Problem 20 - Algebra A Combined Function

Question

Solve each system of equations by the substitution method. $$ \left\\{\begin{array}{l} 3 x+y=-14 \\ 4 x+3 y=-22 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \((-4, -2)\).

Step 1: Solve for y in terms of x

Start with the first equation: \( 3x + y = -14 \). We want to express \( y \) in terms of \( x \). Move \( 3x \) to the other side of the equation by subtracting it from both sides: \( y = -14 - 3x \).

Step 2: Substitute expression for y into the second equation

Take the expression for \( y \) from Step 1 and substitute it into the second equation: \( 4x + 3(-14 - 3x) = -22 \). This gives us: \[ 4x + 3(-14 - 3x) = -22. \]

Step 3: Simplify and solve for x

Distribute the \( 3 \) in the equation from Step 2: \( 4x - 42 - 9x = -22 \). Simplify by combining like terms: \( -5x - 42 = -22 \). Add \( 42 \) to both sides to isolate the term with \( x \): \( -5x = 20 \). Divide both sides by \( -5 \) to find \( x \): \( x = -4 \).

Step 4: Substitute x value back to find y

Use the value of \( x \) found in Step 3 and plug it back into the expression for \( y \) from Step 1. Substitute \( x = -4 \) into \( y = -14 - 3x \): \[ y = -14 - 3(-4). \] Calculate \( y \): \( y = -14 + 12 = -2 \).

Step 5: Write the solution for the system

Now we combine the values of \( x \) and \( y \) found in Steps 3 and 4. The solution to the system of equations is \( x = -4 \) and \( y = -2 \), so the solution is \((-4, -2)\).

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. In the context of linear equations, which are equations of the form \( ax + by = c \), a system can consist of multiple linear equations that share the same variables, like \( x \) and \( y \). The main goal when working with systems of equations is to find the values of the variables that satisfy all equations simultaneously. In the exercise provided, we encounter a system of two equations:

• \( 3x + y = -14 \)
• \( 4x + 3y = -22 \)

To solve this system, we need a method that helps us find the pair \((x, y)\) that fulfills both equations. There are several methods to solve such systems, including substitution, elimination, and graphing. In our case, we use the substitution method, which is particularly effective when one equation is easily solved for a variable, allowing substitution into the other equation.

Solving Linear Equations

Solving linear equations involves finding the value of variables that make the equation true. When dealing with a single linear equation like \( ax + b = c \), the process usually involves isolating the variable on one side of the equation through algebraic techniques, such as addition, subtraction, multiplication, or division.For example, in our system, we start with the equation \( 3x + y = -14 \) and aim to express \( y \) in terms of \( x \) by performing algebraic steps:

• Subtract \( 3x \) from both sides to obtain \( y = -14 - 3x \).

This step is crucial as it sets up a substitution we can use later. Using this expression, we substitute it into the second equation \( 4x + 3y = -22 \), replacing \( y \) to create a new equation in one variable \( x \), which simplifies the system and makes finding the solution more straightforward.

Algebraic Manipulation

Algebraic manipulation refers to the process of rearranging and simplifying equations to solve for unknown variables. This involves the use of basic arithmetic operations such as addition, subtraction, multiplication, and division to both sides of an equation.In our exercise, algebraic manipulation plays a key role in both setting up the substitution and solving the resulting equations. After substituting \( y = -14 - 3x \) into the second equation, we perform the following steps:

• Distribute the \( 3 \) across the terms inside the parentheses to get \( 4x - 42 - 9x = -22 \).
• Combine like terms, resulting in \( -5x - 42 = -22 \).
• Add \( 42 \) to both sides to isolate terms with \( x \), giving us \( -5x = 20 \).
• Divide by \( -5 \) to solve for \( x \), obtaining \( x = -4 \).

After finding \( x \), we substitute it back into \( y = -14 - 3x \) to determine the value of \( y \). This application of algebraic manipulation is critical in systematically approaching and solving the system of linear equations.