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Math  /  Algebra

QuestionSolve the system of equations by elimination. x−y−3z=−16x+y−3z=−43x−2y−3z=−12\begin{array}{l} x-y-3 z=-16 \\ x+y-3 z=-4 \\ 3 x-2 y-3 z=-12 \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 elimination method to solve the system.

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 back into the original equations.

STEP 3

Choose two pairs of equations to eliminate the same variable. Let's eliminate y y .
First, add the first and second equations:
(x−y−3z)+(x+y−3z)=−16+(−4)(x - y - 3z) + (x + y - 3z) = -16 + (-4)
Simplify:
2x−6z=−202x - 6z = -20

STEP 4

Now, eliminate y y from the first and third equations. Multiply the first equation by 2:
2(x−y−3z)=2(−16)2(x - y - 3z) = 2(-16)
Simplify:
2x−2y−6z=−322x - 2y - 6z = -32
Subtract the third equation from this result:
(2x−2y−6z)−(3x−2y−3z)=−32−(−12)(2x - 2y - 6z) - (3x - 2y - 3z) = -32 - (-12)
Simplify:
−x−3z=−20-x - 3z = -20

STEP 5

Now we have a system of two equations:
1. 2x−6z=−20 2x - 6z = -20
2. −x−3z=−20 -x - 3z = -20
Let's solve for x x and z z . Multiply the second equation by 2 to align coefficients:
−2x−6z=−40-2x - 6z = -40
Add this to the first equation:
(2x−6z)+(−2x−6z)=−20+(−40)(2x - 6z) + (-2x - 6z) = -20 + (-40)
Simplify:
−12z=−60-12z = -60
Solve for z z :
z=5z = 5

STEP 6

Substitute z=5 z = 5 back into the first equation of the reduced system:
2x−6(5)=−202x - 6(5) = -20
Simplify:
2x−30=−202x - 30 = -20
Solve for x x :
2x=10⇒x=52x = 10 \quad \Rightarrow \quad x = 5

STEP 7

Substitute x=5 x = 5 and z=5 z = 5 back into the original first equation to solve for y y :
5−y−3(5)=−165 - y - 3(5) = -16
Simplify:
5−y−15=−165 - y - 15 = -16
−y−10=−16-y - 10 = -16
Solve for y y :
−y=−6⇒y=6-y = -6 \quad \Rightarrow \quad y = 6

STEP 8

Verify the solution x=5 x = 5 , y=6 y = 6 , z=5 z = 5 by substituting into all original equations:
1. 5−6−3(5)=−16 5 - 6 - 3(5) = -16 which simplifies to −16=−16-16 = -16
2. 5+6−3(5)=−4 5 + 6 - 3(5) = -4 which simplifies to −4=−4-4 = -4
3. 3(5)−2(6)−3(5)=−12 3(5) - 2(6) - 3(5) = -12 which simplifies to −12=−12-12 = -12
All equations are satisfied.
The solution is:
x=5,y=6,z=5 x = 5, \quad y = 6, \quad z = 5

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