SOLUTION: Solve each system 2x+y+z=14 -x-3y+2z=-2 4x-6y+3z=-5

Algebra ->  Systems-of-equations -> SOLUTION: Solve each system 2x+y+z=14 -x-3y+2z=-2 4x-6y+3z=-5      Log On


   

Question 934937: Solve each system
2x+y+z=14
-x-3y+2z=-2
4x-6y+3z=-5
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

2x%2By%2Bz=14.....eq.1
-x-3y%2B2z=-2.....eq.2
4x-6y%2B3z=-5.....eq.3
_________________________________
start with
2x%2By%2Bz=14.....eq.1
-x-3y%2B2z=-2.....eq.2...both sides multiply by 2
_____________________
2x%2By%2Bz=14.....eq.1
-2x-6y%2B4z=-4.....eq.2
__________________________add
cross%282x%29%2By%2Bz-cross%282x%29-6y%2B4z=14-4
5z-5y=10
5%28z-y%29=10
z-y=10%2F5
z-y=2....solve for z
z=y%2B2...............1a

-x-3y%2B2z=-2.....eq.2...both sides multiply by 4
4x-6y%2B3z=-5.....eq.3
_________________________________
-4x-12y%2B8z=-8.....eq.2
4x-6y%2B3z=-5.....eq.3
_________________________________add
-cross%284x%29-12y%2B8z%2Bcross%284x%29-6y%2B3z=-8-5
-18y%2B11z=-13...solve for z
11z=18y-13
z=18y%2F11-13%2F11.............1b
1a and 1b have equal left sides, so right sides must be equal too
y%2B2=18y%2F11-13%2F11.....solve for y
11y%2B22=18y-13
13%2B22=18y-11y
35=7y
35%2F7=y
highlight%28y=5%29
now find z
z=y%2B2...............1a
z=5%2B2
highlight%28z=7%29
go to one of the given equations, plug in values for y and z and find x
2x%2By%2Bz=14.....eq.1
2x%2B5%2B7=14
2x%2B12=14
2x=14-12
2x=2
highlight%28x=1%29
so, your solutions are: highlight%28x=1%29,highlight%28y=5%29, and
highlight%28z=7%29

this is another way to solve this system:

Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 3 variables







First let A=%28matrix%283%2C3%2C2%2C1%2C1%2C-1%2C-3%2C2%2C4%2C-6%2C3%29%29. This is the matrix formed by the coefficients of the given system of equations.


Take note that the right hand values of the system are 14, -2, and -5 and they are highlighted here:




These values are important as they will be used to replace the columns of the matrix A.




Now let's calculate the the determinant of the matrix A to get abs%28A%29=35. To save space, I'm not showing the calculations for the determinant. However, if you need help with calculating the determinant of the matrix A, check out this solver.



Notation note: abs%28A%29 denotes the determinant of the matrix A.



---------------------------------------------------------



Now replace the first column of A (that corresponds to the variable 'x') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5Bx%5D (since we're replacing the 'x' column so to speak).






Now compute the determinant of A%5Bx%5D to get abs%28A%5Bx%5D%29=35. Again, as a space saver, I didn't include the calculations of the determinant. Check out this solver to see how to find this determinant.



To find the first solution, simply divide the determinant of A%5Bx%5D by the determinant of A to get: x=%28abs%28A%5Bx%5D%29%29%2F%28abs%28A%29%29=%2835%29%2F%2835%29=1



So the first solution is x=1




---------------------------------------------------------


We'll follow the same basic idea to find the other two solutions. Let's reset by letting A=%28matrix%283%2C3%2C2%2C1%2C1%2C-1%2C-3%2C2%2C4%2C-6%2C3%29%29 again (this is the coefficient matrix).




Now replace the second column of A (that corresponds to the variable 'y') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5By%5D (since we're replacing the 'y' column in a way).






Now compute the determinant of A%5By%5D to get abs%28A%5By%5D%29=175.



To find the second solution, divide the determinant of A%5By%5D by the determinant of A to get: y=%28abs%28A%5By%5D%29%29%2F%28abs%28A%29%29=%28175%29%2F%2835%29=5



So the second solution is y=5




---------------------------------------------------------





Let's reset again by letting A=%28matrix%283%2C3%2C2%2C1%2C1%2C-1%2C-3%2C2%2C4%2C-6%2C3%29%29 which is the coefficient matrix.



Replace the third column of A (that corresponds to the variable 'z') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5Bz%5D






Now compute the determinant of A%5Bz%5D to get abs%28A%5Bz%5D%29=245.



To find the third solution, divide the determinant of A%5Bz%5D by the determinant of A to get: z=%28abs%28A%5Bz%5D%29%29%2F%28abs%28A%29%29=%28245%29%2F%2835%29=7



So the third solution is z=7




====================================================================================

Final Answer:




So the three solutions are x=1, y=5, and z=7 giving the ordered triple (1, 5, 7)




Note: there is a lot of work that is hidden in finding the determinants. Take a look at this 3x3 Determinant Solver to see how to get each determinant.