1.

Can anybody solve this equation pls

Answer»

\huge \red{ANSWER}

-4x+3y-z=0.....EQ1

-X-3y+2z=0.......EQ2

x+y+z=27........EQ3

EQ1-EQ2

-4x+3y-z=0

-4x-12y+8z=0×4

-__________________

0+15x-9z=0

15y-9z=0.....EQ4

EQ2+EQ3

-x-3y+2z=0

x+y+z=27

___________________

-2y+3z=27....EQ5

Now,

EQ4+EQ5

15y-9z=0

-6y+9y=81×3

_______________

9y+0=81

Y=81/9

Y=9...EQ6

Sub,EQ6 in EQ4

15y-9z=0

15(9)-9z=0

9z=135

Z=135/9

Z=15...EQ7

sub,EQ6 & EQ7 in EQ3

X+Y+Z=27

X+9+15=27

X+24=27

X=27-24

X=3

THEN THE VALUES OF

X=3,Y=9 AND Z=15

sub, x, y, z values in EQ3

=>X+Y+Z=27

=>3+9+15=27

=>12+15=27

=>27=27

therefore, LHS=RHS

In the same WAY you can substitute in EQ1 & EQ2..


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