How do you solve the system #-5x+3y+6z=4#, #-3x+y+5z=-5#, and #-4x+2y+z=13#?

Question

Madelyn Anderson · Accepted Answer

#[ (1, 0, 0, |, -2), (0, 1, 0,|, 4), (0, 0, 1, |, -3) ]#

#x = -2, y = 4, z = -3# Write #−5x + 3y + 6z =4# into the first row of an augmented matrix: #[ (−5, 3, 6, |, 4) ]# Add #−3x + y + 5z = −5# to the second row of the augmented matrix: #[ (−5, 3, 6, |, 4), (−3, 1, 5,|, −5) ]# Add #−4x + 2y + z = 13# to the third row of the augmented matrix: #[ (−5, 3, 6, |, 4), (−3, 1, 5,|, −5), (−4, 2, 1, |, 13) ]# Subtract row 2 from row 1: #[ (−2, 2, 1, |, 9), (−3, 1, 5,|, −5), (−4, 2, 1, |, 13) ]# Subtract row 3 from row 1: #[ (2, 0, 0, |, -4), (−3, 1, 5,|, −5), (−4, 2, 1, |, 13) ]# Divide row 1 by 2: #[ (1, 0, 0, |, -2), (−3, 1, 5,|, −5), (−4, 2, 1, |, 13) ]# Multiply row 1 by 3 and add to row 2: #[ (1, 0, 0, |, -2), (0, 1, 5,|, -11), (−4, 2, 1, |, 13) ]# Multiply row 1 by 4 and add to row 3: #[ (1, 0, 0, |, -2), (0, 1, 5,|, -11), (0, 2, 1, |, 5) ]# Multiply row 2 by -2 and add to row 3: #[ (1, 0, 0, |, -2), (0, 1, 5,|, -11), (0, 0, -9, |, 27) ]# Divide row 3 by -9: #[ (1, 0, 0, |, -2), (0, 1, 5,|, -11), (0, 0, 1, |, -3) ]# Multiply row 3 by -5 and add to row 2: #[ (1, 0, 0, |, -2), (0, 1, 0,|, 4), (0, 0, 1, |, -3) ]#

Benjamin Achtenberg · Answer

undefined To solve the system of equations -5x + 3y + 6z = 4, -3x + y + 5z = -5, and -4x + 2y + z = 13, you can use the method of substitution or elimination. Here, I'll demonstrate the elimination method:

First, multiply the second equation by 2 and add it to the third equation:
(-3x + y + 5z) * 2 = -10x + 2y + 10z = -10
-4x + 2y + z = 13

Add the modified second equation to the third equation:
(-10x + 2y + 10z) + (-4x + 2y + z) = -10 + 13
-14x + 12z = 3

Now, multiply the first equation by 2 and add it to the modified third equation:
(-5x + 3y + 6z) * 2 = -10x + 6y + 12z = 8
-14x + 12z = 3

Add the modified first equation to the modified third equation:
(-10x + 6y + 12z) + (-14x + 12z) = 8 + 3
-24x + 6y = 11

Now, you have two equations:
-14x + 12z = 3
-24x + 6y = 11

Solve one of the equations for a variable (let's solve the second equation for y):
y = (24x + 11) / 6

Substitute y back into the first equation:
-14x + 12z = 3
-14x + 12z = 3

Now, you have one equation with one variable:
-14x + 12z = 3

Solve for x:
x = (12z - 3) / -14

Finally, substitute the values of x and y back into one of the original equations to solve for z. Let's use the first equation:
-5x + 3y + 6z = 4

Substitute x and y:
(-5 * (12z - 3) / -14) + (3 * ((24 * (12z - 3) / -14) + 11) / 6) + 6z = 4

Solve for z using this equation. Once you have z, substitute back into the equations to find the values of x and y.