Question

If you were to use the substitution method to solve the following system, choose the new system of equations that would result if z was isolated in the first equation.

2x – 5y + z = 15
4x – y + 3z = 15
3x – 2y + 2z = 13
possible answers:
–2x + 14y = –30
–x + 8y = –17
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22x + 16z = 90
11x + 8z = 43
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–18x – 14z = –60
–5x + 8z = 43
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10x – 16y = –30
7x – 12y = –15

Answer 1

From the first equation z = -2x + 5y + 15. Plug that into the remaining two, combine like terms, and get your answer.

Answer 2

i dont understand.. sorry

Answer 3

Okay. Not a problem.

We solve the first equation for z to get \[z = -2x + 5y + 15\]Now let's plug this into the second and third equation to obtain\[\begin{matrix}4x - y +3(-2x+5y+15) = 15 \rightarrow 4x - 6x - y + 15 y + 45 = 15 \\ 3x - 2y + 2(-2x+5y+15) = 13 \rightarrow 3x - 4x - 2y+10y + 30 = 13\end{matrix}\]After combining like terms, we get \[\begin{matrix}-2x +14 y = -30 \\ -x + 8y = -17\end{matrix}\]