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horrorfan [7]
4 years ago
13

does the following system of equations have a solution if so find it. 2x+y+z= 4 x-y+3z= -2 -x+y+z= -2

Mathematics
1 answer:
saul85 [17]4 years ago
5 0
I'd suggest using "elimination by addition and subtraction" here, altho' there are other approaches (such as matrices, substitution, etc.).

Note that if you add the 3rd equation to the second, the x terms cancel out, and you are left with the system

 - y + 3z = -2
   y  + z  =  -2
-----------------
         4z = -4, so z = -1.

Next, multiply the 3rd equation by 2:  You'll get -2x + 2y + 2z = -2.

Add this result to the first equation.  The 2x terms will cancel, leaving you with the system

2y + 2z = -2
  y  +  z  = 4

This would be a good time to subst. -1 for z.  We then get:

-2y - 2 = -2.  Then y must be 0.  y = 0.

Now subst. -1 for z and 0 for y in any of the original equations.

For example, x - (-1) + 3(0) = -2, so x + 1 = -2, or x = -3.

Then a tentative solution is     (-3, -1, 0).

It's very important that you ensure that this satisfies all 3 of the originale quations.
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Answer:

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Hello, I am currently very stuck with this problem and I am unsure as to how I would solve it.
borishaifa [10]

We have the equation

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Let's complete the square, to do it let's add and subtract 25 on the right side

\begin{gathered} 20y=x^2-10-15+25-25 \\  \\ 20y=(x-5)^2-15-25_{} \\  \\ 20y=(x-5)^2-40 \\  \\  \end{gathered}

Now we can have y in function of x

\begin{gathered} y=\frac{1}{20}(x-5)^2-2 \\  \\  \end{gathered}

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y=a(x-h)+k

Where the vertex is

(h,k)

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(5,-2)

Now we can continue and find the focus, the focus is

\mleft(h,k+\frac{1}{4a}\mright)

We have a = 1/20, therefore

\begin{gathered} \mleft(5,-2+5\mright) \\  \\ (5,3) \end{gathered}

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(5,3)

And the last one, the directrix, it's

y=k-\frac{1}{4a}

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\begin{gathered} y=-2-5 \\  \\ y=-7 \end{gathered}

Hence the correct answer is: vertex (5, -2); focus (5, 3); directrix y = -7

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