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Solve the system of equations \( 2 x+3 y=6 \) and \( -3 x-5 y=-12 \) by combining the equations. \[ \begin{array}{l}(2 x+3 y=6) \\ (-3 x-5 y=-12) \\ 2 x+3 y=6 \\ -3 x-5 y=-12\end{array} \] \( \begin{array}{l}\text { o } x+0 y=\end{array} \)

Ask by Campos Weston. in the United States
Jan 21,2025

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The solution to the system of equations is \( x = -6 \) and \( y = 6 \).

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To solve the system of equations given, we can multiply the first equation by 3 and the second equation by 2 to eliminate \(x\): 1. Multiply \(2x + 3y = 6\) by 3: \[ 6x + 9y = 18 \] 2. Multiply \(-3x - 5y = -12\) by 2: \[ -6x - 10y = -24 \] Now, we can add the two new equations: \[ (6x + 9y) + (-6x - 10y) = 18 - 24 \] This simplifies to: \[ -1y = -6 \] Thus: \[ y = 6 \] Now that we have \(y\), we can substitute it back into one of the original equations to find \(x\): Substituting into \(2x + 3y = 6\): \[ 2x + 3(6) = 6 \] \[ 2x + 18 = 6 \] \[ 2x = 6 - 18 \] \[ 2x = -12 \] Thus: \[ x = -6 \] Putting it all together, the solution to the system of equations is: \[ \begin{array}{l} x + 0y = -6 \end{array} \]

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