To solve the system of equations using the elimination method, we can follow these steps:
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Write down the equations: \[ -6x - 8y = -28 \quad \text{(Equation 1)} \] \[ 9x + 5y = -14 \quad \text{(Equation 2)} \]
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We want to eliminate one of the variables. Let's eliminate \(y\). To do this, we can manipulate each equation so that the coefficients of \(y\) are opposites.
We can multiply Equation 1 by 5 and Equation 2 by 8 to make the coefficients of \(y\) match:
Multiply Equation 1 by 5: \[ 5(-6x - 8y) = 5(-28) \] This results in: \[ -30x - 40y = -140 \quad \text{(Equation 3)} \]
Multiply Equation 2 by 8: \[ 8(9x + 5y) = 8(-14) \] This results in: \[ 72x + 40y = -112 \quad \text{(Equation 4)} \]
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Now we can add Equation 3 and Equation 4: \[ (-30x - 40y) + (72x + 40y) = -140 + (-112) \]
Simplifying gives: \[ (72x - 30x) + (-40y + 40y) = -252 \] \[ 42x = -252 \]
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Solve for \(x\): \[ x = \frac{-252}{42} = -6 \]
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Now that we have \(x\), we substitute \(x = -6\) into one of the original equations to find \(y\). We'll use Equation 2: \[ 9(-6) + 5y = -14 \] \[ -54 + 5y = -14 \]
Add 54 to both sides: \[ 5y = -14 + 54 \] \[ 5y = 40 \]
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Solve for \(y\): \[ y = \frac{40}{5} = 8 \]
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The solution to the system of equations is: \[ x = -6, \quad y = 8 \]
Thus, the solution is: \[ \boxed{(-6, 8)} \]