Math, asked by bhargavi1019, 1 year ago

solve the equations 2x+3y+z=9,x+2y+3z=6,3x+y+2z=8​

Answers

Answered by ashishks1912
21

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)

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

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)

Now subtracting (4) by (2),

​6x+9y+3z=27

x+2y+3z=6

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

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

_______________

Multiply the equation (1) into 2

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

Now subtracting (6) by (3),

4x+6y+2z=18

3x+y+2z=8​

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

x+5y=10\hfill (7)

_______________

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

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

Subtracting the equations (5) and (8)

5x+7y=21

5x+25y=50

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

 -18y=-29

y=\frac{29}{18}

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

5x+7(\frac{29}{18})=21

5x=21-\frac{203}{18}

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

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

x=\frac{35}{18}

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

2(\frac{35}{18})+3(\frac{29}{18})+z=9

z=9-\frac{35}{9}-\frac{29}{6}

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

z=\frac{5}{18}

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

x=\frac{35}{18} , y=\frac{29}{18} and z=\frac{5}{18}.

Answered by olivaministriesvenka
0

Step-by-step explanation:

say these answer in rank method

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