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Well this is awkward. I've been through three semesters of calculus and yet this system of linear equations is causing me to brain fart.

$$\begin{align*} &&3y&=4x\\ &&2x+3z&=4y\\ 2x+y+z=4z &\to& 2x+y&=3z\\ &&x+y+z&=1 \end{align*}$$

I can obviously just get the answer with Wolfram Alpha, but I'm really bugged that I can't figure this out. All I've been able to do is fudge around terms and I'm not getting anywhere.

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I see arrows in there. Which are the equations that you are given and which are the ones where you do computation? –  Patrick Da Silva Dec 4 '11 at 1:17
    
One arrow meaning I saw the simplification. $2x+y+z=4z$ is the same as $2x+y=3z$. –  Bob K Dec 4 '11 at 1:19
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As it is, there is no solution (the equations are inconsistent), but there's probably a typo. The question says $3y=4z$ but the link says $3y=4x$. –  p.s. Dec 4 '11 at 1:34
    
Yes, that should be $3y=4x$. I'll update the question. –  Bob K Dec 4 '11 at 1:38
    
Do Gauss-Jordan elimination on the matrix of the system... –  Arturo Magidin Dec 4 '11 at 2:22
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1 Answer

Note: This refers to the system as original given.

Rewrite the equations: $$\begin{array}{rcccccl} &&3y&-&4z&=&0\\ 2x&-&4y&+&3z&=&0\\ 2x&+&y&-&3z&=&0 \\ x&+&y&+&z&=&1 \end{array}$$ Now simply solve the system: $$\begin{align*} \left(\begin{array}{rrr|c} 0 & 3 & -4 & 0\\ 2 & -4 & 3 & 0\\ 2 & 1 & -3 & 0\\ 1 & 1 & 1 & 1 \end{array}\right) &\to \left(\begin{array}{rrr|r} 1 & 1& 1 & 1\\ 0 & 3 & -4 & 0\\ 2 & -4 & 3 & 0\\ 2 & 1 & -3 & 0 \end{array}\right) \to\left(\begin{array}{rrr|r} 1 & 1 & 1 & 1\\ 0 & 3 & -4 & 0\\ 0 & -6 & 1 & -2\\ 0 & -1 & -5 & -2 \end{array}\right)\\ &\to \left(\begin{array}{rrr|r} 1 & 1 & 1 & 1\\ 0 & 1 & 5 & 2\\ 0 & 3 & -4 & 0\\ 0 & -6 & 1 & -2 \end{array}\right) \to \left(\begin{array}{rrr|r} 1 & 1 & 1 & 1\\ 0 & 1& 5 & 2\\ 0 & 0 & -19 & -6\\ 0 & 0 & 31 & 10 \end{array}\right). \end{align*}$$ The last two lines tell us that the system is inconsistent, so the system has no solution.

Alternatively: plugging in $\frac{4}{3}z$ for $y$, we obtain from the second equation $$\begin{align*} 2x - \frac{16}{3}z + 3z &= 0\\ 2x -\frac{7}{3}z &= 0\\ 6x -7z &=0.\end{align*}$$ The third equation gives $$\begin{align*} 2x +\frac{4}{3}z - 3z &= 0\\ 2x -\frac{5}{3}z &=0\\ 6x - 5z&=0 \end{align*}$$ Since $6x=7z$ and $6x=5z$, then $x=z=0$, hence $y=0$, which makes $x+y+z=1$ impossible.


If the system is meant to be $$\begin{array}{rcccccl} -4x&+&3y&&&=&0\\ 2x&-&4y&+&3z&=&0\\ 2x&+&y&-&3z&=&0 \\ x&+&y&+&z&=&1 \end{array}$$ then proceed as above to solve the system. $$\begin{align*} \left(\begin{array}{rrr|r} -4 & 3 & 0 & 0\\ 2 & -4 & 3 & 0\\ 2 & 1 & -3 & 0\\ 1 & 1 & 1 & 1 \end{array}\right) &\to \left(\begin{array}{rrr|r} 1 & 1 &1 & 1\\ -4 & 3 & 0 & 0\\ 2 & -4 & 3 & 0\\ 2 & 1 & -3 & 0 \end{array}\right) \to \left(\begin{array}{rrr|r} 1 & 1 & 1 & 1\\ 0 & -5 & 6 & 0\\ 2 & -4 & 3 & 0\\ 0 & 5 & -6 & 0 \end{array}\right)\\ &\to \left(\begin{array}{rrr|r} 1 & 1 & 1 & 1\\ 0 & -5 & 6 & 0\\ 0 & -6 & 1 & -2 \end{array}\right) \to \left(\begin{array}{rrr|r} 1 & 1 & 1 & 1\\ 0 & 1 & 5 & 2\\ 0 & -5 & 6 & 0 \end{array}\right)\\ &\to \left(\begin{array}{rrr|r} 1 & 1 & 1 & 1\\ 0 & 1 & 5 & 2\\ 0 & 0 & 31 & 10 \end{array}\right) \to \left(\begin{array}{rrr|r} 1 & 0 & -4 & -1\\ 0 & 1 & 5 & 2\\ 0 & 0 & 1 & \frac{10}{31} \end{array}\right)\\ &= \left(\begin{array}{rrr|r} 1 & 0 & 0 & \frac{9}{31}\\ 0 & 1 & 0 & \frac{12}{31}\\ 0 & 0 & 1 & \frac{10}{31} \end{array}\right). \end{align*}$$ So the solution is $x = \frac{9}{31}$, $y=\frac{12}{31}$, $z=\frac{10}{31}$.

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