Question

What does the values of k so that the following system have a solution
x-y-3z=3 ; 2x+z=0 and 2y+7z=k

Answer 1

Answer:

k= 19

Step-by-step explanation:

given three eq

x-y-3z=3_____1

2x+z=0_______2

and taking 1 and 2 eq

x-y-3z=3

2x+ z=0

________

3x-2z=3

_______

x-y-z=2________3

talking eq 1and 3

x-y-3z=3

x-y-z=2

- + - =-

_______

z=1

____

putting the value in eq 2

2x+1=0

x+1=0

x= -1

putting value in eq 1

-1-y -3×1=3

-y-3=3

-y=3+3=6

-y=6

taking third give eq

2y+7z=k

2×6+7×1=k

12+7= k

19= k

Answer

hope it ur answer

Answer 2

Answer:

Correct option is D)

let x+2y+3z=14 ------------ (1)

3x+y+2z=11 ---------(2)

2x+3y+z=11 -----------(3)

multiplying eqn (2) by 2 and subtracting eqn(1) from it we get,

=5x+z=8 -------------(4)

again multiplying eqn (2) by 3 and subtracting eqn (3) from it we get,

= 7x+5z=22 ------------(5)

Now multiply eqn (4) by 5 and subtract eqn (5) from it we get

18x=18

∴x=1

substituting the x in eqn (4) we get the value of z as

= 5(1)+z=8

∴z=8−5=3

and substitute x and z in eqn (1) we get

1+2y+3(3)=14

2y=14−1−9=4

∴y=2

hence the answer is option D