Solve the system of linear equations
2x+y-3z=4
4x+2z=10
-2x+3y-13z=-8
*im so lost (all these numbers are in one problem)
if you divide the 2nd by 2, all your equations start with 2x. That is good, since we could add or subtract them to eliminate the x
2x+y-3z=4
2x+z=10
-2x+3y-13z=-8
subtract 1st from the 2nd
-y + 4z = 6 , #4
add the 2nd and 3rd
3y - 12z = 2 , #5
#4 times 3---> -3y + 12z = 18
#5 as is -----> 3y - 12z = 2
add them: 0 = 20 , which is a contradiction
AHHH, there is no unique solution.
If you are studying equations of planes, you should categorize this as one of the special cases.
well, the easy way is to google Gauss Jordan reduction calculator.
http://www.gregthatcher.com/Mathematics/GaussJordan.aspx
otherwise do it by elimination
First eliminate z from first two:
2x+y-3z = 4 times 2 = 4x+ 2y-6z = 8
4x+0y+2z=10 times 3 = 12x+0y+6z = 30
add the two
16 x + 2 y = 38
or
8 x + y = 19
Now eliminate z from the last 2
multiply second by 13 and last by 2
52 x + 0y + 26 z = 130
-4 x + 6 y -26 z = -16
------------------------- add the two
48 x + 6 y = 114
or
24 x + 3 y = 57
Now you have the two equations in x and y
8 x + 1 y = 19
24 x + 3 y = 57
multiply first by 3
24 x + 3 y = 57
Oh , my, system can not be solved, determinant is zero, two lines the same.
Your solution is illustrated by
http://www.google.com/search?q=intersection+of+3+planes&tbm=isch&imgil=VKeuvId73BKgcM%253A%253Bhttp%253A%252F%252Ft2.gstatic.com%252Fimages%253Fq%253Dtbn%253AANd9GcTP5OK2B8xVEFseHgCh5S25ao2K4pX-fClEpupY9ltfPRK6cZC8NA%253B332%253B265%253BL5ht1c6dmTcU5M%253Bhttp%25253A%25252F%25252Fstweb.peelschools.org%25252Fjfsweb%25252Fmga4u%25252Funit3%25252Flesson7.html&source=iu&usg=__Epsu58j8aAIOLuG__2toxIf0TSk%3D&sa=X&ei=SN6gU9wz2KPIBJaOgPAK&sqi=2&ved=0CCkQ9QEwAg&biw=1680&bih=936#imgdii=_
look at the 2nd picture, top left
Your planes intersect in pairs, each one giving us a parallel line as the intersection of the planes, two at a time.
I understand that dealing with multiple equations and variables can be overwhelming. Don't worry, I'll explain how to solve this system of linear equations step by step.
To solve this system, we'll use the method of substitution. Here's how it's done:
Step 1: Solve one equation for one variable in terms of the other variables.
We start with the second equation, 4x + 2z = 10. We can rearrange it to obtain x in terms of z:
4x = 10 - 2z
x = (10 - 2z) / 4
x = (5 - z) / 2
Step 2: Substitute the expression from step 1 into the other equations.
Now that we have x in terms of z, we can substitute it into the other equations to eliminate x. Let's substitute it into the first equation:
2x + y - 3z = 4
2((5 - z) / 2) + y - 3z = 4
Simplifying, we get:
5 - z + y - 3z = 4
-y - 4z = -1 (Equation A)
Similarly, let's substitute it into the third equation:
-2x + 3y - 13z = -8
-2((5 - z) / 2) + 3y - 13z = -8
Simplifying, we get:
-(5 - z) + 3y - 13z = -8
5 - z + 3y - 13z = -8
2y - 14z = -13 (Equation B)
Now we have a system of two equations (A and B) with two variables, y and z. Let's solve this system.
Step 3: Solve the system of equations A and B.
We can solve this system using any method you're comfortable with, such as substitution or elimination.
Let's use elimination to solve the system:
Multiply equation A by 2 to make the coefficients of y in both equations equal:
-2y - 8z = -2
Now, add equation B and the modified equation A:
(2y - 14z) + (-2y - 8z) = -13 + (-2)
-22z = -15
Divide both sides by -22:
z = 15/22
Step 4: Substitute the value of z into either equation A or B to find the value of y.
Let's use equation A:
-y - 4z = -1
Substitute z = 15/22:
-y - 4(15/22) = -1
Simplify and solve for y:
-y - (60/22) = -1
-y = -1 + (60/22)
-y = (-22 + 60)/22
-y = 38/22
y = -19/11
Step 5: Substitute the values of y and z into x = (5 - z) / 2 to find the value of x.
Substitute y = -19/11 and z = 15/22:
x = (5 - (15/22)) / 2
x = (110/22 - 15/22) / 2
x = (95/22) / 2
x = 95/44
So, the solution to the given system of linear equations is:
x = 95/44, y = -19/11, z = 15/22