July 6th, 2019, 05:35 AM  #1 
Newbie Joined: Feb 2017 From: Peja Posts: 11 Thanks: 0 Math Focus: Jack of all trades, master of none, though often better, than master of one.  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: $\displaystyle mx+2y1=0$ $\displaystyle 2x+my+3=0$ $\displaystyle xy3=0$ One of the many things I tried was to write all three in slopeintercept form like this: $\displaystyle y=(1mx)/2$ $\displaystyle y=(32x)/m$ $\displaystyle y=x3$ 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: $\displaystyle (32x)/m=x3$ $\displaystyle 32x=mx3$ $\displaystyle mx+2x=0$ $\displaystyle x(m+2)=0$ $\displaystyle x=0\text{ and }m = 2$ 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. 
July 6th, 2019, 06:42 AM  #2 
Member Joined: Oct 2018 From: USA Posts: 87 Thanks: 59 Math Focus: Algebraic Geometry  
July 6th, 2019, 08:31 AM  #3 
Newbie Joined: Feb 2017 From: Peja Posts: 11 Thanks: 0 Math Focus: Jack of all trades, master of none, though often better, than master of one. 
Thanks for pointing out my mistake, didn't realize I forgot how to multiply. That aside, I'm still stuck here: $\displaystyle (1−mx)/2=2x6 => x=(2+m)/7$ (equating the first and third equations and solving for x) $\displaystyle (3x3)/(m2)=(2+m)/7$ $\displaystyle (3x3)(2+m)=7(m2)$ $\displaystyle 3m^(2)+3m6=7m14 $ $\displaystyle 3m^(2)4m+8=0$ And here I can already see that this is wrong since the discriminant is negative (and m has to be a real number) $\displaystyle D=(4)^(2)4*3*8=1696=80$ Last edited by skipjack; July 6th, 2019 at 09:43 AM. 
July 6th, 2019, 10:00 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,823 Thanks: 2160 
$1  mx = 2y = 2(x  3) = 2x  6$ 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 $m = 10/3$. 
July 7th, 2019, 10:51 AM  #5 
Newbie Joined: Feb 2017 From: Peja Posts: 11 Thanks: 0 Math Focus: Jack of all trades, master of none, though often better, than master of one. 
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. 
July 7th, 2019, 11:16 AM  #6 
Senior Member Joined: Sep 2015 From: USA Posts: 2,500 Thanks: 1372  
July 7th, 2019, 12:37 PM  #7 
Newbie Joined: Feb 2017 From: Peja Posts: 11 Thanks: 0 Math Focus: Jack of all trades, master of none, though often better, than master of one. 
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).

July 7th, 2019, 06:00 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,823 Thanks: 2160 
What options were given?

July 8th, 2019, 02:37 AM  #9 
Newbie Joined: Feb 2017 From: Peja Posts: 11 Thanks: 0 Math Focus: Jack of all trades, master of none, though often better, than master of one. 
They were: 4/5,1/3,4/3,3/4

July 8th, 2019, 07:23 AM  #10 
Global Moderator Joined: Dec 2006 Posts: 20,823 Thanks: 2160 
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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aka, equations, intersection, linear equation, lines, system, variables 
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