Intersection of three lines - aka a system of three equations with three variables

Feb 2017
11
0
Peja
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.
 
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Feb 2017
11
0
Peja
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)
\(\displaystyle D=(-4)^(2)-4*3*8=16-96=-80\)
 
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skipjack

Forum Staff
Dec 2006
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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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Feb 2017
11
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Peja
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

Math Team
Sep 2015
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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.
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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Feb 2017
11
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Peja
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

Forum Staff
Dec 2006
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What options were given?
 

skipjack

Forum Staff
Dec 2006
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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?