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trasher [3.6K]
2 years ago
7

PLEASE HELP

Mathematics
1 answer:
Anna007 [38]2 years ago
4 0
I don’t really know let me ask my friends 1
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Looks like the system is

\begin{cases}-2x+y+6z=1\\3x+2y+5z=16\\7x+3y-4z=11\end{cases}

We can eliminate y by taking

(3x+2y+5z)-2(-2x+y+6z)=16-2(1)

\implies3x+2y+5z+4x-2y-12z=16-2

\implies7x-7z=14

\implies x-z=2

so that z=x-2, and

(7x+3y-4z)-3(-2x+y+6z)=11-3(1)

\implies7x+3y-4z+6x-3y-18z=11-3

\implies13x-22z=8

Substitute z=x-2 into this last equation and solve for x:

13x-22(x-2)=8

\implies13x-22x+44=8

\implies-9x=-36

\implies x=4

Then

z=x-2

\implies z=4-2

\implies z=2

Plug these values into any one of the original equation to solve for y:

-2x+y+6z=1

\implies-2(4)+y+6(2)=1

\implies-8+y+12=1

\implies y=-3

Hence the solution is x = 4, y = -3, and z = 2.

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