Solve the following system of equations: x + 3y + z = 2, x + 2y + 3z = 7, x + 2y + 5z = 21

To solve the given system of equations, we have:

• 1. x + 3y + z = 2
• 2. x + 2y + 3z = 7
• 3. x + 2y + 5z = 21

First, let’s label our equations for convenience:

1. Equation (1): x + 3y + z = 2
2. Equation (2): x + 2y + 3z = 7
3. Equation (3): x + 2y + 5z = 21

We can eliminate x from equations (2) and (3) by subtracting the equations:

Subtracting Equation (1) from Equation (2):

(x + 2y + 3z) - (x + 3y + z) = 7 - 2

This simplifies to:

-y + 2z = 5 	ag{4}

Next, subtract Equation (1) from Equation (3):

(x + 2y + 5z) - (x + 3y + z) = 21 - 2

This simplifies to:

-y + 4z = 19 	ag{5}

Now we have a new system of equations (4) and (5):

• Equation (4): -y + 2z = 5
• Equation (5): -y + 4z = 19

Next, we can eliminate y by setting equations (4) and (5) equal:

-y + 2z = 5
-y + 4z = 19

Subtracting Equation (4) from Equation (5):

(-y + 4z) - (-y + 2z) = 19 - 5

This simplifies to:

2z = 14

So:

z = 7

Now substitute z = 7 back into Equation (4) to find y:

-y + 2(7) = 5

Solving this gives:

-y + 14 = 5
-y = -9
y = 9

Now, we substitute y = 9 and z = 7 into Equation (1) to find x:

x + 3(9) + 7 = 2

Solving gives:

x + 27 + 7 = 2
x + 34 = 2
x = 2 - 34
x = -32

Thus, the solution to the system of equations is:

• x = -32
• y = 9
• z = 7

In conclusion, the solution set is (x, y, z) = (-32, 9, 7).