# SOLUTION: how do i use cramer's rule for three equations on this problem 5x-6y=7+7z 6x-4y+10z=-34 2x+4y=29+3z

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Question 252724: how do i use cramer's rule for three equations on this problem
5x-6y=7+7z
6x-4y+10z=-34
2x+4y=29+3z

Found 2 solutions by drk, jim_thompson5910:
You can put this solution on YOUR website!
To use Cramers rule you need matricies.
step #1 rewrite all equation so that variables are on one side and in order.
5x - 6y - 7Z = 7
6x - 4y + 10z = -34
2x + 4y -3Z = 29.
--
step #2 - We need to find four determinants:
(1) the determinant of the left side, D.
D = -592
(2) the determinant of x, Dx, when you replace the x column with the answer column
Dx = -1184
(3) the determinant of y, Dy, when you replace the y column with the answer column
Dy = -2368
(4) the determinant of z, Dz, when you replace the z column with the answer column
Dz = 1776
--
step #3 use fractions to find the answer:
(1) Dx / D = x answer ; x = 2
(2) Dy / D = y answer ; y = 4
(3) Dz / D = z answer ; z = -3
I don't know an easier way to explain this. Hope this makes sense.

You can put this solution on YOUR website!
The first goal is to get all of the variable terms to the left side for each equation.

Subtract 7z from both sides.

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Move onto the third equation.

Subtract 3z from both sides.

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So we now have the system

Now let's use Cramer's Rule to solve this system

 Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 3 variables First let . 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 , , and 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 . 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: 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 (since we're replacing the 'x' column so to speak). Now compute the determinant of to get . 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 by the determinant of to get: So the first solution is --------------------------------------------------------- We'll follow the same basic idea to find the other two solutions. Let's reset by letting 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 (since we're replacing the 'y' column in a way). Now compute the determinant of to get . To find the second solution, divide the determinant of by the determinant of to get: So the second solution is --------------------------------------------------------- Let's reset again by letting 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 Now compute the determinant of to get . To find the third solution, divide the determinant of by the determinant of to get: So the third solution is ==================================================================================== Final Answer: So the three solutions are , , and giving the ordered triple (2, 4, -3) 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.

If you need more help or practice with Cramer's Rule, check out this solver.