Find x,y,z
2x-3y+z=-1
2x-y+2z=6
5x-4y+3z=6
Please show work I want to learn!!!
http://www.gregthatcher.com/Mathematics/GaussJordan.aspx
x= 1, y=2, z = 3
probably the most method used for 3by3 equations is called elimination
The object is to eliminate one or more variables from pairs of equations.
Usually we look at the coefficents of each variable and choose the one with the smallest or simplest numbers.
In this case it doesn't matter too much.
I will choose the z's
double the first:
4x - 6y + 2z = -2
the second as is
2x - y + 2z = 6
subtract them ---> 2x - 5y = -8 , #4
triple the first:
6x - 9y + 3z = -3
the third as is:
5x - 4y + 3z = 6
subtract them: --> x - 5y = -9 , #5
#4 - #5
x = 1
back in #4
2 - 5y = -8
-5y = -10
y = 2
back in the first:
2 - 6 + z = -1
z = 3
x=1 , y=2, z=3
check by subbing those values into the original 3 equations,
(it works!)
There are many ways I could have gone here.
We were lucky that both y terms in #4 and #5 were the same
Had they been opposite, I would have added the equations
Had both the x's and y's been different, I would have to multiply each equation by some value so that they would be the same
thanks
To find the values of x, y, and z, we need to solve the system of equations. There are a few methods to solve a system of equations, such as substitution, elimination, or using matrices. Let's solve this system using the elimination method.
Step 1: Multiply the first equation by 2 and the second equation by -2, so that the coefficients of x can cancel each other out when we add the equations together:
2(2x - 3y + z) = 2(-1) -> 4x - 6y + 2z = -2 (equation 1 multiplied by 2)
-2(2x - y + 2z) = -2(6) -> -4x + 2y - 4z = -12 (equation 2 multiplied by -2)
Step 2: Add equation 1 to equation 2:
(4x - 6y + 2z) + (-4x + 2y - 4z) = -2 + (-12)
4x - 4x - 6y + 2y + 2z - 4z = -2 - 12
-4y - 2z = -14 (equation 3)
Step 3: Multiply the first equation by -5 and add it to equation 3:
-5(2x - 3y + z) + (-4y - 2z) = -5(-1) + (-14)
-10x + 15y - 5z - 4y - 2z = 5 - 14
-10x + 15y - 4y - 5z - 2z = 5 - 14
-10x + 11y - 7z = -9 (equation 4)
Step 4: Multiply the second equation by 5 and add it to equation 3:
5(2x - y + 2z) + (-4y - 2z) = 5(6) + (-14)
10x - 5y + 10z - 4y - 2z = 30 - 14
10x - 4y + 8z = 16 (equation 5)
Step 5: Now we have a new system of equations:
-10x + 11y - 7z = -9 (equation 4)
10x - 4y + 8z = 16 (equation 5)
-4y - 2z = -14 (equation 3)
Step 6: Multiply equation 3 by -5/2 (or divide by -2 and multiply by 5):
(-5/2)(-4y - 2z) = (-5/2)(-14)
10y + 5z = 35 (equation 6)
Step 7: Add equation 5 to equation 6:
(10x - 4y + 8z) + (10y + 5z) = 16 + 35
10x - 4y + 10y + 8z + 5z = 51
10x + 6y + 13z = 51 (equation 7)
Step 8: Now we have a new system of equations:
-10x + 11y - 7z = -9 (equation 4)
10x + 6y + 13z = 51 (equation 7)
10y + 5z = 35 (equation 6)
Step 9: Now we can solve equations 4, 6, and 7 simultaneously.
Multiplying equation 4 by 10, we get:
-100x + 110y - 70z = -90 (equation 8)
Now, adding equation 8 to equation 7:
(-100x + 110y - 70z) + (10x + 6y + 13z) = -90 + 51
-100x + 10x + 110y + 6y - 70z + 13z = -39
-90x + 116y - 57z = -39 (equation 9)
Adding equation 9 to equation 6:
(-90x + 116y - 57z) + (10y + 5z) = -39 + 35
-90x + 116y + 5z + 10y - 57z = -4
-90x + 126y - 52z = -4 (equation 10)
Step 10: Now we have a new system of equations:
-90x + 126y - 52z = -4 (equation 10)
10y + 5z = 35 (equation 6)
Solving the equation 6 for z, we get:
z = (35 - 10y)/5
Simplifying, we get:
z = 7 - 2y
Now, substituting z into equation 10, we get:
-90x + 126y - 52(7 - 2y) = -4
-90x + 126y - 364 + 104y = -4
-90x + 230y = 360
Step 11: Solving the above equation for x, we get:
-90x = -360 - 230y
90x = 360 + 230y
x = (360 + 230y)/90
Simplifying, we get:
x = 4 + (23/9)y
So, we have obtained equations for x, y, and z:
x = 4 + (23/9)y
y can take any value
z = 7 - 2y
Therefore, the solution to the system of equations is x = 4 + (23/9)y, y can be any real number, and z = 7 - 2y.