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-   -   Intersection of three lines - aka a system of three equations with three variables (http://mymathforum.com/algebra/346711-intersection-three-lines-aka-system-three-equations-three-variables.html)

 granitba July 6th, 2019 05:35 AM

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.

 Greens July 6th, 2019 06:42 AM

Quote:
 Originally Posted by granitba (Post 611366) $\displaystyle (-3-2x)/m=x-3$ $\displaystyle -3-2x=mx-3$
Should be $3-2x = mx -3m$ since you have to distribute. Maybe this can push you in the right direction.

 granitba July 6th, 2019 08:31 AM

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)
$\displaystyle D=(-4)^(2)-4*3*8=16-96=-80$

 skipjack July 6th, 2019 10:00 AM

$1 - mx = 2y = 2(x - 3) = 2x - 6$ 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 $m = 10/3$.

 granitba July 7th, 2019 10:51 AM

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.

 romsek July 7th, 2019 11:16 AM

Quote:
 Originally Posted by granitba (Post 611386) 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.

 granitba July 7th, 2019 12:37 PM

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).

 skipjack July 7th, 2019 06:00 PM

What options were given?

 granitba July 8th, 2019 02:37 AM

They were: -4/5,1/3,-4/3,3/4

 skipjack July 8th, 2019 07:23 AM

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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