
The solution of the system of linear equations {x+y=92x−z=133y+2z=7.

Answer to Problem 41AYU
Solution:
The solution of the system is x=6, y=3, and z=−1.
Explanation of Solution
Given information:
The system of linear equations {x+y=92x−z=133y+2z=7
Explanation:
The given system of linear equations is
{x+y=92x−z=133y+2z=7.
To solve this system using elimination method, first multiply equation x+y=9 by 2.
Therefore, the equation becomes 2x+2y=18
Subtract the equation 2x+2y=18 from 2x−z=13 to eliminate the variable.
2x−z=13−2x+2y=18−2y−z=−5
Then, multiply the equation −2y−z=−5 by 3 and 3y+2z=7 by 2
Therefore, the equations become −6y−3z=−15 and 6y+4z=14
Add the equations −6y−3z=−15 and 6y+4z=14 to eliminate the variable z.
−6y−3z=−15+6y+4z=14z=−1
Plug z=−1 in equation −6y−3z=−15 to get the variable y.
Therefore, the equation becomes −6y−3(−1)=−15
⇒y=3
And plug y=3 in equation 2x+2y=18 to get the variable x.
Therefore, the equation becomes 2x+2(3)=18
⇒x=6.
Therefore, the solution of the system of equation {x+y=92x−z=133y+2z=7 is x=6, y=3, z=−1.
Chapter 11 Solutions
Precalculus
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