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Is it possible to solve this?
#1
I feel kinda dumb for not knowing what to do here, but I know my math knowledge is somewhat lacking, and I figure people here know what they are doing, so..

This isn't homework or anything, I just wrote random numbers down on a dry erase board and tried to find an answer. I'm looking for the value of y.

(7x-3y)^2-6 = 7

So far I've done..

49x^2-42xy+9y^2 = 13
49x^2-y(42x)+y(9y)-13 = 0
49x^2-13 = y(42x)-y(9y)

But I'm not sure where to go from here, or if it's even possible to.

Thanks for the help.
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#2
There's a math help thread you know!

Code:
(7x-3y)^2-6 = 7
(7x-3y)^2 = 13
7x - 3y = sqrt(13)
-3y = sqrt(13) / (7x)
y = -[sqrt(13) / (21x)]
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#3
Oh wow, thanks.

I was just over thinking it.

Didn't realize there was a math help thread, sorry. It's right on the front page too, blah.

Thanks again though.
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#4
KajitiSouls Wrote:There's a math help thread you know!

Code:
(7x-3y)^2-6 = 7
(7x-3y)^2 = 13
7x - 3y = sqrt(13)
-3y = sqrt(13) / (7x)
y = -[sqrt(13) / (21x)]

Shouldn't it be
y = ([sqrt(13)] - 7x)/-3
?

It should go:
Code:
(7x-3y)^2-6 = 7
(7x - 3y)^2 = 13
7x - 3y = sqrt(13)
-3y = sqrt(13) - 7x
y = ([sqrt(13)] - 7x)/-3

Since:

7x - 3y = -3y + 7x
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#5
Hm.

Wouldn't it be..

Code:
(7x-3y)^2-6 = 7
7x-3y = sqrt(13)
-3y = sqrt(13)-7x
y = -((sqrt(13)-7x)/3)

Or am I doing something wrong?
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#6
Taiketo Wrote:Hm.

Wouldn't it be..

Code:
(7x-3y)^2-6 = 7
7x-3y = sqrt(13)
-3y = sqrt(13)-7x
y = -((sqrt(13)-7x)/3)

Or am I doing something wrong?

I think this is right.
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#7
Taiketo Wrote:Hm.

Wouldn't it be..

Code:
(7x-3y)^2-6 = 7
7x-3y = sqrt(13)
-3y = sqrt(13)-7x
y = -((sqrt(13)-7x)/3)

Or am I doing something wrong?

Yeah. I just distributed the -3.

(I didn't say that 7 divided by 3 was 21....YOU DIDN'T SEE ANYTHING! -Fixes it-)
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#8
[Image: ylgdw5g.png]
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#9
y0y0y0y0shi0 Wrote:Yeah. I just distributed the -3.

(I didn't say that 7 divided by 3 was 21....YOU DIDN'T SEE ANYTHING! -Fixes it-)

CRAP. I JUST REALIZED I MULTIPLIED THE SQRT(13) BY -3 TOO.
Frown

-Fixes again-
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#10
you are forgetting that sqrt[(7x-3y)²] = |7x-3y|, so you would have:

y = (7x-sqrt(13))/3
or
y = (7x+sqrt(13))/3
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#11
Shidoshi Wrote:you are forgetting that sqrt[(7x-3y)²] = |7x-3y|, so you would have:

y = (7x-sqrt(13))/3
or
y = (7x+sqrt(13))/3

Oh crap, can't solve it that way.

[Image: yflpcjb.png]

ugly. Does not exist at
[Image: yfy55ev.png]

(Which is a really low number, btw)
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#12
Uh, I think you put the wrong number in the denominator. You put what would be C instead of A.

Wouldn't it actually be... (Sorry it's so messy, I don't really understand how to use mathurl)..

Code:
(-42x +/- sqrt((42x)^2-4*9*(13-49x^2))) / (2 * 9)
(-42x +/- sqrt(1764x^2-36(13-49x^2))) / 18
(-42x +/- sqrt(1764x^2-468+1764x^2)) / 18
(-42x +/- sqrt(2(1764x^2)-468)) / 18
(-42x +/- sqrt(3528x^2-468)) / 18

Not sure if I can go further.
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#13
Um. You foiled it wrong. It's supposed to be 49x^2 - 42xy + 9y^2.

So:
y = (-42x +/- sqrt((42x)^2 - 4*-9*(13 - 49x^2))) / (2 * -9)
= (-42x +/- sqrt(1764x^2 + 468 - 1764x^2)) / -18
= (-42x +/- sqrt(468)) / -18
= (-42x +/- 6 sqrt(13)) / -18
= (7x +/- sqrt(13)) / 3

... which is the same answer you get from completing the square, if you account for the +/-:

(7x - 3y)^2 = 13
7x - 3y = +/-sqrt(13)
-3y = -7x +/- sqrt(13)
y = (7x +/- sqrt(13)) / 3
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#14
y0y0y0y0shi0 Wrote:Shouldn't it be
y = ([sqrt(13)] - 7x)/-3
?

It should go:
Code:
(7x-3y)^2-6 = 7
(7x - 3y)^2 = 13
7x - 3y = sqrt(13)
-3y = sqrt(13) - 7x
y = ([sqrt(13)] - 7x)/-3

Since:

7x - 3y = -3y + 7x

Oh gawd now I'm scared of myself o.o

I'm also ashamed of Taiketo for not catching that mistake. Your syntax in the last line is wrong though.
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#15
Oh, sweet.

I was thinking you'd get the same answer, but I wasn't sure why I wasn't.

EDIT : Yeah, I feel dumb for not noticing it. Well, I mean... I noticed your mistake, but by the time I posted someone else had also. I didn't notice the other mistake, in that I took the a, b, c, etc for the quadratic formula from the post above me without checking them, which is why I was using 9 instead of -9.

Thanks for the continued help everyone, heh.
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#16
KajitiSouls Wrote:Oh gawd now I'm scared of myself o.o

I'm also ashamed of Taiketo for not catching that mistake. Your syntax in the last line is wrong though.

Bleh. I hate syntax crap.
Frown
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#17
Taiketo Wrote:Oh, sweet.

I was thinking you'd get the same answer, but I wasn't sure why I wasn't.

EDIT : Yeah, I feel dumb for not noticing it. Well, I mean... I noticed your mistake, but by the time I posted someone else had also. I didn't notice the other mistake, in that I took the a, b, c, etc for the quadratic formula from the post above me without checking them, which is why I was using 9 instead of -9.

Thanks for the continued help everyone, heh.

Such a Renaissance man, Sha. <3
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#18
Interesting that I did not catch that one. Harr.
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#19
Devil's Sunrise Wrote:Does not exist at
[Image: yfy55ev.png]
It perfectly well exists at that condition.
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#20
MaplePorn Wrote:It perfectly well exists at that condition.

If you notice, my whole calculation's horribly wrong. It's supposed to be +9y² and the denominator isn't supposed to be the constant c, but the constant a.
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