Need Help

[homeschool girl]

Find [MATH]x-y[/MATH] if the pair [MATH](x,y)[/MATH] satisfies

[MATH]x^4=y^4+18\sqrt3,[/MATH]
[MATH]x^2+y^2=6,[/MATH]
and

[MATH]x+y=3.[/MATH]
I can find x and y that satisfies two of the equations, but not all three.

 

[Dr.Peterson]

Please show us the x and y you found (and how you did it, in case you are wrong).

I solved the last two equations, then just checked in the first, and it worked.

 

[homeschool girl]

I used the last equation to get x=3-y then substituted it in the second one to get

[MATH](3-y)^2+y^2=6[/MATH]
[MATH](3-y)(3-y)+y^2=6[/MATH]
[MATH]9 -3y-3y +y^2+y^2=6[/MATH]
[MATH]9 -6y +2y^2=6[/MATH]
[MATH]2y^2-6y+9=6[/MATH]
and then I tried solving it and got -4,-6.
(i don't remember how and i lost my notes)

 

[Dr.Peterson]

I used the last equation to get x=3-y then substituted it in the second one to get

[MATH](3-y)^2+y^2=6[/MATH]
[MATH](3-y)(3-y)+y^2=6[/MATH]
[MATH]9 -3y-3y +y^2+y^2=6[/MATH]
[MATH]9 -6y +2y^2=6[/MATH]
[MATH]2y^2-6y+9=6[/MATH]
and then I tried solving it and got -4,-6.
(i don't remember how and i lost my notes)

Good; you came close (right equation), but went astray at this last step. I presume you subtracted 6 from both sides as the first step in solving. But did you check your solutions before moving on?

[MATH]2(-4)^2-6(-4)+9=2(16)-(-24)+9 = 65 \ne 6[/MATH]​

[MATH]2(-6)^2-6(-6)+9=2(36)-(-36)+9 = 117 \ne 6[/MATH]​

I can't tell what mistake you made; but try again. (You'll have to complete the square or use the quadratic formula, because it can't be factored.)

By the way, I always check whatever I can as I go, and if anything doesn't work like this, I check every step or just redo my work. Finding and fixing errors is a large part of doing math!

 

[JeffM]

I used the last equation to get x=3-y then substituted it in the second one to get

[MATH](3-y)^2+y^2=6[/MATH]
[MATH](3-y)(3-y)+y^2=6[/MATH]
[MATH]9 -3y-3y +y^2+y^2=6[/MATH]
[MATH]9 -6y +2y^2=6[/MATH]
[MATH]2y^2-6y+9=6[/MATH]
and then I tried solving it and got -4,-6.
(i don't remember how and i lost my notes)

What you have shown is correct.

[MATH]2y^2 - 6y + 3 = 0.[/MATH]
Look at the discriminant of that quadratic [MATH](-\ 6)^2 - 4(2)(3) = 36 - 24 = 12.[/MATH]
So there is at least one distinct real root, but it will contain [MATH]\sqrt{12} = \sqrt{4 * 3} = 2\sqrt{3}.[/MATH]
So your answers of - 4 and - 6 must be wrong. (Of course you can find that out Dr. Peterson's way as well, but his way takes a bit more work.)

When you get your answer for y, be very careful deriving x.

 

[pka]

Find [MATH]x-y[/MATH] if the pair [MATH](x,y)[/MATH] satisfies
[MATH]x^4=y^4+18\sqrt3,[/MATH][MATH]x^2+y^2=6,[/MATH] and
[MATH]x+y=3.[/MATH]

I have a different idea: \(x^4-y^4=18\sqrt3,~ x^2+y^2=6,~\&~x+y=3\)
Factor \((x^2+y^2)(x^2-y^2)=6(x^2-y^2)=18(x-y)=18\sqrt3\)
OR. \(x-y=\sqrt3\).
Try to finish.

Spoiler

="See solution"]