can someone help me with this math question?Solve the system of equations given below.

x+3z=12
-x-2y+z=10
3x+5y+2z=-7

I am really stumped. Please help! Thanks in advance!

You want to take your 3 equations in 3 variables to 2 equations with the same 2 variables. Since the first equation has x and z, lets see if we can eliminate the y by combining the 2nd and 3rd equations.

To eliminate y you need the coefficients to be equal and opposite. So you would want a -10 and a + 10 since you are dealing with a -2 and 5.

Multiply the second equation by 5 to get
-5x-10y + 5z = 50
Multiply the 3rd equation by 2 to bet
6x + 10y + 4z = -14

Now, combine those two equations and you will bet x + 9z =36
Now, combine this equation with the first equation. If you want to eliminate x, then multiply the first equation by negative 1.

-x -3z = -12 combine this with x+9z = 36

The x will be eliminated.
6z =48 Solve for z.

Once you have z, put it into the first equation and solve for x.

Then put your value for x and z into the second equation to find y.

Once you have (x,y,z) you should check those value in all three equations.

from the first: x = 12-3z

plug that into the 2nd:
-12+3z-2y+z = 10
4z - 2y = 22
2z - y = 11 , #4

plut x = 12-3z into the 3rd
36-9z +5y + 2z = -7
-7z + 5y = -43 , #5

#4 times 5 ---> 10z - 5y = 55
add this to #5
3z = 12
z = 4
in #4: 8-y = 11
y = -3
back in x = 12-3z
x = 12-12 = 0

x=0 , y = -3, z = 4

As you can see, there are many ways to do this.
Just wiggle the variables back and forth a bit, look for simple combinations

Of course! I can help you solve the system of equations step-by-step.

Step 1: Let's start by simplifying the system of equations by eliminating one variable at a time. We'll use the method of elimination.

Step 2: Add the first and second equations together to eliminate "x":
(x + 3z) + (-x - 2y + z) = 12 + 10
3z - 2y + 2z = 22
5z - 2y = 22 - (Equation 1)

Step 3: Add the first and third equations together to eliminate "x":
(x + 3z) + (3x + 5y + 2z) = 12 - 7
4x + 8y + 5z = 5 - (Equation 2)

Step 4: Add twice the second equation to the third equation to eliminate "y":
2(-x - 2y + z) + (3x + 5y + 2z) = 20 + (-7)
-2x - 4y + 2z + 3x + 5y + 2z = 13
x + y + 4z = 13 - (Equation 3)

Step 5: Now we have a system of three equations:
5z - 2y = 22 - (Equation 1)
4x + 8y + 5z = 5 - (Equation 2)
x + y + 4z = 13 - (Equation 3)

We will now solve this system using a method called substitution.
From Equation 3, we have:
y = 13 - x - 4z

Substitute this value of "y" into Equations 1 and 2:
5z - 2(13 - x - 4z) = 22
4x + 8(13 - x - 4z) + 5z = 5

Simplify and solve these equations to find the values of "x" and "z".

I'll be here if you need help with further steps!

Of course! I'd be happy to help you solve this system of equations. There are several methods to solve systems of equations, but I will guide you through the process of elimination.

Step 1: Take the first two equations and eliminate one variable by adding or subtracting them. In this case, we can eliminate the "x" variable. Add the first equation to the negation of the second equation:

(x + 3z) + (-x - 2y + z) = (12) + (-10)
3z - 2y + 2z = 2

Simplifying this equation gives us:
5z - 2y = 2 ------ Equation A

Step 2: Now, let's eliminate the "x" variable between the first and third equations. Multiply the first equation by 3 and add it to the third equation:

3(x + 3z) + (3x + 5y + 2z) = 3(12) + (-7)
3x + 9z + 3x + 5y + 2z = 36 - 7
6x + 11z + 5y = 29

Simplifying this equation gives us:
6x + 11z + 5y = 29 ------ Equation B

Step 3: Now we have two equations (Equation A and Equation B) with the variables "y" and "z" only. We can solve this system of equations by further eliminating one of the variables.

Multiply Equation A by 5, and let's subtract it from Equation B:

(5z - 2y) - 5(5z - 2y) = 2 - 5(2)
5z - 2y - 25z + 10y = 2 - 10
-20z + 8y = -8

Simplifying this equation gives us:
-20z + 8y = -8 ------ Equation C

Step 4: Now we have two equations (Equation C and Equation B) with the variables "y" and "z" only. We can solve this system of equations using the same process.

Multiply Equation C by 6 and Equation B by 20 to get:
(-20z + 8y) * 6 = -8 * 6
(6 * 20z + 6 * 8y) = 29 * 20

Simplifying those equations gives us:
-120z + 48y = -48
120z + 220y = 580

Step 5: Add both equations together to eliminate the "z" variable:

(-120z + 48y) + (120z + 220y) = -48 + 580
48y + 220y = 532
268y = 532
y = 532/268
y = 2

Step 6: Substitute the value of "y" back into Equation C or B to solve for "z":

-20z + 8(2) = -8
-20z + 16 = -8
-20z = -8 - 16
-20z = -24
z = (-24)/(-20)
z = 6/5 or 1.2

Step 7: Substitute the values of "y" and "z" into any of the original equations to solve for "x". Let's use the first equation:

x + 3(6/5) = 12
x + 18/5 = 12
x = 12 - 18/5
x = (60 - 18)/5
x = 42/5 or 8.4

So the solution to the system of equations is:
x = 8.4
y = 2
z = 1.2

And that's how you solve the system of equations!