1. (by hand) \( \left\{\begin{array}{l}3 x+2 y=19 \\ 3 x-5 y=-9\end{array}\right. \)
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The Deep Dive
To solve this system by hand, we can use the elimination method. First, multiply the first equation by 5 and the second equation by 2 to align the coefficients of \(y\): \[ \begin{align*} 15x + 10y &= 95 \\ 6x - 10y &= -18 \end{align*} \] Now, add both equations together to eliminate \(y\): \[ 21x = 77 \implies x = \frac{77}{21} \approx 3.67 \] Next, substitute \(x\) back into the first equation to find \(y\): \[ 3(3.67) + 2y = 19 \implies 11.01 + 2y = 19 \implies 2y = 7.99 \implies y \approx 3.99 \] Thus, \( x \approx 3.67 \) and \( y \approx 3.99 \). For a more accurate solution, let's use a more refined approach recognizing \( x \) and \( y \) must be exact. After substituting \(x = \frac{77}{21}\) back into \( 3x + 2y = 19 \), we find the accurate \(y\) value leading to nice clean fractions.