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Artist 52 [7]
3 years ago
5

What is the solution to this system of equations?

Mathematics
1 answer:
Ede4ka [16]3 years ago
8 0

Answer:

First choice.

Step-by-step explanation:

You could plug in the choices to see which would make all the 3 equations true.

Let's start with (x=2,y=-6,z=1):

2x+y-z=-3

2(2)+-6-1=-3

4-6-1=-3

-2-1=-3

-3=-3 is true so the first choice satisfies the first equation.

5x-2y+2z=24

5(2)-2(-6)+2(1)=24

10+12+2=24

24=24 is true so the first choice satisfies the second equation.

3x-z=5

3(2)-1=5

6-1=5

5=5 is true so the first choice satisfies the third equation.

We don't have to go any further since we found the solution.

---------Another way.

Multiply the first equation by 2 and add equation 1 and equation 2  together.

2(2x+y-z=-3)

4x+2y-2z=-6 is the first equation multiplied by 2.

5x-2y+2z=24

----------------------Add the equations together:

9x+0+0=18

9x=18

Divide both sides by 9:

x=18/9

x=2

Using the third equation along with x=2 we can find z.

3x-z=5 with x=2:

3(2)-z=5

6-z=5

Add z on both sides:

6=5+z

Subtract 5 on both sides:

1=z

Now using the first equation along with 2x+y-z=-3 with x=2 and z=1:

2(2)+y-1=-3

4+y-1=-3

3+y=-3

Subtract 3 on both sides:

y=-6

So the solution is (x=2,y=-6,z=1).

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Comment

It's a good thing that the domain is confined or the graph is. That function is undefined if x < - 3 

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One more and then we'll start drawing conclusions. If x = 9 then f(9) = y =  6. 
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OK I think we should be ready to look at answers. There's nothing there that makes the answer anything but a. Let's find out what the problem is with the rest of the choices.

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D
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