6
$\begingroup$

$\begin{align*} x + 2y - z &= -3\\ 2x - 4y + z &= -7\\ -2x + 2y - 3z &= 4 \end{align*}$

I multiply (1) by $-2$ and get $-2x-4y+2z=6$ and then add it to (2) which gives me $-8y+3z = -1$.

I add (2) and (3) and get $-2y-2z=-3$

I take the first result and get $y=\frac{-1-3z}{8}$ putting that into $-2y-2z=-3$ I get $\frac{-2(-1-3z)}{8-2z}=-3$ which gives me $2+6z-16z = -24$, which gives me $10z=-26$ which I know isn't right already. I messed up the math somewhere, but I have done this problem about a dozen times and always get the wrong answer no matter what.

  • 1
    From $-8y+3y=-1$ you get first $-8y=-1-3y$. Then $y=\frac{-1-3y}{-8}=\frac{1+3y}{8}$, not $y=\frac{-1-3y}{8}$, because you have to divide $-8y=-1-3y$ by $-8$ and not by $8$.2011-04-17

2 Answers 2

3

You should get $y=\frac{1+3z}{8}$, not $y=\frac{-1-3z}{8}$.

The answer should be x=-3, y=1/2, z=1.

  • 0
    I just messed up basic math again, this is why I fail all my tests, I took my time, double checked everything, attempted to prove my order of operations was correct by making up a nother series of numbers and multiplying it both ways to see which gave me the proper asnwer, then using that on my equation.2011-04-17
3

Your mistake is when you solve for $y$ from the new (2). You have $-8y + 3z = -1$ from which you get $-8y = -1 -3z.$ If you now divide by $-8$, you get $y = \frac{-1-3z}{-8} = \frac{-(1+3z)}{-8} = \frac{1+3z}{8}.$

I should also point out that you need to be careful with the order. After you add a multiple of the first equation to the second, your system now looks like $\begin{align*} x + 2y - z &= -3\\ - 8y + 3z &= -1\\ -2x + 2y - 3z &= 4 \end{align*}$ so you are not adding the second equation to the third to eliminate $-2x$. Rather, you first add equation (2) to equation (3), and then you add -2 times the first equation to the second.