Solve the system of equations for each variable x, y, and z. Please show your work. (Hint: try the elimination method)
-x - 5y - 5z = 2
4x - 5y + 4z = 19
x + 5y - z = -20
To solve the system of equations using the elimination method, we will eliminate one variable at a time.
First, let's eliminate the y variable by multiplying the second equation by 5 and adding it to the first equation:
(5)(-x - 5y - 5z) + (5)(4x - 5y + 4z) = (5)(2) + (5)(19)
-5x - 25y - 25z + 20x - 25y + 20z = 10 + 95
15x - 50z - 50y = 105
Simplifying this equation gives us: 15x - 50y - 50z = 105
Now, let's eliminate the y variable by multiplying the third equation by 5 and adding it to the first equation:
(5)(-x - 5y - 5z) + (5)(x + 5y - z) = (5)(2) + (5)(-20)
-5x - 25y - 25z + 5x + 25y - 5z = 10 - 100
-20z = -90
Dividing both sides of the equation by -20, we get:
z = 4.5
Now, substitute z = 4.5 back into the simplified equation from step 2:
15x - 50(4.5) - 50(4.5) = 105
15x - 225 - 225 = 105
15x - 450 = 105
15x = 555
x = 555/15
x = 37
Finally, substitute the values of x = 37 and z = 4.5 into one of the original equations, such as the third equation:
37 + 5y - 4.5 = -20
41.5 + 5y = -20
5y = -20 - 41.5
5y = -61.5
y = -61.5/5
y = -12.3
Therefore, the solution to the system of equations is:
x = 37
y = -12.3
z = 4.5
To solve the system of equations using the elimination method, we will eliminate one variable at a time. Let's start by eliminating the y variable.
1. -x - 5y - 5z = 2
2. 4x - 5y + 4z = 19
3. x + 5y - z = -20
To eliminate the y variable, we will multiply equation 1 by 5 and equation 3 by -5. This will allow us to add equations 1 and 3 together to eliminate the y terms.
-5*(-x - 5y - 5z) = -5*2
-5x - 25y - 25z = -10
-5*(x + 5y - z) = -5*(-20)
-5x - 25y + 5z = 100
Now, we can add equations 1 and 3 together:
(-5x - 25y - 25z) + (-5x - 25y + 5z) = -10 + 100
-10x - 50z = 90 ----(Equation 4)
Next, we will multiply equation 2 by 5 and equation 3 by 4 to eliminate the y terms:
5*(4x - 5y + 4z) = 5*19
20x - 25y + 20z = 95
4*(x + 5y - z) = 4*(-20)
4x + 20y - 4z = -80
Now, we can add equations 2 and 3 together:
(20x - 25y + 20z) + (4x + 20y - 4z) = 95 + (-80)
24x - 5z = 15 ----(Equation 5)
We now have a system of two equations with two variables (Equations 4 and 5).
To eliminate the x variable, we will multiply equation 4 by 24, and equation 5 by 10:
24*(-10x - 50z) = 24*90
-240x - 1200z = 2160
10*(24x - 5z) = 10*15
240x - 50z = 150
Now, we can add equations 4 and 5 together:
(-240x - 1200z) + (240x - 50z) = 2160 + 150
-1250z = 2310
To solve for z, divide both sides of the equation by -1250:
z = 2310 / -1250
z = -1.848
Now substitute the value of z into equation 4:
-10x - 50(-1.848) = 90
-10x + 92.4 = 90
-10x = 90 - 92.4
-10x = -2.4
To solve for x, divide both sides of the equation by -10:
x = -2.4 / -10
x = 0.24
Substitute the values of x and z into equation 5:
24(0.24) - 5z = 15
5.76 - 5z = 15
-5z = 15 - 5.76
-5z = 9.24
To solve for z, divide both sides of the equation by -5:
z = 9.24 / -5
z = -1.848
Therefore, the solutions to the system of equations are:
x = 0.24
y = unknown
z = -1.848
To solve the system of equations, we will use the elimination method. The goal is to add or subtract the equations so that one of the variables is eliminated, allowing us to solve for the remaining variables.
First, let's set up the system of equations:
Equation 1: -x - 5y - 5z = 2
Equation 2: 4x - 5y + 4z = 19
Equation 3: x + 5y - z = -20
We will start by eliminating the variable y.
Multiply Equation 1 by 5 and Equation 3 by -5:
-5x - 25y - 25z = 10
-5x - 25y + 5z = 100
Now we can subtract Equation 1 from Equation 2 to eliminate y:
(4x - 5y + 4z) - (-5x - 25y + 5z) = 19 - 100
4x - 5y + 4z + 5x + 25y - 5z = -81
9x + 20y - z = -81 (Equation 4)
Next, we will eliminate the variable y between Equation 2 and Equation 4.
Multiply Equation 2 by 4 and Equation 4 by 5:
16x - 20y + 16z = 76
45x + 100y - 5z = -405
Now subtract Equation 4 from Equation 2 to eliminate y:
(16x - 20y + 16z) - (45x + 100y - 5z) = 76 - (-405)
16x - 20y + 16z - 45x - 100y + 5z = 481
-29x - 120y + 21z = 481 (Equation 5)
At this point, we have eliminated y and are left with two equations (Equation 4 and Equation 5) involving x, y, and z.
Now we will eliminate y between Equation 4 and Equation 5.
Multiply Equation 4 by 6 and Equation 5 by 1:
54x + 120y - 6z = -486
-29x - 120y + 21z = 481
Add Equation 4 and Equation 5 to eliminate y:
(54x + 120y - 6z) + (-29x - 120y + 21z) = -486 + 481
54x + 120y - 6z - 29x - 120y + 21z = -5
25x + 15z = -5 (Equation 6)
Now we have two equations involving only x and z (Equation 5 and Equation 6).
To eliminate z, multiply Equation 5 by 3 and Equation 6 by 7:
-87x - 360y + 63z = 1447
175x + 105z = -35
Now we can add Equation 5 and Equation 6 to eliminate z:
(-87x - 360y + 63z) + (175x + 105z) = 1447 - 35
-87x - 360y + 63z + 175x + 105z = 1412
88x + 168z = 1412 (Equation 7)
Now we have one equation involving only x and z (Equation 7).
To solve for x, we need to isolate x in Equation 7.
Rearrange Equation 7:
88x = 1412 - 168z
Divide both sides by 88:
x = (1412 - 168z) / 88
Simplify:
x = (353 - 42z) / 22 (Equation 8)
Now that we have the value of x in terms of z, we can substitute this expression into one of the original equations to solve for y and z.
Let's substitute Equation 8 into Equation 6:
25(353 - 42z)/22 + 15z = -5
To simplify, let's clear the fractions by multiplying both sides of the equation by 22:
25(353 - 42z) + 330z = -110
Expand and combine like terms:
8825 - 1050z + 330z = -110
-720z = -8935
z = -8935 / -720
z = 12.43
Now substitute the value of z back into Equation 8 to solve for x:
x = (353 - 42 * 12.43) / 22
x = (353 - 520.86) / 22
x = -167.86 / 22
x = -7.63
Finally, substitute x = -7.63 and z = 12.43 into one of the original equations to solve for y:
Equation 1: -x - 5y - 5z = 2
-(-7.63) - 5y - 5 * 12.43 = 2
7.63 - 5y - 62.15 = 2
-5y = 56.52
y = 56.52 / -5
y = -11.30
Therefore, the solution to the system of equations is:
x = -7.63, y = -11.30, z = 12.43.