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Chapter 3: Q26. (page 148)

26: Solve each system of equations.

x+4y-z=63x+2y+3z=162x-y+z=3

Short Answer

Expert verified

The solution isx=1,y=2,z=3.

Step by step solution

01

Step-1 – Use elimination to make a system of two equations in two variables.

Given equations are-:

x+4y-z=6......(1)3x+2y+3z=16.......(2)2x-y+z=3.......(3)

Multiplying the first equation by 3 and subtracting with second equation, we get

12y-2y-3z-3z=18-1610y-6z=2........(4)

Again, multiplying the first equation by 2and subtracting with third equation, we get

8y+y-2z-z=12-39y-3z=9........(5)

02

Step-2 – Solve the system of two equations.

10y-6z=2.......(4)9y-3z=9........(5)

Multiplying the fifth equation by 2 and subtracting from fourth equation.

10y-18y=2-18-8y=-16y=2

Putting the value of yin the fourth equation, we get

10y-6z=210(2)-6z=2-6z=2-20-6z=-18z=3

03

Step-3 – Substituting the value according to the coordinates in one of the equations with three variables.

x+4y-z=6x+4(2)-3=6x+8-3=6x=1

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