Problem 16 - A Graphical Approach to College Algebra

Question

Solve each system by substitution. $$\begin{aligned}&y=4 x-6\\\&2 x+5 y=-8\end{aligned}$$

Step-by-Step Solution

Answer

\((x, y) = (1, -2)\)

Step 1: Identify the Equations

We are given the following system of equations: 1. \( y = 4x - 6 \) 2. \( 2x + 5y = -8 \)

Step 2: Substitute y in the Second Equation

Substitute \( y = 4x - 6 \) into the second equation, \( 2x + 5y = -8 \). This gives us:\[ 2x + 5(4x - 6) = -8 \]

Step 3: Simplify the Equation

Distribute the 5 in the equation:\[ 2x + 20x - 30 = -8 \]Combine like terms:\[ 22x - 30 = -8 \]

Step 4: Solve for x

Add 30 to both sides of the equation:\[ 22x = 22 \]Now divide both sides by 22:\[ x = 1 \]

Step 5: Substitute x back into First Equation

Use \( x = 1 \) and substitute it into \( y = 4x - 6 \):\[ y = 4(1) - 6 \]This simplifies to:\[ y = 4 - 6 = -2 \]

Step 6: State the Solution

The solution to the system of equations is:\((x, y) = (1, -2)\)

Key Concepts

Substitution Method

The substitution method is a reliable technique used to solve systems of equations, especially when one equation is already solved for a variable. In our problem, this method works perfectly because the equation for \( y \) is already isolated. Here’s how the substitution method simplifies the process:

• Take the equation where a variable is isolated. For instance, we have \( y = 4x - 6 \).
• Substitute this expression into the other equation. This allows you to focus on a single variable at a time, making it easier to solve.
• After substitution, simplify the equation as much as possible, solve for the variable, and then backtrack to find other unknowns.

Using substitution can greatly reduce the complexity of solving systems of equations, especially when variables appear straightforward. It transforms a system of equations into a single equation with one variable, making solutions more accessible.

Linear Equations

Linear equations represent the simplest form of algebraic equations, involving no exponents or nonlinear terms. In its essence, a linear equation depicts a straight line when graphed. These equations can take many forms, but they usually look like \( ax + by = c \). For our exercise, the equations are:

• \( y = 4x - 6 \) which is already rearranged in slope-intercept form \( y = mx + b \), making it very easy for substitution.
• \( 2x + 5y = -8 \) which is in standard form.

Linear equations are often used to model real-world situations due to their simplicity. They’re easy to manipulate and are the building blocks for more complex algebra. Understanding and handling these equations with methods like substitution is essential for solving mathematical and real-life problems alike.

Algebraic Solutions

Algebraic solutions involve finding the values of unknown variables that satisfy the given equations. In our system, we started by substituting the value of \( y \) from one equation into another. By solving linear equations algebraically, we transition through the following steps:

• Identify and isolate one of the variables using substitution, solving the transformed equation to find one variable.
• Once identified, use this value to find the associated value of the other variable through back-substitution.
• Finally, verify the solution by plugging the values back into the original equations to ensure they hold true.

Using algebraic solutions not only provides the exact values required but also reinforces understanding of the logical steps involved. The answers we arrive at, like \((x, y) = (1, -2)\), give a clear and comprehensive conclusion to whether the solution satisfies both equations equally.