elimination method: 3(2x)^.5-4(3y)^.5=-6, 2(2x)^.5+3(3y)^.5=

[eusazwi]

Trying to solve for x and y using the elimination method:

3(2x)^0.5 – 4(3y)^0.5 = -6
2(2x)^0.5 + 3(3y)^0.5 = 13

 

[Denis]

Re: elimination method

Well, get going this way:
multiply 1st equation by -2, then 2nd equation by 3

By the way, clearer if you show, as example, a^.5 as sqrt(a)

 

[eusazwi]

Re: elimination method

This was my attempt:

3[3(2x)^.5 - 4(3y)^.5 = -6]
4[2(2x)^.5 + 3(3y)^.5 = 13]

9(2x)^.5 - 12(3y)^.5 = -18
8(2x)^.5 + 12(3y)^.5 = -52

17*(2x)^.5 = -70


17*(2x)^.5 = -70
2x^.5 = -70/17
(2)(1/2)log x = - log 4.12
log x = -.615
(stopped here, knew something wasn't right)

 

[stapel]

3(2x)^0.5 – 4(3y)^0.5 = -6
2(2x)^0.5 + 3(3y)^0.5 = 13

Let sqrt[2x] = X and sqrt[3y] = Y. :idea:

Then you have:

. . . . .3X - 4Y = -6
. . . . .2X + 3Y = 13

Since the Y-terms have opposite signs, I'll cancel (eliminate) them.

Note: This is a personal choice. This is by no means the "right" method. Any mathematically-valid steps that lead to the correct solution(s) will be a "right" method. If you choose a different method that leads sensibly to the correct solution(s), then your method is "right", too! :wink:

Multiplying the first row by 3 and the second row by 4, I get:

. . . . .9X - 12Y = -18
. . . . .8X + 12Y = 52

Adding down gives:

. . . . .17X = 34

...so X = 34/17 = 2. Then Y, from the second (original) equation, is:

. . . . .3Y = 13 - 2X = 13 - 2[2] = 13 - 4 = 9

...so Y = 9/3 = 3.

So now you only need to solve:

. . . . .\(\displaystyle \sqrt{2x\,}\, =\, 2\)

. . . . .\(\displaystyle \sqrt{3y\,}\, =\, 3\)

I'll bet you can finish from here!

Eliz.

 

[Denis]

Re: elimination method

This was my attempt:
3[3(2x)^.5 - 4(3y)^.5 = -6]
4[2(2x)^.5 + 3(3y)^.5 = 13]

9(2x)^.5 - 12(3y)^.5 = -18
8(2x)^.5 + 12(3y)^.5 = -52 : WHY is 13 * 4 = -52 ?!

17*(2x)^.5 = -70

 

[eusazwi]

Eliz you are the BOMB (that's slang for great). I now understand.

Thanks,
Dean