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mina [271]
3 years ago
9

Solve the system by elimination

Mathematics
1 answer:
Alja [10]3 years ago
3 0

First of all we will eliminate x from our equations. In order to do that we will use our first and second equation and then we will use second and third equation.

-2x+2y+3z=0...(1)\\-2x-y+z=-3....(2)

Upon subtracting 2 equation from 1 we will get,

3y+2z=3....(4)

Now we will use second and third equation to eliminate x.

-2x-y+z=-3....(2)\\2x+3y+3z=5....(3)

Adding 2nd and 3rd equation we will get,

2y+4z=2....(5)

Now we will find out y from our 4th and 5th equation.

2*(3y+2z)=2*3....(4)\\2y+4z=2....(5)

Upon subtracting 5th equation from 4th equation we will get,

4y=4\\y=1

Now let us find out z by substituting y's value in 5th equation.

2*1+4z=2\\4z=2-2\\z=0

Now we will find x from by substituting y and z's value in equation 1.

-2x+2*1+3*0=0\\-2x+2=0\\2=2x\\x=1

Therefore, x=1, y=1 and z=0 is the solution of the given system.

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1. all real numbers less than or equal to 4 = </= 4.

Answer = YES!

2. all real numbers less than or equal to -3 = </= -3.

Answer = YES!

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Answer = YES!

4. all real numbers greater than or equal to -3 = >/= -3

Answer = NO!

Step-by-step explanation:

1. all real numbers less than or equal to 4 = </= 4.

1 + 1 = 2

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2 + 2 = 4

2. all real numbers less than or equal to -3 = </= -3.

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3. all real numbers greater than or equal to 4

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