y = 2}}}

{{{x=1+2(2)=1+4=5}}}

So one set of solutions is {{{x=5}}} and {{{y=2}}} giving us the ordered pair (or point of intersection) (5,2)

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Plug in {{{y = -6}}}

{{{x=1+2(-6)=1-12=-11}}}

This means that another set of solutions is {{{x=-11}}} and {{{y=-6}}} giving us the ordered pair (or point of intersection) (-11,-6)

Here's a graph to visually confirm the answer:

<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/Algebra%20dot%20com/graph3.png">

Graph of {{{x^2 - 3y^2 = 13}}} (red) and {{{x - 2y = 1}}} (blue) with the intersection points (-11,-6) and (5,2) (black points)```