Problem 82 - Algebra 2

Question

Solve each system of equations. $$ \begin{array}{l}{3 x-2 y=-3} \\ {3 x+y=3}\end{array} $$

Step-by-Step Solution

Answer

\(x = \frac{1}{3}\), \(y = 2\)

Step 1: Verify Equations

Ensure both equations are correctly written. The system of equations is: 1. \(3x - 2y = -3\) 2. \(3x + y = 3\) We need to solve this system to find the values of \(x\) and \(y\).

Step 2: Eliminate Variable \(x\)

Subtract the second equation from the first to eliminate the \(x\) variable. \((3x - 2y) - (3x + y) = -3 - 3\). This simplifies to \(-3y = -6\).

Step 3: Solve for \(y\)

Solve the equation \(-3y = -6\) for \(y\) by dividing both sides by -3: \[ y = \frac{-6}{-3} = 2 \] So, \(y = 2\).

Step 4: Substitute \(y\) in Second Equation

Substitute \(y = 2\) into the second equation \(3x + y = 3\): \[ 3x + 2 = 3 \] Subtract 2 from both sides to isolate the term with \(x\): \[ 3x = 1 \]

Step 5: Solve for \(x\)

Divide both sides of the equation \(3x = 1\) by 3 to solve for \(x\): \[ x = \frac{1}{3} \]

Step 6: Verify Solution

Substitute \(x = \frac{1}{3}\) and \(y = 2\) back into both original equations to ensure they are satisfied:1. Equation 1: \(3(\frac{1}{3}) - 2(2) = 1 - 4 = -3\) 2. Equation 2: \(3(\frac{1}{3}) + 2 = 1 + 2 = 3\) Both equations are true. Therefore, \(x = \frac{1}{3}\) and \(y = 2\) are the correct solutions.

Key Concepts

Linear Equations

Linear equations are algebraic expressions in which each term is either a constant or the product of a constant and a single variable. They are called linear because they graph as straight lines in a Cartesian plane. For instance, the equations in this exercise are both linear:

• \(3x - 2y = -3\)
• \(3x + y = 3\)

Each equation represents a line, and solving a system of linear equations means finding the point(s) at which these lines intersect. These points, if they exist, represent the solution(s) to the system. It's crucial because linear equations are the foundation of many algebraic concepts used in higher mathematics and various applications.

Elimination Method

The elimination method is a technique used to solve a system of equations. By manipulating the equations, you aim to eliminate one of the variables, allowing you to solve for the other. Here's how it works in our exercise:

• Start with the two equations: \(3x - 2y = -3\) and \(3x + y = 3\).
• To eliminate \(x\), subtract the second equation from the first: \((3x - 2y) - (3x + y) = -3 - 3\).
• This simplification results in the equation \(-3y = -6\), effectively removing \(x\) from the equation.

The beauty of the elimination method is its ability to simplify complex systems of equations by focusing on one variable at a time, making it easier to find a solution.

Substitution Method

The substitution method involves finding the value of one variable in terms of the other and substituting it back into one of the original equations. In this exercise, after determining that \(y = 2\), we use substitution to find \(x\):

• Take the value of \(y\) from the solution we found: \(y = 2\).
• Substitute it into the second equation: \(3x + y = 3\).
• Resulting in: \(3x + 2 = 3\).
• Solve for \(x\) by isolating it: \(3x = 1\), and divide by 3 to find \(x = \frac{1}{3}\).

Substitution is a helpful method particularly when one variable is already isolated or can be easily solved for, making the process straightforward and efficient.

Solution Verification

Solution verification is an essential step to ensure the values found for the variables satisfy both original equations. After solving for \(x\) and \(y\), check if these values make the original equations true:

• For equation one: \(3(\frac{1}{3}) - 2(2) = -3\). Calculate: \(1 - 4 = -3\), which holds true.
• For equation two: \(3(\frac{1}{3}) + 2 = 3\). Calculate: \(1 + 2 = 3\), which also holds true.

Both checks confirm that \(x = \frac{1}{3}\) and \(y = 2\) are indeed the correct solutions. Verification ensures accuracy, which is crucial, especially in more complex systems where small errors can propagate.