Question

Solve the linear equation:6x+2y-5=13;3x+3y-2z=13;7x+5y-3z=26​

Answer 1

$x = 2$

$y = 3$

$z = 1$

Answer 2

The value is x=2,y=3, and z=1.

Given:

6x+2y-5=13

3x+3y-2z=13

7x+5y-3z=26​

To find:

The values of x,y,z.

Solution:

The linear equations are

$6x + 2y - 5z = 13 ----(1)\\3x + 3y - 2z = 13----(2)\\7x + 5y - 3z = 26----(3)$

Solve equation (3)for the variable  y

$5y = -7x + 3z + 26\\y = -7x/5 + 3z/5 + 26/5$

Plug this in for variable  y  in equation [1]

$6x + 2*(-7x/5+3z/5+26/5) - 5z = 13\\16x/5 - 19z/5 = 13/5\\16x - 19z = 13$

Plug this in for variable  y  in equation [2]

$3x + 3(-7x/5+3z/5+26/5) - 2z = 13\\-6x/5 - z/5 = -13/5\\-6x - z = -13$

Solve equation [2] for the variable  z

$z = -6x + 13$

Plug this in for variable  z  in equation [1]

$16x - 19(-6x+13) = 13\\130x = 260$

Solve equation [1] for the variable  x

$130x = 260 \\x = 2$

By now we know this much :

$x = 2\\y = -7x/5+3z/5+26/5\\z = -6x+13$

Use the  x  value to solve for  z

$z = -6(2)+13 = 1$

Use the  x  and  z  values to solve for  y

$y = -(7/5)(2)+(3/5)(1)+26/5 \\\\ y=3$

Thus,  {x,y,z} = {2,3,1}

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