Problem 24 - Precalculus with Limits

Question

Solving a System by Substitution In Exercises \(15-24\) , solve the system by the method of substitution. $$\left\\{\begin{array}{r}{-\frac{2}{3} x+y=2} \\ {2 x-3 y=6}\end{array}\right.$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x=3\), and \(y=4\).

Step 1: Rearrange the first equation for y

Rewrite the equation \(-\frac{2}{3} x + y = 2\) as \(y = \frac{2}{3}x + 2\).

Step 2: Substitute y in the second equation

Replace \(y\) of second equation \(2x - 3y = 6\) with \(\frac{2}{3}x + 2\). This gives: \(2x - 3(\frac{2}{3}x + 2) = 6\).

Step 3: Solve for x

Now, simply calculate \(x\) from previously obtained equation. This might need several calculation steps like distributing and combining like terms. The solution is \(x=3\).

Step 4: Substitute x in the rearranged first equation

Replace \(x\) in equation \(y = \frac{2}{3}x + 2\) with \(3\) to find the value of \(y\). This gives: \(y = \frac{2}{3}*3 + 2\).

Step 5: Solve for y

Now, calculate y-value from the equation. The solution is \(y=4\).

Key Concepts

Substitution Method

The substitution method is a valuable technique for solving systems of equations. It involves substituting one equation into another to eliminate a variable, making it easier to solve. Here's how it works:

1. **Solve one equation for one variable**: First, you should choose one of the equations and solve it for one of the variables. In our exercise, we rearranged the first equation to solve for \(y\), resulting in \(y = \frac{2}{3}x + 2\).

2. **Substitute the expression**: Next, we take the expression we found for \(y\) and substitute it into the other equation. This substitution helps eliminate a variable, simplifying the problem to one equation with one variable.

3. **Solve the resulting equation**: Finally, solve the simplified equation by using basic algebra to find the value of the remaining variable.

Substitution is especially useful for solving systems that are already set up for straightforward rearrangement, like those with a coefficient of 1 for one of the variables.

Solving Linear Equations

Solving linear equations is a foundational skill in algebra. When we talk about solving linear equations, we mean finding the value of the variable that makes the equation true.

This involves several steps:

• **Simplify the equation**: Combine like terms on each side as needed.
• **Isolate the variable**: Use inverse operations such as addition, subtraction, multiplication, or division to get the variable by itself. This might involve moving terms across the equation by using opposite operations.

In our example, once the substitution was made, the equation \(2x - 3(\frac{2}{3}x + 2) = 6\) was simplified and solved for \(x\). After finding \(x = 3\), we substituted back to find \(y = 4\).

Always remember to check your solutions by substituting back into the original equations to ensure both equations are satisfied.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to solve for variables. It's like a toolkit of methods used throughout algebra to work with equations effectively.

Key techniques of algebraic manipulation include:

• **Rearranging equations**: As seen in our exercise, we rearranged \(-\frac{2}{3}x + y = 2\) to solve for \(y\) as a function of \(x\).
• **Distribution**: This is used when dealing with parentheses, such as in the step \(2x - 3(\frac{2}{3}x + 2) = 6\), distributing the \(-3\) across terms within the parentheses.
• **Combining like terms**: Simplifies expressions by adding or subtracting similar terms.
• **Using inverse operations**: These help in isolating variables, like dividing both sides by a coefficient to solve for a variable directly.

Algebraic manipulation is crucial, as it provides the methods needed to simplify complex expressions into solvable equations. Mastery of these tools makes solving systems of equations, such as through substitution, a manageable task.