Problem 10 - A Graphical Approach to Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

The solution is \((x, y) = (1, -2)\).

Step 1: Identify substitution

We are given a system of equations: \( y = 4x - 6 \) and \( 2x + 5y = -8 \). Notice the first equation expresses \( y \) in terms of \( x \): \( y = 4x - 6 \). This expression can be substituted into the second equation in place of \( y \).

Step 2: Substitute \( y \) into the second equation

Take the expression for \( y \) from the first equation, \( y = 4x - 6 \), and substitute it into the second equation \( 2x + 5y = -8 \). This gives us: \( 2x + 5(4x - 6) = -8 \).

Step 3: Solve for \( x \)

Expand and solve the equation from Step 2. Start by distributing \( 5 \) into \( 4x - 6 \), giving \( 20x - 30 \). Substitute this back to get \( 2x + 20x - 30 = -8 \). Combine like terms: \( 22x - 30 = -8 \). Add 30 to both sides to obtain \( 22x = 22 \). Divide both sides by 22, resulting in \( x = 1 \).

Step 4: Substitute \( x \) back into the equation for \( y \)

Use the value of \( x = 1 \) in the expression for \( y \) from the first equation: \( y = 4x - 6 \). Substitute \( x = 1 \) to get \( y = 4(1) - 6 = 4 - 6 = -2 \).

Step 5: Verify the solution

Check the solution \((x, y) = (1, -2)\) against the second original equation \( 2x + 5y = -8 \). Substitute to verify correctness: \( 2(1) + 5(-2) = 2 - 10 = -8 \). The left-hand side equals the right-hand side, confirming that the solution satisfies both equations.

Key Concepts

Substitution Method

The substitution method is a popular technique for solving systems of equations. It involves expressing one variable in terms of the other, then substituting this expression into another equation. This approach reduces the number of variables, making it easier to solve the equations.

In the given problem, the first equation is already solved for one variable: \( y = 4x - 6 \). This is a perfect starting point for substitution. It allows us to replace \( y \) in the second equation, \( 2x + 5y = -8 \), with \( 4x - 6 \). This turns the equation into one with only the variable \( x \), simplifying the process of finding its value.

The substitution method is particularly useful when one equation is already solved for one of the variables, or can be easily manipulated to do so.

• Identify the equation where a variable is isolated.
• Substitute this into the other equation.
• Solve the resulting single-variable equation.

By solving for one variable first, it becomes possible to find the value of the other variable efficiently.

Linear Equations

Linear equations are mathematical expressions that describe relationships between variables with constant coefficients. They are called 'linear' because their graph is a straight line. A linear equation in two variables typically takes the form \( ax + by = c \), where \( a \) and \( b \) are coefficients, and \( c \) is a constant.

In systems of linear equations like the ones given, each equation represents a line. The solution to the system is the point where these lines intersect. This characteristic helps visualize why solving systems leads to a solution that satisfies all included equations simultaneously. Once both equations are graphed, the intersection point reveals the values of \( x \) and \( y \) that work for both.

Understanding the nature of linear equations facilitates solving them because:

• They maintain a consistent ratio, making algebraic manipulation predictable.
• Their graphs give a visual insight into solutions.
• They ensure that systems with two equations usually have one unique solution if the lines intersect, are parallel when no solution exists, or coincide when infinite solutions exist.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions to solve equations. It is a crucial skill in solving systems of equations by substitution or any other method.

When substituting an expression into another equation, it's necessary to carefully manage the algebraic structure. In the example, \( 5(4x - 6) \) was expanded to \( 20x - 30 \), which allowed the combination of like terms: \( 2x + 20x \) leading to \( 22x \). These steps demonstrate how algebraic manipulation transforms complex expressions into simpler, solvable forms.

Effective algebraic manipulation encompasses:

• Expanding expressions (e.g., distributing a number across terms within parentheses).
• Combining like terms to consolidate expressions.
• Rearranging equations to isolate variables.
• Using inverse operations like addition and multiplication to both sides of an equation to maintain equality.

Mastering these techniques requires practice but simplifies the process of finding solutions.