Solve

2x-y-5z=1
x+2y+3z=9
3x+6y+z=11

2x-3y=-13
-5x+2y=16

1. Eq1: 2x -y - 5z = 1.

Eq2: x + 2y + 3z = 9.
Eq3: 3x + 6y + z = 11.
Multiply Eq1 by 3 and Eq2 by 5:
6x - 3y -15z = 3
5x + 10y + 15z = 45
Add Eq1 and Eq2:
Eq4: 11x + 7y = 48

Eq2: x + 2y + 3z = 9
Multiply Eq3 by -3:
Eq3: -9x - 18y - 3z = -33
Add Eq2 and Eq3 and get:
Eq5: -8x - 16y = -24
Eq5: -x - 2y = -3

Eq4: 11x + 7y = 48
Eq5: -x - 2y = -3.
Multiply Eq5 by 11 and add to Eq4:
Eq5: -11x - 22y = -33
-15y = 15
Y = -1.
In Eq5, substitute -1 for y:
-x - 2*(-1) = -3
-x + 2 = -3
-x = -5
X = 5.
In Eq1, sub. 5 for X and -1 for Y:
2*5 - (-1) - 5z = 1
10 + 1 - 5z = 1
-5z = 1 - 11 = -10
Z = 2.
Solution set = (x,y,z) = (5,-1,2).

2. Eq1: 2x - 3y = -13.
Eq2: -5x + 2y = 16.
Multiply Eq1 by 5 and Eq2 by 2:
10x - 15y = -65.
-10x + 4y = 32.
Add the 2 Eqs:
-11y = -33
Y = 3.

In Eq1, substitute 3 for Y:
2x - 3*3 = -13
2x = -13 + 9 = -4
X = -2.
Solution set = (-2,3).

To solve the system of equations, we can use a method called elimination or substitution. Let's start with the elimination method.

First, let's rewrite the system of equations:

Equation 1: 2x - y - 5z = 1
Equation 2: x + 2y + 3z = 9
Equation 3: 3x + 6y + z = 11

Let's multiply Equation 2 by 2 and Equation 3 by -1 to create opposites for the y-term:

Equation 2 (multiplied by 2): 2x + 4y + 6z = 18
Equation 3 (multiplied by -1): -3x - 6y - z = -11

Now, we will add Equation 1 + Equation 2 + Equation 3 and Equation 2 + Equation 1 to eliminate the y-term:

2x - y - 5z + 2x + 4y + 6z + -3x - 6y - z = 1 + 18 - 11
2x + 2x - 3x - y + 4y - 6y - 5z + 6z - z = 1 + 18 - 11

Simplifying, we get:

-2x - y = 8 (Equation 4)
3x - 4y + z = 8 (Equation 5)

Now, we have a system of two equations with two variables:

Equation 4: -2x - y = 8
Equation 5: 3x - 4y + z = 8

We can solve this system using either substitution or elimination.

Let's try solving by substitution using Equation 4:

Solve Equation 4 for x:
-2x = 8 + y
x = (8 + y)/(-2)

Now substitute x in Equation 5:

3((8 + y)/(-2)) - 4y + z = 8
(24 + 3y) / (-2) - 4y + z = 8
(24 + 3y) - 4y*(-2) + z*(-2) = 8*(-2)
24 + 3y + 8y - 2z = -16

Combine like terms:

11y - 2z = -40 (Equation 6)

Now we have two equations:

Equation 4: -2x - y = 8
Equation 6: 11y - 2z = -40

To eliminate y, we can multiply Equation 4 by 11 and add it to Equation 6:

-22x - 11y = 88
11y - 2z = -40
--------------
-22x - 2z = 48 (Equation 7)

Now we have two equations:

Equation 7: -22x - 2z = 48
Equation 6: 11y - 2z = -40

To eliminate z, we can multiply Equation 7 by -5 and add it to Equation 6:

110x + 10z = -240
11y - 2z = -40
--------------
110x + 11y = -280 (Equation 8)

Now we have two equations:

Equation 8: 110x + 11y = -280
Equation 6: 11y - 2z = -40

To eliminate y, we can multiply Equation 8 by -1 and add it to Equation 6:

-110x - 11y = 280
11y - 2z = -40
--------------
-2z + 280 = 0

Simplifying, we get:

-2z = -280

Dividing the equation by -2, we find:

z = 140

Now we substitute z = 140 back into Equation 6:

11y - 2z = -40
11y - 2(140) = -40
11y - 280 = -40
11y = 240
y = 240 / 11

Finally, we substitute the values of y and z into Equation 4 to find x:

-2x - y = 8
-2x - (240 / 11) = 8
-2x = 8 + (240 / 11)
-2x = (88 + 240) / 11
-2x = 328 / 11
x = (328 / 11) / (-2)

Simplifying, we get:

x = -328 / 22

Hence, the solution to the system of equations is:
x = -328 / 22
y = 240 / 11
z = 140