Problem 2 - Algebra 2

Question

Solve each system of equations by completing a table. \(3 x+2 y=10\) \(2 x+3 y=10\)

Step-by-Step Solution

Answer

The solution is \((x, y) = (3, 0.5)\).

Step 1: Define Variables and Equations

We are given the system of equations: \(3x + 2y = 10\) and \(2x + 3y = 10\). Our goal is to find values of \(x\) and \(y\) that satisfy both equations.

Step 2: Solve for y in the First Equation

Starting with the equation \(3x + 2y = 10\), let's express \(y\) in terms of \(x\).First, subtract \(3x\) from both sides: \(2y = -3x + 10\).Next, divide each term by 2: \(y = \frac{-3x}{2} + 5\).

Step 3: Solve for y in the Second Equation

Using the equation \(2x + 3y = 10\), solve for \(y\) in terms of \(x\).Subtract \(2x\) from both sides: \(3y = -2x + 10\).Divide each term by 3: \(y = \frac{-2x}{3} + \frac{10}{3}\).

Step 4: Create a Table of Values

We will create a table to find matching \(x, y\) pairs that satisfy both equations:| \(x\) | \(y_1 = \frac{-3x}{2} + 5\) | \(y_2 = \frac{-2x}{3} + \frac{10}{3} \)||----|-------------|----------------|| 0 | 5 | \(\frac{10}{3}\) || 2 | 2 | \(\frac{8}{3}\) || 4 | -1 | \(\frac{2}{3}\) || 3 | 0.5 | 0.5 |Try different values for \(x\) and calculate \(y_1\) and \(y_2\) until they match.

Step 5: Identify the Solution

From the table, we observe that when \(x = 3\), both equations result in \(y = 0.5\). Therefore, the solution is \((x, y) = (3, 0.5)\).

Key Concepts

Solving by Substitution

To solve a system of equations, substitution is a method where you solve one equation for one variable and then substitute that expression into the other equation. This way, you reduce the system to a single equation with one variable, making it easier to find solutions. In our exercise, we began by solving the first equation, which was \(3x + 2y = 10\), for \(y\). We expressed \(y\) in terms of \(x\) as \(y = \frac{-3x}{2} + 5\). This expression was then substituted into the second equation to find compatible \(x\) and \(y\) values. The substitution method can be particularly useful when one of the equations is already solved for one of the variables, or can be rearranged easily. In practice:

• Solve for one variable in one of the equations.
• Substitute this expression into the other equation.
• Solve the resulting equation.

Using substitution helps reduce complexity and focus on the relationships between the variables.

Linear Equations

Linear equations form the basis of algebraic systems and consist of simple relationships between variables. A linear equation in two variables, like \(3x + 2y = 10\), describes a straight line on a Cartesian coordinate plane. This is because the degree of the equation is one, as evidenced by the variables not being raised to any power other than one.The solution to a system of linear equations is the point where their lines intersect. If you solve each equation for \(y\), you can easily graph these equations to visually find this intersection point. In our exercise, we transformed both equations to: \(y = \frac{-3x}{2} + 5\) and \(y = \frac{-2x}{3} + \frac{10}{3}\). Finding a common \(x\) and \(y\) that satisfies both equations implies finding the intersection of their corresponding lines.Understanding linear equations helps in:

• Recognizing patterns in data.
• Solving problems involving rates of change.
• Modeling real-world situations such as mixtures or finances.

Table of Values

A table of values is a practical tool to find solutions to equations when graphical or algebraic methods become cumbersome. It allows you to systematically plug in values for one variable and compute the outcome for another, making it easier to spot a solution.In our exercise, we used this approach by evaluating both \(y_1 = \frac{-3x}{2} + 5\) and \(y_2 = \frac{-2x}{3} + \frac{10}{3}\) for different \(x\) values. Constructing a table lets you quickly compare outcomes:

• See which \(x, y\) pairs satisfy both equations.
• Visually compare the two different \(y\) values for each \(x\).
• Easily identify the point that satisfies both equations, \((x, y) = (3, 0.5)\) in this exercise.

The table of values method provides a clear, step-by-step view and is a wonderful fallback when other methods seem difficult to apply.