Solving simultaneous linear equations in three variables

Algebra Level 1

Find the solution to the following system of linear equations:

x + 3 y + 3 z = − 8 x + 3 y + 4 z = − 1 x + 4 y + 3 z = − 4. \begin{aligned} x+3y+3z &=-8 \\ x+3y+4z &=-1 \\ x+4y+3z &=-4. \end{aligned}

x = 5 , y = 7 , z = − 27 x=5, y=7, z=-27 x = 0 , y = − 33 , z = 14 x=0, y=-33, z=14 x = − 41 , y = 4 , z = 7 x=-41, y=4, z=7 x = 3 , y = 10 , z = − 21 x=3, y=10, z=-21

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Lawahar Qasid by Lawahar Qasid

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2 solutions

Subtracting the first equation from the second equation we get ( x + 3 y + 4 z ) − ( x + 3 y + 3 z ) = z = − 1 − ( − 8 ) = − 1 + 8 = 7 \color{#D61F06}{(x+3y+4z)-(x+3y+3z)=z=-1-(-8)=-1+8=7} .Subtracting the first equation from the third equation we get ( x + 4 y + 3 z ) − ( x + 3 y + 3 z ) = y = − 4 − ( − 8 ) = − 4 + 8 = 4 \color{#20A900}{(x+4y+3z)-(x+3y+3z)=y=-4-(-8)=-4+8=4} .Putting the values of z z and y y in the first equation we get x + 3 ( 4 ) + 3 ( 7 ) = x + 12 + 21 = x + 33 = − 8 → = − 8 − 33 = − 41 \color{#3D99F6}{x+3(4)+3(7)=x+12+21=x+33=-8\rightarrow=-8-33=-41} .So the solution family is: { x = − 41 y = 4 z = 7 \boxed{\begin{cases}x=-41\\y=4\\z=7\end{cases}}

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K P
Feb 28, 2019

Using elementary row transformations and then premultiplying the inverse of the coefficient matrix thus obtained on both sides of the system of given linear equations (matrix form) gives solution as:- [ x y z] = [-41 4 7]. :) I got column matrix though🤣🤣. Steps of elementary row transformation are given below:- Step1. row3=row3 + row2×(-1). Step2. row2=row2 + row1×(-1). Step3.row1=row1+ row3×(-3). Step4.row2=row2+row3. Step5.row3=row3 +row2×(-1). Step6.row1=row1 +row3×6. Step7.row3=row3×(-1).

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