Answer:
r ≈ -7.57, y = -2, z = 3
Explanation:
To solve the system of linear equations:
5r + 2y - 3z = -17 -4r + 5y - 6z = 4 7r - 6y + 7z = -20
We can use various methods such as substitution, elimination, or matrix manipulation. Let's use the elimination method.
First, let's multiply the second equation by 5 and add it to the first equation to eliminate the variable y:
(5*(-4r + 5y - 6z)) + (5r + 2y - 3z) = 5*4 -20r + 25y - 30z + 5r + 2y - 3z = 20 -15r + 27y - 33z = 20 (Resultant Equation)
Now, we can multiply the second equation by 7 and add it to the third equation to eliminate the variable y:
(7*(-4r + 5y - 6z)) + (7r - 6y + 7z) = 7*4 -28r + 35y - 42z + 7r - 6y + 7z = 28 -21r + 29y - 35z = 28 (Resultant Equation)
We now have two new equations:
-15r + 27y - 33z = 20 -21r + 29y - 35z = 28
Subtracting the first new equation from the second new equation eliminates the variable y:
-21r + 29y - 35z - (-15r + 27y - 33z) = 28 - 20 -21r + 29y - 35z + 15r - 27y + 33z = 8 -6r + 2y - 2z = 8 -3r + y - z = 4 (Resultant Equation)
Now we have a new equation:
-3r + y - z = 4
We can solve for y in terms of r and z in this equation and then substitute that value of y into the first original equation to solve for the values of r and z.