To eliminate a variable when working with a system of linear equations, you can use either the substitution method or the elimination method. In this case, we'll focus on the elimination method, which is particularly effective when both equations are simplified.
Given the two equations:
- \( 5x + 3y = 4 \)
- \( -2x - 8y = 6 \)
Steps to use the elimination method:
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Align the equations: Write the equations one below the other for clarity.
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Multiply equations (if necessary): You can multiply one or both equations by a coefficient that will help you eliminate one of the variables. Here, we can manipulate the second equation to align coefficients for \( y \).
We can multiply the first equation by \( 8 \) so that the coefficients of \( y \) match in terms of magnitude:
\[ 8(5x + 3y = 4) \implies 40x + 24y = 32 \]
The second equation stays the same: \[ -2x - 8y = 6 \]
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Multiply the second equation: Now, let's multiply the second equation by \( 3 \) to match the \( y \) coefficients: \[ 3(-2x - 8y = 6) \implies -6x - 24y = 18 \]
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Rewrite the equations:
- \( 40x + 24y = 32 \) (from the modified first equation)
- \( -6x - 24y = 18 \) (from the modified second equation)
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Add the equations: Now, add the two new equations together to eliminate \( y \): \[ (40x + 24y) + (-6x - 24y) = 32 + 18 \] This simplifies to: \[ 34x = 50 \]
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Solve for \( x \): \[ x = \frac{50}{34} = \frac{25}{17} \]
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Substitute back to find \( y \): Use either original equation to solve for \( y \). Using the first equation: \[ 5\left(\frac{25}{17}\right) + 3y = 4 \] \[ \frac{125}{17} + 3y = 4 \] \[ 3y = 4 - \frac{125}{17} = \frac{68 - 125}{17} = \frac{-57}{17} \] \[ y = \frac{-57}{51} = \frac{-19}{17} \]
The solution to the system is \( x = \frac{25}{17} \) and \( y = \frac{-19}{17} \).
Thus, the strategy used here to eliminate a variable successfully was to manipulate the equations through multiplication to create a system where adding them together would eliminate one variable.
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