Question 201016
I'll do the first two to get you going in the right direction.



a)


"The sum of the squares of two numbers is 128" translates to {{{x^2+y^2=128}}}



"The product of the numbers is 64" translates to {{{xy=64}}}



{{{xy=64}}} Start with the second equation.



{{{y=64/x}}} Divide both sides by "x".



{{{x^2+y^2=128}}} Move onto the first equation



{{{x^2+(64/x)^2=128}}} Plug in {{{y=64/x}}}



{{{x^2+4096/x^2=128}}} Square {{{64/x}}} to get {{{4096/x^2}}}



{{{x^4+4096=128x^2}}} Multiply EVERY term by the LCD {{{x^2}}} to clear out the fractions.



{{{x^4+4096-128x^2=0}}} Subtract {{{128x^2}}} from both sides.



{{{x^4-128x^2+4096=0}}} Rearrange the terms.



Let {{{z=x^2}}}. So {{{z^2=x^4}}}



{{{z^2-128z+4096=0}}} Replace {{{x^4}}} with {{{z^2}}}. Replace {{{x^2}}} with {{{z}}}



Notice that the quadratic {{{z^2-128z+4096}}} is in the form of {{{Az^2+Bz+C}}} where {{{A=1}}}, {{{B=-128}}}, and {{{C=4096}}}



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



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



{{{z = (-(-128) +- sqrt( (-128)^2-4(1)(4096) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-128}}}, and {{{C=4096}}}



{{{z = (128 +- sqrt( (-128)^2-4(1)(4096) ))/(2(1))}}} Negate {{{-128}}} to get {{{128}}}. 



{{{z = (128 +- sqrt( 16384-4(1)(4096) ))/(2(1))}}} Square {{{-128}}} to get {{{16384}}}. 



{{{z = (128 +- sqrt( 16384-16384 ))/(2(1))}}} Multiply {{{4(1)(4096)}}} to get {{{16384}}}



{{{z = (128 +- sqrt( 0 ))/(2(1))}}} Subtract {{{16384}}} from {{{16384}}} to get {{{0}}}



{{{z = (128 +- sqrt( 0 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{z = (128 +- 0)/(2)}}} Take the square root of {{{0}}} to get {{{0}}}. 



{{{z = (128 + 0)/(2)}}} or {{{z = (128 - 0)/(2)}}} Break up the expression. 



{{{z = (128)/(2)}}} or {{{z =  (128)/(2)}}} Combine like terms. 



{{{z = 64}}} or {{{z = 64}}} Simplify. 



So the only solution (in terms of "z") is {{{z = 64}}}



Now recall that we let {{{z=x^2}}}. So this means that {{{64=x^2}}} and that {{{x=8}}} or {{{x=-8}}}


--------------------------


Now let's find "y" when {{{x=8}}}:



{{{y=64/x}}} Go back to the first isolated equation.



{{{y=64/8}}} Plug in {{{x=8}}}



{{{y=8}}} Reduce.



So when {{{x=8}}}, {{{y=8}}} giving the ordered pair (8,8)




--------------------------


Now let's find "y" when {{{x=-8}}}:



{{{y=64/x}}} Go back to the first isolated equation.



{{{y=64/(-8)}}} Plug in {{{x=-8}}}



{{{y=-8}}} Reduce.



So when {{{x=-8}}}, {{{y=-8}}} giving the ordered pair (-8,-8)



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Answer:

  

So the two ordered pair solutions are (8,8) and (-8,-8)



<hr>


b) 



"The difference between the squares of two numbers is 3" translates to {{{x^2-y^2=3}}}



"Twice the square of the first number increased by the square of the second number is 9" translates to {{{2x^2+y^2=9}}}



{{{x^2-y^2=3}}} Start with the first equation.



{{{x^2=3+y^2}}} Add {{{y^2}}} to both sides.



{{{2x^2+y^2=9}}} Move onto the second equation.



{{{2(3+y^2)+y^2=9}}} Plug in {{{x^2=3+y^2}}}



{{{6+2y^2+y^2=9}}} Distribute



{{{2y^2+y^2=9-6}}} Subtract 6 from both sides.



{{{3y^2=3}}} Combine like terms.



{{{y^2=3/3}}} Divide both sides by 3.



{{{y^2=1}}} Reduce



{{{y=sqrt(1)}}} or {{{y=-sqrt(1)}}} Take the square root of both sides (don't forget the "plus/minus")


 
{{{y=1}}} or {{{y=-1}}} Take the square root of 1 to get 1


--------------------------



Now that we know "y", we can use these values to find "x".



Let's find "x" when {{{y=1}}}:



{{{x^2=3+y^2}}} Start with the given equation.



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



{{{x^2=3+1}}} Square 1 to get 1



{{{x^2=4}}} Combine like terms.



{{{x=2}}} or {{{x=-2}}} Take the square root of both sides



So the first two ordered pair solutions are (2,1) and (-2,1)



--------------------------


{{{x^2=3+y^2}}} Start with the given equation.



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



{{{x^2=3+1}}} Square 1 to get 1



{{{x^2=4}}} Combine like terms.



{{{x=2}}} or {{{x=-2}}} Take the square root of both sides



So the next two ordered pair solutions are (2,-1) and (-2,-1)


========================================================


Answer:



So the four ordered pair solutions are:


(2,1), (-2,1), (2,-1) and (-2,-1)



<hr>


c) 


I'll let you do this one on your own. Let me know if you have questions.



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