# What's the solution to this system of equations?

$\tag{1}x - y - 1 = 0$

$\tag{2}x + y + z = 3$

$\tag{3}x - y - z = 0$

I got $x = 3/2$, $y = 3/2$ and $z = 0$ but the answer is $P(1.5,0.5,1)$

http://s16.postimage.org/pj1b0tp6s/lol.jpg

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You can easily enough check that your answer is wrong by substituting into the original equations; similarly, you can check that the other answer is correct. There are many ways to get the correct answer; here’s one simple one. You have this system

\left\{\begin{align*} &x-y-1=0\\ &x+y+z=3\\ &x-y-z=0 \end{align*}\right.

From the first equation you have $x-y=1$; substituting this into the third equation yields $1-z=0$, or $z=1$. The first and second equations can then be rewritten as

\left\{\begin{align*} &x-y=1\\ &x+y=2\;; \end{align*}\right.

adding them yields $2x=3$, or $x=\frac32$, and therefore $y=2-\frac32=\frac12$.

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You can plug in to see yours does not satisfy the first.

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Hint $\$ By $(3)$ and $(1),$ $\rm\: z = x-y = 1,\:$ not $\rm\:z = 0.\:$ Further $\rm\,(2)+(3)\,$ yields $\rm\,2\,x = 3,\,$ so $\rm\:x = 3/2.\:$ Finally by $(1)$ we have $\rm\:y = x-1 = 1/2,\:$ and these values also satisfy $(2)$ and $(3)$.

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