Problem 32 - Understanding Elementary Algebra with Geometry

Question

In Exercises \(29-62,\) solve the system of equations using any method you choose. $$\left\\{\begin{aligned} 3 x+6 y &=2 \\ -3 x-3 y &=1 \end{aligned}\right.$$

Step-by-Step Solution

Answer

x = - \frac{2}{3}, y = \frac{1}{3}

Step 1: Simplify the equations

Simplify each equation to make the calculations easier. Divide the first equation by 3: \( x + 2y = \frac{2}{3} \). Divide the second equation by -3: \( x + y = - \frac{1}{3} \)

Step 2: Subtract the second simplified equation from the first

Subtract the equation \( x + y = - \frac{1}{3} \) from the equation \( x + 2y = \frac{2}{3} \) to eliminate \( x \): \( (x + 2y) - (x + y) = \frac{2}{3} - \frac{1}{3} \) This simplifies to: \( y = \frac{1}{3} \)

Step 3: Solve for x

Substitute \( y = \frac{1}{3} \) back into one of the simplified equations, for example \( x + y = - \frac{1}{3} \): \( x + \frac{1}{3} = - \frac{1}{3} \) Subtract \( \frac{1}{3} \) from both sides: \( x = - \frac{2}{3} \)

Key Concepts

solving linear equations

Solving linear equations is the process of finding the values of variables that satisfy given equations. These equations typically have the form: \(Ax + By = C\), where \(A\), \(B\), and \(C\) are constants, and \(x\) and \(y\) are variables. By solving such equations, we determine the values of \(x\) and \(y\) that make the equation true.

There are different methods to solve linear equations, such as graphing, substitution, and elimination. Choosing the right method often depends on the simplicity and form of the equations.

In our exercise, we are dealing with a system of linear equations. To solve the system, we need to find a common solution for both equations simultaneously. This means we need values for \(x\) and \(y\) that satisfy both given equations.

substitution method

The substitution method involves isolating one variable in one of the equations and substituting its expression into the other equation. This helps to find the value of one variable at a time.

Let's break it down:

• First, solve one of the equations for one variable in terms of the other.
• Substitute this expression into the other equation.
• Solve the resulting equation for the remaining variable.
• Substitute the value back into the expression found in the first step to find the other variable.

In the provided exercise, we initially simplified the equations to make substitution easier:

Simplified first equation: \(x + 2y = \frac{2}{3}\)
Simplified second equation: \(x + y = - \frac{1}{3}\)
By substituting the simplified second equation into the first, we eliminated one variable and solved for the other.

simplification of equations

Simplifying equations is a crucial step when solving algebraic systems. It involves reducing equations to their simplest form to make calculations more manageable and accurate.

In our exercise, the original equations were as follows:
\[\begin{cases}3x + 6y = 2 \-3x - 3y = 1\end{cases}\]
To simplify:

• Divide the first equation by 3: \( x + 2y = \frac{2}{3} \)
• Divide the second equation by -3: \( x + y = -\frac{1}{3} \)

This simplification made the equations easier to handle. Once simplified, we used subtraction to eliminate one variable and then solved for the remaining variable.

By simplifying the given equations, we make the entire process of solving much more straightforward and less prone to errors. It's a fundamental technique in algebra that helps in clear and concise problem-solving.