SOLUTION: how do you solve 10x+2y+1z=885 5x+4y+1z=865 9x+6y+6z=1410

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Question 272861: how do you solve
10x+2y+1z=885
5x+4y+1z=865
9x+6y+6z=1410
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'm going to solve this system by using Cramer's Rule.

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







First let A=%28matrix%283%2C3%2C10%2C2%2C1%2C5%2C4%2C1%2C9%2C6%2C6%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 885, 865, and 1410 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=132. 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.



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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=7920. 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=%287920%29%2F%28132%29=60



So the first solution is x=60




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We'll follow the same basic idea to find the other two solutions. Let's reset by letting A=%28matrix%283%2C3%2C10%2C2%2C1%2C5%2C4%2C1%2C9%2C6%2C6%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=18480.



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=%2818480%29%2F%28132%29=140



So the second solution is y=140




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Let's reset again by letting A=%28matrix%283%2C3%2C10%2C2%2C1%2C5%2C4%2C1%2C9%2C6%2C6%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=660.



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=%28660%29%2F%28132%29=5



So the third solution is z=5




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Final Answer:




So the three solutions are x=60, y=140, and z=5 giving the ordered triple (60, 140, 5)




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.