Question

Solve the system by elimination:

x + 5y - 4z = -10
2x - y + 5z = -9
2x - 10y - 5z = 0

Answer 1

let us pick 2 off the 3 functions there, and cancel out , say "z"

x + 5y - 4z = -10 <---- this one
2x - y + 5z = -9 <---- and this one
2x - 10y - 5z = 0

\(\large \begin{array}{llll}
x + 5y - 4z = -10&\times 5\implies & 5x + 25y - 20z = -50\\
2x - y + 5z = -9&\times 4\implies & 8x - 4y + 20z = -36\\
\hline\\
\textit{sum them up}&\implies&13x-21y+0z=-86\\ \quad \\
\color{blue}{13x-21y=-86}


\end{array}\)

Answer 2

let us pick ANOTHER 2 functions off the 3, and again, cancel out "z"

x + 5y - 4z = -10
2x - y + 5z = -9 <---- this one
2x - 10y - 5z = 0 <---- and this one

\(\large \begin{array}{llll}
2x - y + 5z = -9&\implies & 2x - y + 5z = -9\\
2x - 10y - 5z = 0&\implies & 2x - 10y - 5z = 0\\
\hline\\
\textit{sum them up}&\implies&4x-11y+0z=-9\\ \quad \\
\color{blue}{4x-11y=-9}


\end{array}\)

now notice, you have obtained, 2 RESULTANT functions, with only 2 variables, "x" and "y"

in the same manner, eliminate either of the variables, and get the other variable

Answer 3

so use the 2 RESULTANT functions to get the value of one variable

13x - 2y = -86
4x -11y = -9

Answer 4

Thank you this is so helpful!!

Answer 5

yw

Answer 6

hmmm

Answer 7

25y - 4y = ... 21y positive... .ok... so.... minor sum issue there...

\(\large \begin{array}{llll}
x + 5y - 4z = -10&\times 5\implies & 5x + 25y - 20z = -50\\
2x - y + 5z = -9&\times 4\implies & 8x + 4y + 20z = -36\\
\hline\\
\textit{sum them up}&\implies&13x+21y+0z=-86\\ \quad \\
\color{blue}{13x+21y=-86}


\end{array}\)

Answer 8

well.. \(\large \begin{array}{llll}
x + 5y - 4z = -10&\times 5\implies & 5x + 25y - 20z = -50\\
2x - y + 5z = -9&\times 4\implies & 8x - 4y + 20z = -36\\
\hline\\
\textit{sum them up}&\implies&13x+21y+0z=-86\\ \quad \\
\color{blue}{13x+21y=-86}


\end{array}\)

rather

Answer 9

so the 2 functions will be
13x + 21y = -86
4x -11y = -9


rechecking... I noticed ... should have been +21y

Answer 10

Oh okay thank you!

Answer 11

yw

Answer 12

so z=0, right?

Answer 13

yes, reason why I multiplied by 5 and 4, to have them be -20z and +20z and thus -20z+20z = 0, thus effectively cancelling it out, or ELIMINATING it

Answer 14

notice in the 2nd paiir, I didn't have to do that, because I had a +5z and -5z, and they cancel each other right off

Answer 15

Okay yeah I was just making sure, thanks!!

Answer 16

np