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Math  /  Algebra

Question15 points)Solve the system analytically. 3) 4x+5y+z=−315x−2y−z=−52x+y+5z=−29\begin{array}{l} 4 x+5 y+z=-31 \\ 5 x-2 y-z=-5 \\ 2 x+y+5 z=-29 \end{array}

Studdy Solution

STEP 1

1. We are given a system of three linear equations with three variables: x x , y y , and z z .
2. The goal is to find the values of x x , y y , and z z that satisfy all three equations simultaneously.
3. We will use the method of elimination or substitution to solve the system analytically.

STEP 2

1. Eliminate one variable from two pairs of equations.
2. Solve the resulting system of two equations with two variables.
3. Substitute back to find the third variable.
4. Verify the solution by substituting the values into the original equations.

STEP 3

First, let's eliminate z z from the first two equations. We have:
1) 4x+5y+z=−31 4x + 5y + z = -31
2) 5x−2y−z=−5 5x - 2y - z = -5
Add equations (1) and (2) to eliminate z z :
(4x+5y+z)+(5x−2y−z)=−31+(−5) (4x + 5y + z) + (5x - 2y - z) = -31 + (-5)
9x+3y=−36 9x + 3y = -36
Simplify:
3x+y=−12 3x + y = -12

STEP 4

Now, eliminate z z from the second and third equations. We have:
2) 5x−2y−z=−5 5x - 2y - z = -5
3) 2x+y+5z=−29 2x + y + 5z = -29
Multiply equation (2) by 5 and equation (3) by 1, then add them to eliminate z z :
5(5x−2y−z)+1(2x+y+5z)=5(−5)+1(−29) 5(5x - 2y - z) + 1(2x + y + 5z) = 5(-5) + 1(-29)
25x−10y−5z+2x+y+5z=−25−29 25x - 10y - 5z + 2x + y + 5z = -25 - 29
27x−9y=−54 27x - 9y = -54
Simplify:
3x−y=−6 3x - y = -6

STEP 5

Now we have a simpler system of two equations:
1) 3x+y=−12 3x + y = -12
2) 3x−y=−6 3x - y = -6
Add these two equations to eliminate y y :
(3x+y)+(3x−y)=−12+(−6) (3x + y) + (3x - y) = -12 + (-6)
6x=−18 6x = -18
Solve for x x :
x=−3 x = -3

STEP 6

Substitute x=−3 x = -3 back into one of the simpler equations to find y y . Use equation (1):
3(−3)+y=−12 3(-3) + y = -12
−9+y=−12 -9 + y = -12
Solve for y y :
y=−3 y = -3

STEP 7

Substitute x=−3 x = -3 and y=−3 y = -3 back into one of the original equations to find z z . Use equation (1):
4(−3)+5(−3)+z=−31 4(-3) + 5(-3) + z = -31
−12−15+z=−31 -12 - 15 + z = -31
Solve for z z :
z=−4 z = -4

STEP 8

Verify the solution by substituting x=−3 x = -3 , y=−3 y = -3 , and z=−4 z = -4 into the original equations:
1) 4(−3)+5(−3)+(−4)=−31 4(-3) + 5(-3) + (-4) = -31
2) 5(−3)−2(−3)−(−4)=−5 5(-3) - 2(-3) - (-4) = -5
3) 2(−3)+(−3)+5(−4)=−29 2(-3) + (-3) + 5(-4) = -29
Check each equation:
1) −12−15−4=−31 -12 - 15 - 4 = -31 (True)
2) −15+6+4=−5 -15 + 6 + 4 = -5 (True)
3) −6−3−20=−29 -6 - 3 - 20 = -29 (True)
All equations are satisfied.
The solution is:
x=−3, y=−3, z=−4 x = -3, \, y = -3, \, z = -4

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