Problem 5 - A Graphical Approach to Precalculus with Limits

Question

Solve each system by substitution $$\begin{aligned}6 x-y &=5 \\\y &=x\end{aligned}$$.

Step-by-Step Solution

Answer

The solution is \( x = 1 \) and \( y = 1 \).

Step 1: Identify the Equations

The system of equations given is: \( 6x - y = 5 \) and \( y = x \). Our goal is to solve the system using substitution.

Step 2: Substitute for y

Since \( y = x \), substitute \( x \) for \( y \) in the first equation. This gives us: \( 6x - x = 5 \).

Step 3: Simplify the Equation

Simplify the equation from the substitution: \( 6x - x = 5 \) becomes \( 5x = 5 \).

Step 4: Solve for x

Solve the equation \( 5x = 5 \) by dividing both sides by 5, which gives \( x = 1 \).

Step 5: Substitute x back into y = x

Since we have found \( x = 1 \) and we know \( y = x \), substitute \( x = 1 \) back into \( y = x \) to find \( y = 1 \).

Step 6: Solution of the System

The solution of the system is \( x = 1 \) and \( y = 1 \).

Key Concepts

Substitution Method

The substitution method is a strategic approach used to solve systems of equations where you solve one equation for one variable and then substitute that expression into the other equation. It is particularly useful when one of the equations is already simplified in terms of one variable. For example, if you are given a system of equations like:

• \( 6x - y = 5 \)
• \( y = x \)

Since the second equation is solved for \( y \), it gives us an immediate substitution opportunity. This means we can directly plug \( y = x \) into the first equation without further manipulation. The goal here is to reduce the system to a single equation with one unknown variable, which is easier to solve. This simplicity makes the substitution method a powerful tool, especially when dealing with linear equations. The key is to make sure that the variable is isolated before substitution.

Solving Linear Equations

Solving linear equations is a fundamental skill in algebra that involves finding values for the variables that make the equation true. In the context we are discussing, after substitution, you will end up with a linear equation: \( 6x - x = 5 \).

Linear equations generally take the form \( ax + b = c \). The variable involved is usually raised to the power of one, creating a straight line when graphed. To solve such equations, the emphasis is on isolating the variable.

• Combine like terms, which simply means collecting all terms involving the same variables or constants.
• In our example, after combining terms, the equation becomes \( 5x = 5 \).
• To isolate \( x \), you'll divide both sides by the coefficient of \( x \), which is 5 in this case, resulting in \( x = 1 \).

Remember, the goal of solving a linear equation is to have the variable alone on one side of the equation. This allows you to state the solution directly. Always perform consistent operations on both sides to maintain equality.

Algebraic Manipulation

Algebraic manipulation involves a collection of techniques used to transform one algebraic expression into another equivalent expression. This is crucial when solving equations because it helps simplify equations and isolate variables. Let's delve deeper into some key techniques used in our exercise scenario:

• **Substitution:** Replacing a variable with an equivalent expression to reduce the complexity of the system.
• **Combining Like Terms:** When we substitute \( y = x \) into \( 6x - y = 5 \), we're combining the terms involving \( x \) by realizing \( 6x - x \) simplifies to \( 5x \).
• **Isolating Variables:** We move terms across the equation to get \( x \) by itself, such as dividing \( 5x = 5 \) by 5.

These steps often involve operations such as addition, subtraction, multiplication, or division, and they rely on properties of equalities to maintain the balance of the equations. Mastering algebraic manipulation is essential for mathematical problem-solving and for tackling more complex equations confidently.