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

#### granitba

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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#### Greens

$$\displaystyle (-3-2x)/m=x-3$$
$$\displaystyle -3-2x=mx-3$$
Should be $3-2x = mx -3m$ since you have to distribute. Maybe this can push you in the right direction.

• 1 person

#### granitba

Thanks for pointing out my mistake, didn't realize I forgot how to multiply.

That aside, I'm still stuck here:

$$\displaystyle (1âˆ’mx)/2=2x-6 => x=(2+m)/7$$ (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
$1 - mx = 2y = 2(x - 3) = 2x - 6$ 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 $m = 10/3$.

• 1 person

#### 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.

#### romsek

Math Team
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.

• 1 person

#### granitba

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
What options were given?

#### granitba

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

#### skipjack

Forum Staff
Perhaps the 1/3 option was a typo for 10/3. Is the problem from a textbook? If so, what book?