Here's how to use elimination to solve a system of two linear equations step-by-step, filling in the answers for the placeholders indicated:
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Identify elimination as the best method for the problem.
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Find a pair of equations with equal or opposite coefficients in the system. This may involve multiplying one or both of the equations by a suitable number to achieve this.
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Add or subtract one equation from the other to eliminate one variable.
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Solve the one-variable equation. This means you'll be left with an equation that contains just one variable, which you can solve for that variable.
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Substitute the value of that variable into the other equation to find the other variable.
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Solve for the second variable. This will give you the value of the second variable in the system of equations.
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Check your answer. Use substitution and/or graphing to ensure the solution satisfies both original equations.
Example:
Let's use elimination with a specific example to clarify the steps:
Given the system of equations:
- \( 2x + 3y = 6 \)
- \( 4x - 3y = 10 \)
Step 1
Identify elimination as the best method for the problem.
Step 2
We have equal and opposite coefficients for \(y\) (3 and -3).
Step 3
Add both equations: \[ (2x + 3y) + (4x - 3y) = 6 + 10 \] This simplifies to: \[ 6x = 16 \]
Step 4
Solve the one-variable equation: \[ x = \frac{16}{6} = \frac{8}{3} \]
Step 5
Substitute \(x = \frac{8}{3}\) back into one of the original equations. We'll use the first equation: \[ 2\left(\frac{8}{3}\right) + 3y = 6 \]
Step 6
Solve for \(y\): \[ \frac{16}{3} + 3y = 6 \] Subtract \(\frac{16}{3}\) from both sides: \[ 3y = 6 - \frac{16}{3} \] Convert 6 to a fraction: \[ 3y = \frac{18}{3} - \frac{16}{3} = \frac{2}{3} \] Thus, \[ y = \frac{2}{9} \]
Step 7
Check your answer. Substitute \(x = \frac{8}{3}\) and \(y = \frac{2}{9}\) back into both original equations to verify they hold true.
Following these steps will effectively use elimination to solve a system of linear equations successfully.
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