1.) Solve this system of linear equations. Write the final answer as an ordered pair (x, y, -3).
x - y + 2z = 22
3y - 8z = -9
z = -3
2.) Solve this system of linear equations. Write the final answer as an ordered pair (x, y, z).
2x + 4y + z = 1
x - 2y - 3z = 2
x + y - z = -1
using substitution,
z = -3, so
3y-8(-3) = -9
3y = -33
y = -11
so,
x-(-11)+2(-3) = 22
x+11-6 = 22
x = 17
So the solution is (17,-11,-3)
The other is not quite so simple, but
since z = 1-2x-4y, we have
x-2y-3(1-2x-4y) = 2
x+y-(1-2x-4y) = -1
or,
7x+10y = 5
3x+5y = 0
So, since 5y = -3x, 10y = -6x and we have
7x-6x = 5
x = 5
so, y = -3
and z = 3
solution is (5,-3,3)
To solve the system of linear equations, we'll use the method of substitution. Here's how you can solve each of the given systems:
1.) Solve this system of linear equations. Write the final answer as an ordered pair (x, y, -3).
Given equations:
x - y + 2z = 22 ------- (1)
3y - 8z = -9 ------- (2)
z = -3 ------- (3)
To solve this system, let's substitute the value of "z" from equation (3) into equations (1) and (2):
Substituting z = -3 into equation (1):
x - y + 2(-3) = 22
Simplifying the equation:
x - y - 6 = 22
Substituting z = -3 into equation (2):
3y - 8(-3) = -9
Simplifying the equation:
3y + 24 = -9
Now, we have a system of two equations with two variables:
x - y = 28 -------- (4) (after rearranging equation (1))
3y = -33 -------- (5) (after rearranging equation (2))
To solve equations (4) and (5), let's solve for y in equation (5):
y = -33/3
y = -11
Now, we can substitute the value of y = -11 into equation (4):
x - (-11) = 28
x + 11 = 28
x = 28 - 11
x = 17
So, the solution to the system of linear equations is (x, y, -3) = (17, -11, -3).
2.) Solve this system of linear equations. Write the final answer as an ordered pair (x, y, z).
Given equations:
2x + 4y + z = 1 ----------- (1)
x - 2y - 3z = 2 ----------- (2)
x + y - z = -1 ----------- (3)
To solve this system, let's use the method of substitution:
Let's solve equation (3) for x:
x = -1 - y + z ----------- (4)
Now, substitute equation (4) into equations (1) and (2):
Substituting x = -1 - y + z into equation (1):
2(-1 - y + z) + 4y + z = 1
Simplifying the equation:
-2 - 2y + 2z + 4y + z = 1
Combining like terms:
-2 - 2y + 4y + 2z + z = 1
Simplifying further:
-2 + 2y + 3z = 1
2y + 3z = 3 -------------- (5)
Substituting x = -1 - y + z into equation (2):
(-1 - y + z) - 2y - 3z = 2
Simplifying the equation:
-1 - y + z - 2y - 3z = 2
Focusing on y terms:
-1 - 3y - 2z = 2
-3y - 2z = 3 --------------- (6)
Now, we have a system of two equations with two variables:
2y + 3z = 3 ------------- (5)
-3y - 2z = 3 ------------- (6)
Let's solve equations (5) and (6) to find the values of y and z:
Multiplying equation (5) by 3 and equation (6) by -2, we can eliminate y by adding the equations:
6y + 9z = 9 ------------- (7)
6y + 4z = -6 --(8)------- (8)
Subtracting equation (7) from equation (8), we get:
(6y + 4z) - (6y + 9z) = -6 - 9
-5z = -15
Simplifying further, we find:
z = -15 / -5
z = 3
Now, substitute the value of z = 3 into equation (5):
2y + 3(3) = 3
Simplifying the equation:
2y + 9 = 3
2y = 3 - 9
2y = -6
y = -6 / 2
y = -3
Finally, substitute the values of y = -3 and z = 3 into equation (4) to solve for x:
x = -1 - (-3) + 3
x = -1 + 3 + 3
x = 5
The solution to the system of linear equations is (x, y, z) = (5, -3, 3).