solve this system of equations what is x y and z
x+y+z=6
2x-y+3z=9
2x+z=5
To solve this system of equations, we can use the method of substitution or elimination.
Let's use the method of elimination:
1) Multiply the third equation by 2:
2(2x + z) = 2(5)
4x + 2z = 10
2) Subtract the first equation from the second equation:
(2x - y + 3z) - (x + y + z) = 9 - 6
x - 2y + 2z = 3
3) Subtract the first equation from the third equation:
(2x + z) - (x + y + z) = 5 - 6
x - y = -1
x = y - 1
Now we have two equations:
Equation 1: x = y - 1
Equation 2: x - 2y + 2z = 3
Substitute Equation 1 into Equation 2:
(y - 1) - 2y + 2z = 3
y - 2y + 2z = 3 + 1
-z = 4 - 3
-z = 1
z = -1
Substitute z = -1 into Equation 2:
x - 2y + 2(-1) = 3
x - 2y - 2 = 3
x - 2y = 5
Now we have two equations:
Equation 3: x = y - 1
Equation 4: x - 2y = 5
Substitute Equation 3 into Equation 4:
(y - 1) - 2y = 5
y - 2y = 5 + 1
-y = 6
y = -6
Substitute y = -6 into Equation 3:
x = -6 - 1
x = -7
Therefore, the solution to the system of equations is x = -7, y = -6, and z = -1.
To solve this system of equations, we can use the method of elimination or substitution.
Let's start by using the method of substitution:
From the third equation, we can express x in terms of z:
2x + z = 5
2x = 5 - z
x = (5 - z)/2
Now we substitute the expression for x in the first equation:
x + y + z = 6
((5 - z)/2) + y + z = 6
Simplifying this equation gives us:
5 - z + 2y + 2z = 12
2y + 3z = 7 --> Equation (1)
Next, we substitute the expression for x in the second equation:
2x - y + 3z = 9
2((5 - z)/2) - y + 3z = 9
5 - z - y + 3z = 9
Simplifying this equation gives us:
-y + 2z = 4 --> Equation (2)
Now we have two equations:
2y + 3z = 7 --> Equation (1)
-y + 2z = 4 --> Equation (2)
We can solve this system of equations using simultaneous substitution or elimination.
Let's use elimination method to cancel out y. Multiply equation (2) by 2 and add it to equation (1):
-2y + 4z = 8
2y + 3z = 7
Adding these equations gives us:
7z = 15
z = 15/7
Now, substitute the value of z back into equation (2) to solve for y:
-y + 2 * (15/7) = 4
-y + (30/7) = 4
-y = 4 - (30/7)
-y = (28 - 30)/7
-y = -2/7
y = 2/7
Finally, substitute the values of y and z back into equation (1) to solve for x:
2 * (2/7) + 3 * (15/7) = 7
4/7 + 45/7 = 7
(49/7) = 7
x = 1
Therefore, the solution to the system of equations is x = 1, y = 2/7, and z = 15/7.
To solve the system of equations, we can use the method of substitution or elimination. Let's use the method of elimination to eliminate variables one by one.
First, let's eliminate the variable x. Multiply equation 1 by 2 and equation 3 by -1 to create opposite coefficients for x.
2(x + y + z) = 2(6)
-1(2x + z) = -1(5)
This simplifies to:
2x + 2y + 2z = 12
-2x - z = -5
Adding these two equations together eliminates the x variable:
(2x + 2y + 2z) + (-2x - z) = 12 + (-5)
2y + z = 7 --> Equation 4
Now let's eliminate the variable x again. Multiply equation 3 by -2:
-2(2x + z) = -2(5)
-4x - 2z = -10
Adding this equation to equation 2 eliminates the x variable:
(2x - y + 3z) + (-4x - 2z) = 9 + (-10)
-y + z = -1 --> Equation 5
Now we have two equations to solve simultaneously:
Equation 4: 2y + z = 7
Equation 5: -y + z = -1
We can solve these equations by adding them together to eliminate the z variable:
(2y + z) + (-y + z) = 7 + (-1)
y = 6
Now substitute the value of y into equation 5 to find the value of z:
-y + z = -1
-(6) + z = -1
z = 5
Finally, substitute the values of y and z back into equation 1 (or any of the earlier equations) to find the value of x:
x + y + z = 6
x + 6 + 5 = 6
x + 11 = 6
x = -5
Therefore, the solution to the system of equations is:
x = -5, y = 6, z = 5.