Solve the system of linear equations. \[ \begin{array}{r}x+y+z=6 \\ 2 x-y+z=3 \\ 3 x+y-z=2\end{array} \]
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To solve the system of equations, we can use the elimination method or substitution. Let's put the equations together for clarity: 1. \( x + y + z = 6 \) (Equation 1) 2. \( 2x - y + z = 3 \) (Equation 2) 3. \( 3x + y - z = 2 \) (Equation 3) First, we can eliminate \( z \) by manipulating these equations. Let's express \( z \) from Equation 1: From Equation 1: \[ z = 6 - x - y \] Now we can substitute \( z \) in Equations 2 and 3. Substituting into Equation 2: \[ 2x - y + (6 - x - y) = 3 \] Combine like terms: \[ 2x - y + 6 - x - y = 3 \implies x - 2y + 6 = 3 \implies x - 2y = -3 \implies x = 2y - 3 \quad (Equation 4) \] Now substitute \( z \) in Equation 3: \[ 3x + y - (6 - x - y) = 2 \] Combine like terms: \[ 3x + y - 6 + x + y = 2 \implies 4x + 2y - 6 = 2 \implies 4x + 2y = 8 \implies 2x + y = 4 \quad (Equation 5) \] Now we have a new system with the two equations: 1. \( x = 2y - 3 \) (from Equation 4) 2. \( 2x + y = 4 \) (Equation 5) Substitute \( x \) from Equation 4 into Equation 5: \[ 2(2y - 3) + y = 4 \implies 4y - 6 + y = 4 \implies 5y - 6 = 4 \implies 5y = 10 \implies y = 2 \] Now substitute \( y = 2 \) back into Equation 4 to find \( x \): \[ x = 2(2) - 3 = 4 - 3 = 1 \] Now substitute \( x = 1 \) and \( y = 2 \) back into Equation 1 to find \( z \): \[ 1 + 2 + z = 6 \implies z = 6 - 3 = 3 \] Thus, the solution to the system of equations is: \[ \boxed{(1, 2, 3)} \]
