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Solve each system of equations.

x+y+z=−12x+4y+z=1x+2y−3z=−3

Short Answer

Expert verified

The solution of the given equation is x=−4,y=2andz=1.

Step by step solution

01

– Use elimination method to make a system of two equations in two variables

x+y+z=−12x+4y+z=1x+2y−3z=−3

....(1)....(2)....(3)

Multiplying equation (1) by 2, we get

2x+2y+2z=−2

....(4)

Subtracting (1) from (4), we get

2x+2y+2z+2−2x+4y+z−1=02x+2y+2z+2−2x−4y−z+1=0−2y+z+3=02y−z−3=02y−z=3

....(5)

Subtracting (3) from (1), we get

x+y+z+1−x+2y−3z+3=0x+y+z+1−x−2y+3z−3=0−y+4z−2=0y−4z+2=0y−4z=−2

....(6)

02

– Solve the system of two equations

Multiplying (6) by 2, we get

2y−8z=−4

....(7)

Subtracting (7) from (5), we get

2y−z−3−2y−8z+4=02y−z−3−2y+8z−4=07z−7=07z=7z=1

Substituting the value of in (5), we get

2y−z=32y−1=32y=3+12y=4y=2

03

– Evaluate the value of x

Substituting the value of zand yin (1), we get

x+y+z=−1x+2+1=−1x+3=−1x=−1−3x=−4

Thus, the values are x=−4,y=2andz=1.

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