Solve the system. Give answer as \( (x, y, z) \) \( -6 x-5 y+5 z=-23 \) \( 1 x-10 y-5 z=38 \) \( 6 x-20 y+5 z=43 \) \( (x, y, z)=( \) Question Help: Message instructor Submit Question Jump to Answer

To solve the given system of equations, we can use the elimination or substitution methods. Let's rewrite the equations for clarity: 1) \(-6x - 5y + 5z = -23\) 2) \(x - 10y - 5z = 38\) 3) \(6x - 20y + 5z = 43\) First, we can manipulate these equations to eliminate one variable. Let's start with equations (1) and (2). We can simplify the process by manipulating equations (2) and (3) first. From equation (2): \[ z = \frac{x - 10y - 38}{-5} \] Now, substitute \( z \) from equation (2) into equations (1) and (3): **Substituting into equation (1):** \[ -6x - 5y + 5\left(\frac{x - 10y - 38}{-5}\right) = -23 \] Expanding this, \[ -6x - 5y - (x - 10y - 38) = -23 \] Combine like terms: \[ -6x - 5y - x + 10y + 38 = -23 \implies -7x + 5y + 38 = -23 \implies -7x + 5y = -61 \] Rearranged, that's: \[ 7x - 5y = 61 \quad \text{(Equation 4)} \] **Now for equation (3):** \[ 6x - 20y + 5\left(\frac{x - 10y - 38}{-5}\right) = 43 \] Expanding this, \[ 6x - 20y - (x - 10y - 38) = 43 \] Combine like terms: \[ 6x - 20y - x + 10y + 38 = 43 \implies 5x - 10y + 38 = 43 \implies 5x - 10y = 5 \] Rearranged, that's: \[ x - 2y = 1 \quad \text{(Equation 5)} \] Now we can solve the new equations (4) and (5): From equation (5): \[ x = 2y + 1 \] Substituting \( x \) in equation (4): \[ 7(2y + 1) - 5y = 61 \] Expanding and solving for \( y \): \[ 14y + 7 - 5y = 61 \implies 9y + 7 = 61 \implies 9y = 54 \implies y = 6 \] Using \( y \) to find \( x \): \[ x = 2(6) + 1 = 12 + 1 = 13 \] Substituting \( x \) and \( y \) back into equation (2) to find \( z \): \[ 13 - 10(6) - 5z = 38 \implies 13 - 60 - 5z = 38 \implies -47 - 5z = 38 \implies -5z = 85 \implies z = -17 \] Final solution: \[ (x, y, z) = (13, 6, -17) \]