Problem 28 - Precalculus

Question

Solve each system of linear equations by elimination. $$\begin{array}{r} 6 x-2 y=3 \\ -3 x+2 y=-2 \end{array}$$

Step-by-Step Solution

Answer

The solution is \(x = \frac{1}{3}\) and \(y = -\frac{1}{2}\).

Step 1: Align and Identify

First, ensure both equations are aligned and review them for variables that might eliminate each other when added or subtracted. The system presented is: \(6x - 2y = 3\) and \(-3x + 2y = -2\). Notice that the coefficients of \(y\) are \(-2\) and \(2\); adding these equations will eliminate \(y\).

Step 2: Add Equations to Eliminate a Variable

Add the two equations together to eliminate the \(y\) variable. This gives you: \[(6x - 2y) + (-3x + 2y) = 3 + (-2)\]Simplifying the left we get \[6x - 2y - 3x + 2y = 1\]The \(-2y\) and \(2y\) cancel out, resulting in: \[3x = 1\]

Step 3: Solve for x

Now that \(y\) is eliminated, solve for \(x\) in the equation \(3x = 1\). Divide both sides by 3: \[x = \frac{1}{3}\]

Step 4: Substitute x to Find y

Substitute \(x = \frac{1}{3}\) back into one of the original equations to find \(y\). Using the first equation \(6x - 2y = 3\), substitute \(x\):\[6(\frac{1}{3}) - 2y = 3\]Simplify to get\[2 - 2y = 3\]

Step 5: Solve for y

Rearrange \(2 - 2y = 3\) to determine \(y\):\[-2y = 3 - 2\]\[-2y = 1\]Divide both sides by \(-2\):\[y = -\frac{1}{2}\]

Step 6: Verify Solution

Verify the solution by substituting \(x = \frac{1}{3}\) and \(y = -\frac{1}{2}\) back into the second original equation \(-3x + 2y = -2\):\[-3(\frac{1}{3}) + 2(-\frac{1}{2}) = -2\] This gives: \[-1 - 1 = -2\] which holds true, confirming the solution is correct.

Key Concepts

Elimination Method

The elimination method is a primary technique used for solving systems of linear equations. This method involves manipulating the equations such that one of the variables gets eliminated. It's often preferred when the coefficients allow for straightforward cancellation of variables. This can happen by direct addition or subtraction of equations.

• Start by aligning the equations, ensuring that corresponding variables and constants are in the same columns.
• Look for variables with coefficients that can cancel each other out when the equations are added or subtracted.

The task is to eliminate one variable first, reducing the system to a single equation with one unknown, which simplifies solving the equation. In our example, the coefficient of the variable \(y\) in the system \(6x - 2y = 3\) and \(-3x + 2y = -2\) sets us up perfectly to add the equations. This results in one less variable, paving the way to readily solve for the remaining variable.

Solve for x and y

Once a variable has been eliminated using the elimination method, you solve for the remaining variable. This gives you one variable's value, which you then substitute back into one of the original equations to find the other variable's value. Let's break it down:

• After eliminating \(y\), we're left with \(3x = 1\), which quickly resolves to \(x = \frac{1}{3}\) after dividing both sides by 3.
• Now, substitute \(x = \frac{1}{3}\) into one of the original equations— we used \(6x - 2y = 3\)—allowing us to solve for \(y\).
• Substituting and solving lands us with \(y = -\frac{1}{2}\).

The systematic solving for each variable ensures clarity and precision, leading us logically from one step to the next.

Verification of Solution

Verifying the solution is the crucial final step when using the elimination method. This ensures both values found for \(x\) and \(y\) satisfy the original equations:

• Substitute both \(x = \frac{1}{3}\) and \(y = -\frac{1}{2}\) back into the other equation apart from the one used to find \(y\).
• In our problem, we checked \(-3x + 2y = -2\) by substituting the values for \(x\) and \(y\).
• Calculate: \(-3(\frac{1}{3}) + 2(-\frac{1}{2}) = -2\), simplifying to \(-1 - 1 = -2\), confirms that both sides of the equation are equal.

Verification steps are essential as they prove the solution is accurate and assures you that the steps you took were correct, transforming a teachable moment into a robust learning tool.