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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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The equation — E = mc2 — means "energy equals mass times the speed of light squared." It shows that energy (E) and mass (m) are interchangeable.

Hope this helped

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