Solve the system by elimination. x+5y-4z=-10, 2x-y+5z=-9, 2x-10y-5z=0. A.(5, -1, 0) B.(-5, 1, 0) C.(-5, -1, 0) D.(-5, -1, -2)
Alright, alright! Let's eliminate those variables and solve this system using a bit of clown magic!
We'll start by eliminating the "x" variable. Let's take the first equation and multiply it by 2:
2(x + 5y - 4z) = 2(-10)
This simplifies to:
2x + 10y - 8z = -20
Now we can subtract the second equation (2x - y + 5z = -9) from this new equation (2x + 10y - 8z = -20):
(2x + 10y - 8z) - (2x - y + 5z) = -20 - (-9)
If we simplify this, we get:
11y - 13z = -11
Now, let's multiply the third equation (2x - 10y - 5z = 0) by 2:
2(2x - 10y - 5z) = 2(0)
This simplifies to:
4x - 20y - 10z = 0
Finally, subtract this equation (4x - 20y - 10z = 0) from the new equation we just found (11y - 13z = -11):
(11y - 13z) - (4x - 20y - 10z) = -11 - 0
If we simplify this, we get:
31y - 3z = -11
Now we have a new equation. Let's call it Equation 4: 31y - 3z = -11
Now, let's solve the system formed by Equation 4 (31y - 3z = -11) and the original second equation (2x - y + 5z = -9).
I'm going to introduce a circus lion to help us out here. Let's call him Leo.
(Approaching the lion) Hey, Leo! We need your help solving this system!
Leo the Lion: (Roars)
Okay, let's use Leo to eliminate the "y" variable. We'll multiply Equation 4 (31y - 3z = -11) by 2:
2(31y - 3z) = 2(-11)
This simplifies to:
62y - 6z = -22
Now, let's add this equation (62y - 6z = -22) to the original second equation (2x - y + 5z = -9):
(2x - y + 5z) + (62y - 6z) = -9 + (-22)
If we simplify this, we get:
2x + 61y - z = -31
Now, we have a new equation! Let's call it Equation 5: 2x + 61y - z = -31
Now, we're going to introduce a new member to our clown magic show, a dancing elephant! Let's call her Ella.
(Approaching the elephant) Ella, we need your dancing skills to eliminate the "z" variable!
Ella the Elephant: (Starts dancing)
Great! Let's multiply Equation 1 (x + 5y - 4z = -10) by 6:
6(x + 5y - 4z) = 6(-10)
This simplifies to:
6x + 30y - 24z = -60
Now, let's add this equation (6x + 30y - 24z = -60) to Equation 5 (2x + 61y - z = -31):
(2x + 61y - z) + (6x + 30y - 24z) = -31 + (-60)
If we simplify this, we get:
8x + 91y - 25z = -91
Now, we have another new equation! Let's call it Equation 6: 8x + 91y - 25z = -91
Alright, it's time to present our final act! Let's subtract Equation 5 (2x + 61y - z = -31) from Equation 6 (8x + 91y - 25z = -91):
(8x + 91y - 25z) - (2x + 61y - z) = -91 - (-31)
If we simplify this, we get:
6x + 30y - 24z = -60
Wait a minute...this equation is just Equation 1 multiplied by 2! It means that Equation 1 is a combination of the other equations.
So, we can conclude that the system of equations is dependent (or infinitely many solutions). In other words, there is not a unique solution for this system!
Now, it's time for some clown fun! Feel free to choose any answer option you want, because they're all equally valid (or equally invalid, depending on how you see it):
A. (5, -1, 0)
B. (-5, 1, 0)
C. (-5, -1, 0)
D. (-5, -1, -2)
Remember, in the world of clowns, anything is possible!
To solve the system of equations by elimination, we need to eliminate one variable at a time. Let's start by eliminating the x variable.
Multiply the first equation by 2 and the third equation by -1 to create opposite coefficients for x:
2(x + 5y - 4z) = 2(-10)
-1(2x - 10y - 5z) = -1(0)
This simplifies to:
2x + 10y - 8z = -20
-2x + 10y + 5z = 0
Now, add these two equations together:
(2x + 10y - 8z) + (-2x + 10y + 5z) = -20 + 0
This simplifies to:
20y - 3z = -20
Now we have one equation with only the variables y and z. Let's express one of these variables in terms of the other.
Next, let's eliminate the x variable from the second equation by multiplying it by 2:
2(2x - y + 5z) = 2(-9)
This simplifies to:
4x - 2y + 10z = -18
Now, adding this equation to the third equation:
(4x - 2y + 10z) + (2x - 10y - 5z) = -18 + 0
This simplifies to:
6x - 12y + 5z = -18
Now, divide this equation by 6 to simplify it:
x - 2y + (5/6)z = -3
Finally, we have a system of two equations:
20y - 3z = -20 (1)
x - 2y + (5/6)z = -3 (2)
To eliminate y, multiply equation (2) by 10 and equation (1) by 3:
10(x - 2y + (5/6)z) = 10(-3)
3(20y - 3z) = 3(-20)
This simplifies to:
10x - 20y + 5z = -30
60y - 9z = -60
Now, add these two equations together:
(10x - 20y + 5z) + (60y - 9z) = -30 + (-60)
This simplifies to:
10x + 40y - 4z = -90
Now we have an equation with only the variable x.
Let's summarize the three equations we have so far:
20y - 3z = -20 (1)
60y - 9z = -60 (2)
10x + 40y - 4z = -90 (3)
Now, we have a system of three equations with three variables. We can solve this system using various methods, such as substitution or Gaussian elimination.
However, none of the answer choices provided satisfy all three equations, which means there is no solution to the system.
You can start out like this to eliminate x:
x+5y-4z = -10
2x-y+5z = -9
2x-10y-5z = 0
double row 1 and subtract it from rows 2,3:
2x+10y-8z = -20
.....-11y+13z = 11
.....-20y+3z = 20
Now proceed to eliminate y using row 2, and eliminate z using row 3
There are also handy online calculators that show the steps, such as
onlinemschool.c om/math/assistance/equation/gaus/