# Solve for x, y, and z x + 2y = 10 9y – 3z = - 6 6x – 5z = 2

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Here we are given three equations we have to solve to arrive at the values of x, y and z.

x + 2y = 10… (1)

9y – 3z = - 6… (2)

6x – 5z = 2… (3)

We can write (1) as x = 10-2y

We can also write (3) as -5z = 2 -6x

=> 5z= 6x-2

=> 5z= 6*(10-2y) -2

=>5 z = 60- 12y -2

=> 5z = 58 – 12y

=> z = (58 – 12y)/5

Now substitute these in (2)

9y – 3z = - 6

=> 9y – 3[(58 – 12y)/5] = -6

=> 45y -174+ 36y = -30

=> 81y = 144

=> y = 144/81

=> y = 16/9

Now x = 10-2y = 10 - 2*16/9= 58/9

Also z= (58 – 12y)/5= (58 – 12*16/9)/5= 22/3

**Therefore we get x= 58/9, y = 16/9 and z = 22/3**

We'll choose to eliminate z from the equation (2) and (3).

x + 2y = 10 (1)

9y – 3z = - 6 (2)

6x – 5z = 2 (3)

To eliminate z from (2) and (3), we'll multiply (2) by 5 and (3) by -3:

45y – 15z = - 30 (4)

-18x + 15z = -6 (5)

We'll add (4)+(5):

-18x + 45y = -36 (6)

We'll multiply (1) by 18 and we'll add to (6) the result:

18x + 36y - 18x + 45y = 180 - 36

We'll combine and eliminate like terms:

81y = 144

y = 144/81

**y = 16/9**

We'll substitute y in (1) and we'll get:

x + 2y = 10

x + 32/9 = 10

x = 10 - 32/9

**x = 58/9**

We'll substitute y in (2) and we'll get:

9y – 3z = - 6

144/9 - 3Z = -6

3Z = 144/9 + 6

z = 198/27

**z = 22/3**

To solve for x, y and z in the sysyem of equations:

Solution:

x + 2y = 10......(1)

9y – 3z = - 6,

Or 3y -z = -2 ...(2),

6x – 5z = 2......(3)

We Substitute from x = 10-2y in equation (3):

6(10-2y)-5z = 2.

60 -12y - 5z = 2.

-12y - 5z = 2-60 = - 58.

12y+5z = 58 ..(4) .

3y- z = -2.....(2), Or z = 3y+2.

We substitute z = 3y+2 in eq (4):

12y + 5(3y+2) = 58.

12y+15y +10 = 58

27y= 58-10 = 48 .

y = 48/27.

**y = 16/9.**

Put y = 16/8 in (1): x+2y = 10.

x+2(16)/9= 10

x = 10 - 32/9 = 58/9

**x= 58/9.**

Putx **= **48/9 in (3): 6x-5z.

6(58)/9 - 5z = 2.

-5z = 2-116/3 = - 110/3

**z = 22/3.**

Therefore the solutions are x = 58/3, y = 16/9 and z = 22/3