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July 6th, 2019, 05:35 AM   #1
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Cool Intersection of three lines - aka a system of three equations with three variables

So I'm supposed to find the value of m for which these three lines intersect at the same point:

$\displaystyle mx+2y-1=0$
$\displaystyle 2x+my+3=0$
$\displaystyle x-y-3=0$

One of the many things I tried was to write all three in slope-intercept form like this:

$\displaystyle y=(1-mx)/2$
$\displaystyle y=(-3-2x)/m$
$\displaystyle y=x-3$

Then I thought, I can just equate the second and third equation since I know for sure that the lines do intersect, and that left me with this mess:

$\displaystyle (-3-2x)/m=x-3$
$\displaystyle -3-2x=mx-3$
$\displaystyle mx+2x=0$
$\displaystyle x(m+2)=0$
$\displaystyle x=0\text{ and }m = -2$

Now I'm pretty damn sure that this is not the solution, in fact I'm convinced that my whole method of going into this is wrong.

Last edited by skipjack; July 6th, 2019 at 07:42 AM.
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July 6th, 2019, 06:42 AM   #2
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Originally Posted by granitba View Post
$\displaystyle (-3-2x)/m=x-3$
$\displaystyle -3-2x=mx-3$
Should be $3-2x = mx -3m$ since you have to distribute. Maybe this can push you in the right direction.
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July 6th, 2019, 08:31 AM   #3
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Thanks for pointing out my mistake, didn't realize I forgot how to multiply.

That aside, I'm still stuck here:

$\displaystyle (1−mx)/2=2x-6 => x=(2+m)/7$ (equating the first and third equations and solving for x)
$\displaystyle (3x-3)/(m-2)=(2+m)/7$
$\displaystyle (3x-3)(2+m)=7(m-2)$
$\displaystyle 3m^(2)+3m-6=7m-14 $
$\displaystyle 3m^(2)-4m+8=0$
And here I can already see that this is wrong since the discriminant is negative (and m has to be a real number)
$\displaystyle D=(-4)^(2)-4*3*8=16-96=-80$

Last edited by skipjack; July 6th, 2019 at 09:43 AM.
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July 6th, 2019, 10:00 AM   #4
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$1 - mx = 2y = 2(x - 3) = 2x - 6$ implies $(m + 2)x = 7$.
The first two equations imply $(m + 2)x + (m + 2)y + 2 = 0$,
so $(m + 2)x + (m + 2)(x - 3) + 2 = 0$.
Hence $7 + 7 - 3(m + 2) + 2 = 0$, which leads to $m = 10/3$.
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July 7th, 2019, 10:51 AM   #5
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Oh, that's one of the many ways I tried to solve it, and I also got 10/3.
Problem is, 10/3 is not an option.
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July 7th, 2019, 11:16 AM   #6
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Quote:
Originally Posted by granitba View Post
Oh, that's one of the many ways I tried to solve it, and I also got 10/3.
Problem is, 10/3 is not an option.
I'm showing one solution

$\left\{m= \dfrac{10}{3},x= \dfrac{21}{16},y= -\dfrac{27}{16}\right\}$

option or not $m=\dfrac{10}{3}$ is the answer.
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July 7th, 2019, 12:37 PM   #7
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Thanks for the input, I guess the authors got it wrong, even WolframAlpha shows 10/3 to be the result (I thought it was some input error on my side).
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July 7th, 2019, 06:00 PM   #8
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What options were given?
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July 8th, 2019, 02:37 AM   #9
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They were: -4/5,1/3,-4/3,3/4
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July 8th, 2019, 07:23 AM   #10
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Perhaps the 1/3 option was a typo for 10/3. Is the problem from a textbook? If so, what book?
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