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Solve the system by elimination

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(Based on todays review)

  • ApusApus


    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.