Question

Solve by system of equations:5x+3y+z=-8x-3y+2z=2014x-2y+3z=20

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ANSWERED
asked 2021-05-28

Solve by system of equations:
\(5x+3y+z=-8\)
\(x-3y+2z=20\)
\(14x-2y+3z=20\)

Answers (1)

2021-05-29

Step 1
Given system of equations are
\(5x+3y+z=-8\)...(1)
\(x-3y+2z=20\)...(2)
\(14x-2y+3z=20\)...(3)
Step 2
By subtracting 5 times equation (2) from equation (1), we get
\(5x+3y+z-5(x-3y+2z)=-8-100\)
\(18y-9z=-108\)
\(2y-y=-12\)...(4)
Step 3
By subtracting 14 times equation (2) from equation (3), we get
\(14x-2y+3z-14(x-3+2z)=20-280\)
\(40y-25z=-260\)
\(8y-5z=-52\)...(5)
Step 4
By subtracting 4 times equation (4) from equation (5), we get
\(8y-5z-4(2y-z)=-52+48\)
\(-z=-4\)
\(z=4\)
Step 5
By substituting \(z=4\) in equation (4), we get value of y and then substituting values of y and z in equation (2), we get value of x
\(2y-z=-12\)
\(2y-4=-12\)
\(2y=-8\)
\(y=-4\)
\(x-3y+2z=20\)
\(x-3(-4)+2(4)=20\)
\(x+12+8=20\)
\(x=0\)
Step 6
Answer: Solution is \(x=0\),\(y=-4\) and \(z=4\)

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