Question 2157
Lets number each equation so we can refer to each one with only a number (instead of writing the whole equation).
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1)  x + 2y + z = 6
2)  x +  y     = 4
3) 3x +  y + z = 8
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Now lets eliminate 1 (one) variable from from all the equations.  Lets start with only 2 equations at a time.  I will eliminate x from #1 & #2 first.
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(-1) * #1 = -x - 2y - z = -6
       #2 =  x +  y     = 4
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Then add these, and we will get equation #4, like this
#4 = -y - z = -2
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Now, lets eliminate the SAME variable (x) from equation #1 & #3 (since before, we eliminated x from #1 & #2)
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(-3) * #1 = -3x - 6y - 3z = -18
       #3 =  3x +  y +  z =   8
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Add these, and we get equation #5, which is
#5 = -5y - 2z = -10
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All we have now is 2 equations with 2 variables (#4 & #5).  We can use "substitution to solve for y in #4 and plug it in #5.
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#4 = y = 2 - z
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Now plug in #4, which we solved for y, into #5 and simplify.
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-5(2 - z) - 2z = -10
<b>z = 0</b>
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Now, plug in z=0 in either #4 or #5.  I chosed #4 because it looks easier.
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-y - 0 = -2
<b>y = 2</b>
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Now, since we have y & z, we can plug these values into either of the original equations (#1, #2, or #3).  I chosed #1 because it looks easier.
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x + 2(2) + 0 = 6
<b>x = 2</b>
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Now we have our final answer, <b> (2, 2, 0)</b>, as in (x, y, z).
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CHECK
#1. 2 + 2(2) + 0 = 6
#2. 2 + 2 = 4
#3. 3(2) + 2 + 0 = 8
MS