solve x+y+z=15,2x+2y+z=10,x-y-z=-9 by elimination

asked by guest
on Jan 15, 2025 at 1:59 pm



You asked:

Solve the system of equations \(x + y + z = 15\), \(2 x + 2 y + z = 10\), \(x - y - z = -9\) by elimination.

MathBot Answer:

\[\begin{aligned}x &= 3\\y &= -8\\z &= 20\end{aligned}\]

\[\begin{aligned}x + y + z &= 15\\2 x + 2 y + z &= 10\\x - y - z &= -9\end{aligned}\]

Eliminate \(x\) from equation 2 by multiply both sides of equation 1 by \(-2\), and adding the results to both sides of equation 2.

Eliminate \(x\) from equation 3 by multiply both sides of equation 1 by \(-1\), and adding the results to both sides of equation 3.

\[\begin{aligned}x + y + z &= 15\\- z &= -20\\- 2 y - 2 z &= -24\end{aligned}\]

Swap equations 2 and 3.

\[\begin{aligned}x + y + z &= 15\\- 2 y - 2 z &= -24\\- z &= -20\end{aligned}\]

Solve equation 3 for \(z\), and substitute the solution into the remaining equations.

\[\begin{aligned}x + y &= -5\\- 2 y &= 16\\z &= 20\end{aligned}\]

Solve equation 2 for \(y\), and substitute the solution into the remaining equations.

\[\begin{aligned}x &= 3\\y &= -8\\z &= 20\end{aligned}\]

Solve equation 1 for the remaining unknown \(x\).

\[\begin{aligned}x &= 3\\y &= -8\\z &= 20\end{aligned}\]