A system of equations is given. \( \begin{array}{l}-2 y=10-5 x \\ -3 y=-4 x+15\end{array} \) Solve for \( (x, y) \) using the elimination method. Show all work.
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Let's solve the given system of equations using the elimination method step by step. The system of equations is: 1. \(-2y = 10 - 5x\) 2. \(-3y = -4x + 15\) First, we can rearrange both equations into standard form \(Ax + By = C\). 1. From \(-2y = 10 - 5x\), we can rewrite it as: \[5x - 2y = 10\] 2. From \(-3y = -4x + 15\), we rearrange it to: \[4x - 3y = 15\] Now we have: \[ \begin{array}{l} 5x - 2y = 10 \quad (1) \\ 4x - 3y = 15 \quad (2) \end{array} \] Next, we will manipulate these equations to eliminate one variable. Let's multiply equation (1) by 3 and equation (2) by 2 to align the coefficients of \(y\): Multiplying equation (1) by 3: \[ 15x - 6y = 30 \quad (3) \] Multiplying equation (2) by 2: \[ 8x - 6y = 30 \quad (4) \] Now we will subtract equation (4) from equation (3): \[ (15x - 6y) - (8x - 6y) = 30 - 30 \] This simplifies to: \[ 15x - 8x = 0 \] \[ 7x = 0 \] Therefore, solving for \(x\): \[ x = 0 \] Now, substitute \(x = 0\) back into either of the original equations to find \(y\). We can use equation (1): \[ 5(0) - 2y = 10 \] \[ -2y = 10 \] \[ y = -5 \] Thus, the solution to the system of equations is: \[ (x, y) = (0, -5) \]
