# How to go about solving this system of equations

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
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

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 $x = \frac{9}{31}$, $y=\frac{12}{31}$, $z=\frac{10}{31}$.