{2x+y-z=11 {3x+2y-4z=15 {4x+3y+7z=19
To solve the system of equations:
- \(2x + y - z = 11\)
- \(3x + 2y - 4z = 15\)
- \(4x + 3y + 7z = 19\)
You can use the method of elimination or substitution. Here, I'll demonstrate using elimination:
Step 1: Eliminate \(y\) from equations 1 and 2.
Multiply equation 1 by 2 to align the coefficients of \(y\):
\[ 4x + 2y - 2z = 22 \]
Subtract equation 2 from this new equation:
\[ (4x + 2y - 2z) - (3x + 2y - 4z) = 22 - 15 \implies x + 2z = 7 \]
Step 2: Eliminate \(y\) from equations 2 and 3.
Multiply equation 2 by 3 and equation 3 by 2:
\[ 9x + 6y - 12z = 45 \] \[ 8x + 6y + 14z = 38 \]
Subtract the second equation from the first:
\[ (9x + 6y - 12z) - (8x + 6y + 14z) = 45 - 38 \implies x - 26z = 7 \]
Step 3: Solve the two new equations:
From the results of steps 1 and 2, we have:
- \(x + 2z = 7\)
- \(x - 26z = 7\)
Subtract the second from the first:
\[ (x + 2z) - (x - 26z) = 7 - 7 \implies 28z = 0 \implies z = 0 \]
Plug \(z = 0\) into \(x + 2z = 7\):
\[ x + 2(0) = 7 \implies x = 7 \]
Step 4: Find \(y\) using equation 1:
Substitute \(x = 7\) and \(z = 0\) into equation 1:
\[ 2(7) + y - 0 = 11 \implies 14 + y = 11 \implies y = -3 \]
Solution:
\(x = 7\), \(y = -3\), \(z = 0\).
