Math, asked by priyarwt1112, 5 months ago

2x + 3y + z = 9
4x + y = 7
x - 3y - 7z = 6​

Answers

Answered by archana7148
6

Answer:

Karna Kya hai

complete the question

the question is incomplete

Answered by imrockstar0001
1

GIVEN :

The equations are 2x+3y+z=9, x+2y+3z=6, 3x+y+2z=8

TO FIND :

The values of x, y, and z in the given equations.

SOLUTION :

Given that the equations are

2x+3y+z=9\hfill (1)2x+3y+z=9\hfill(1)

x+2y+3z=6\hfill (2)x+2y+3z=6\hfill(2)

3x+y+2z=8\hfill (3)3x+y+2z=8\hfill(3)

Now solving the equations (1) ,(2) and (3) by using Elimination method.

Multiply the equation (1) into 3

6x+9y+3z=27\hfill (4)6x+9y+3z=27\hfill(4)

Now subtracting (4) by (2),

6x+9y+3z=27

x+2y+3z=6

(-)_(-)_(-)_(-)___

5x+7y=21\hfill (5)5x+7y=21\hfill(5)

_______________

Multiply the equation (1) into 2

4x+6y+2z=18\hfill (6)4x+6y+2z=18\hfill(6)

Now subtracting (6) by (3),

4x+6y+2z=18

3x+y+2z=8

(-)_(-)_(-)_(-)___

x+5y=10\hfill (7)x+5y=10\hfill(7)

_______________

Now multiply the equation (7) into 5 we get,

5x+25y=50\hfill (8)5x+25y=50\hfill(8)

Subtracting the equations (5) and (8)

5x+7y=21

5x+25y=50

(-)_(-)__(-)____

-18y=-29

∴ y=\frac{29}{18}y=

18

29

Substitute the value of y in equation (5) we get

5x+7(\frac{29}{18})=215x+7(

18

29

)=21

5x=21-\frac{203}{18}5x=21−

18

203

5x=\frac{378-203}{18}5x=

18

378−203

x=\frac{175}{18}\times \frac{1}{5}x=

18

175

×

5

1

∴ x=\frac{35}{18}x=

18

35

Substitute the values of x and y in equation (1) we get

2(\frac{35}{18})+3(\frac{29}{18})+z=92(

18

35

)+3(

18

29

)+z=9

z=9-\frac{35}{9}-\frac{29}{6}z=9−

9

35

6

29

z=\frac{162-70-87}{18}z=

18

162−70−87

∴ z=\frac{5}{18}z=

18

5

∴ the values of x , y and z are \frac{35}{18},\frac{29}{18},\frac{5}{18}

18

35

,

18

29

,

18

5

respectively.

∴ x=\frac{35}{18}x=

18

35

, y=\frac{29}{18}y=

18

29

and z=\frac{5}{18}z=

18

5

.

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