My Math Forum Intersection of three lines - aka a system of three equations with three variables

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 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: $\displaystyle mx+2y-1=0$ $\displaystyle 2x+my+3=0$ $\displaystyle x-y-3=0$ 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.
July 6th, 2019, 06:42 AM   #2
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Quote:
 Originally Posted by granitba $\displaystyle (-3-2x)/m=x-3$ $\displaystyle -3-2x=mx-3$
Should be $3-2x = mx -3m$ since you have to distribute. Maybe this can push you in the right direction.

 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: $\displaystyle (1−mx)/2=2x-6 => x=(2+m)/7$ (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.
 July 6th, 2019, 10:00 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,965 Thanks: 2214 $1 - mx = 2y = 2(x - 3) = 2x - 6$ 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 $m = 10/3$. Thanks from granitba
 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
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Quote:
 Originally Posted by granitba 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.

 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: 20,965 Thanks: 2214 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: 20,965 Thanks: 2214 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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