Solving Pair of Linear Equations by Elimination Method

We are given the following pair of linear equations:

1. \(\dfrac{1}{2}x+\dfrac{2}{3}y=-1\)
2. \(x-\dfrac{1}{3}y=3\)

We will use the elimination method to solve this system of equations. The goal is to eliminate one variable by making its coefficients equal in both equations and then adding or subtracting the equations.

Step 1: Clearing Fractions from the Equations

To make the equations easier to work with, let's clear the fractions by multiplying each equation by the least common multiple (LCM) of the denominators.

• For Equation 1, the denominators are 2 and 3. The LCM of 2 and 3 is 6. Multiply Equation 1 by 6:

\(6 \times \left(\dfrac{1}{2}x+\dfrac{2}{3}y\right) = 6 \times (-1)\)

\(6 \times \dfrac{1}{2}x + 6 \times \dfrac{2}{3}y = -6\)

\(3x + 4y = -6\) (Let's call this Equation 3)

• For Equation 2, the denominator is 3. The LCM is 3. Multiply Equation 2 by 3:

\(3 \times \left(x-\dfrac{1}{3}y\right) = 3 \times 3\)

\(3x - 3 \times \dfrac{1}{3}y = 9\)

\(3x - y = 9\) (Let's call this Equation 4)

Now we have a simpler system of linear equations:

1. \(3x + 4y = -6\) (Equation 3)
2. \(3x - y = 9\) (Equation 4)

Step 2: Eliminating One Variable

Observe Equations 3 and 4. The coefficient of \(x\) is 3 in both equations. We can eliminate \(x\) by subtracting Equation 4 from Equation 3.

Subtract Equation 4 from Equation 3:

\((3x + 4y) - (3x - y) = -6 - 9\)

\(3x + 4y - 3x + y = -15\)

\((3x - 3x) + (4y + y) = -15\)

\(0x + 5y = -15\)

\(5y = -15\)

Step 3: Solving for the Remaining Variable

Now we have an equation with only one variable, \(y\). Solve for \(y\):

\(5y = -15\)

\(y = \dfrac{-15}{5}\)

\(y = -3\)

Step 4: Substituting the Value to Find the Other Variable

Substitute the value of \(y = -3\) into either Equation 3 or Equation 4 to find the value of \(x\). Let's use Equation 4:

\(3x - y = 9\)

\(3x - (-3) = 9\)

\(3x + 3 = 9\)

Subtract 3 from both sides:

\(3x = 9 - 3\)

\(3x = 6\)

Divide by 3:

\(x = \dfrac{6}{3}\)

\(x = 2\)

Step 5: State the Solution

The solution to the pair of linear equations is \(x = 2\) and \(y = -3\).

Verification (Optional but Recommended)

Let's check the solution \(x=2, y=-3\) in the original equations:

• Equation 1: \(\dfrac{1}{2}(2) + \dfrac{2}{3}(-3) = 1 + (-2) = 1 - 2 = -1\). This matches the right side of Equation 1.
• Equation 2: \(2 - \dfrac{1}{3}(-3) = 2 - (-1) = 2 + 1 = 3\). This matches the right side of Equation 2.

The solution is correct.

Revision Table: Linear Equations and Elimination

When solving a pair of linear equations using the elimination method, remember these key points:

• Make sure the equations are in a standard form, often \(Ax + By = C\).
• If there are fractions or decimals, clear them by multiplying by the appropriate number.
• Choose a variable to eliminate (either \(x\) or \(y\)).
• Multiply one or both equations by constants so that the coefficients of the variable to be eliminated are equal in magnitude.
• If the signs of the coefficients are the same, subtract one equation from the other. If the signs are opposite, add the equations.
• Solve the resulting equation for the remaining variable.
• Substitute the value found back into one of the original or modified equations to find the value of the other variable.
• Always verify your solution in the original equations.

Additional Information: Solving Systems of Equations

Besides the elimination method, other methods can be used to solve a system of linear equations, such as:

• Substitution Method: Solve one equation for one variable in terms of the other variable, and then substitute this expression into the other equation.
• Graphical Method: Graph both equations on the same coordinate plane. The point of intersection of the two lines represents the solution to the system. This method can sometimes be less precise than algebraic methods.
• Matrix Method (using Cramer's Rule or inverse matrices): This method is typically introduced at higher levels of mathematics and is very efficient for larger systems.

Choosing the best method depends on the specific equations given. The elimination method is particularly useful when coefficients are easy to match or eliminate.