\( C \left\{ \left. \begin{array} { c | l } { m + 2 n = 1 } \\ { - 3 m + n = - 10 } \end{array} \right. \begin{array} { l } { m = } \\ { n = } \end{array} \right. \)
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Bonus Knowledge
To solve the system of equations, we can start with the two equations: 1. \( m + 2n = 1 \) 2. \( -3m + n = -10 \) Let’s solve for \( n \) in terms of \( m \) using the first equation: From \( m + 2n = 1 \), we get: \[ 2n = 1 - m \] \[ n = \frac{1 - m}{2} \] Now, substitute \( n \) into the second equation: \[ -3m + \frac{1 - m}{2} = -10 \] To eliminate the fraction, multiply the entire equation by 2: \[ -6m + (1 - m) = -20 \] \[ -6m + 1 - m = -20 \] \[ -7m + 1 = -20 \] \[ -7m = -21 \] \[ m = 3 \] Now plug \( m = 3 \) back into the equation for \( n \): \[ n = \frac{1 - 3}{2} = \frac{-2}{2} = -1 \] Thus, the solution is: \( m = 3, n = -1 \).
