Problem 18 - Algebra 1: Concepts and Skills

Question

Use the substitution method to solve the linear system. $$ \begin{aligned} &2 c-d=-2\\\ &4 c+d=20 \end{aligned} $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(c = 3\) and \(d = 8\).

Step 1: Isolate a variable

Choose one of the equations to begin. A good place to start is the first equation. Solve this equation for one variable in terms of the other variable. Here, let's solve for \(d\) in terms of \(c\) from the first equation: \(d = 2c + 2\)

Step 2: Perform substitution

Substitute \(2c + 2\) for \(d\) in the second equation, resulting in an equation with one variable: \(4c + (2c + 2) = 20\)

Step 3: Solve the equation

Solve the substituted equation for the remaining variable \(c\). Doing so gives \(6c = 18\), therefore, \(c = 18/6 = 3\).

Step 4: Solve for the other variable

Substitute \(c = 3\) back into the first equation \(2c - d = -2\) to solve for \(d\). That gives \(d = 2 * 3 + 2 = 8\).

Key Concepts

Linear Equations

Linear equations are equations of the first degree, meaning their highest power of any variable is one. In the context of solving systems of equations, linear equations typically involve two or more variables that can be plotted as straight lines on a graph. Every solution to a linear equation represents a point on this line.
Linear equations appear in many forms, but their simplest form is the standard linear equation format: \( ax + by = c \). Here, \( a \) and \( b \) are coefficients, \( x \) and \( y \) are variables, and \( c \) is a constant.
In more complex applications, linear equations can be used to model real-world situations, such as budgeting, calculating distances, or as building blocks in more elaborate mathematical concepts.

Solving Systems of Equations

Solving systems of equations involves finding values for the variables that satisfy each equation in the system simultaneously. The substitution method is a systematic way to tackle such problems and is particularly useful when you have simple linear equations at hand.

• First, solve one of the equations for one variable in terms of the other variable(s). For example, the equation can be manipulated to express \( d \) in terms of \( c \): \( d = 2c + 2 \).
• Substitute this expression into the other equation to reduce the number of variables. And solve for the remaining variable, like \( c \).
• Use the value of this variable to find the other variable from either of the original equations.

This method is systematic, ensuring that your calculations stay organized and manageable, even when dealing with more complex systems.

Algebraic Manipulation

Algebraic manipulation refers to the process of rearranging and simplifying algebraic expressions to solve equations. In the context of the substitution method, algebraic manipulation is crucial.
Let's break it down:

• Rearranging equations: You'll need to manipulate the original equations to isolate one variable, making it easier to substitute into the other equation.
• Simplifying expressions: After substitution, combine like terms and simplify the resulting equation to find the value of the unknown variable. For example, \( 4c + (2c + 2) \) simplifies to \( 6c + 2 \).
• Solving for variables: Once the equation is simplified, solve for the remaining variable by performing basic arithmetic operations, such as division \( c = \frac{18}{6} \).
• Verifying solutions: After calculating the variables' values, plug them back into the original equations to verify that they satisfy both equations.

Algebraic manipulation is a powerful tool that enables you to work through complex equations and systems efficiently and accurately.