Problem 74 - Intermediate Algebra

Question

Solve each system. $$ \left\\{\begin{array}{l} a+c+2 d=-4 \\ b-2 c=1 \\ a+2 b-c=-2 \\ 2 a+b+3 c-2 d=-4 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( a = -1 \), \( b = -5 \), \( c = -3 \), \( d = 0 \).

Step 1: Label the Equations

Label each equation in the system for easy reference: \( (1) \) \( a + c + 2d = -4 \), \( (2) \) \( b - 2c = 1 \), \( (3) \) \( a + 2b - c = -2 \), \( (4) \) \( 2a + b + 3c - 2d = -4 \).

Step 2: Solve for b from Equation (2)

Using equation (2), which is \( b - 2c = 1 \), solve for \( b \): \( b = 2c + 1 \).

Step 3: Substitute b in Equation (3)

Substitute \( b = 2c + 1 \) from Step 2 into equation (3), which is \( a + 2b - c = -2 \): - Start with \( a + 2(2c + 1) - c = -2 \).- Simplify to \( a + 4c + 2 - c = -2 \), which simplifies further to \( a + 3c = -4 \).

Step 4: Express a in terms of c

From the equation \( a + 3c = -4 \), solve for \( a \): \( a = -3c - 4 \).

Step 5: Substitute a and b in Equation (4)

Substitute \( a = -3c - 4 \) and \( b = 2c + 1 \) into equation (4), \( 2a + b + 3c - 2d = -4 \):- Start with \( 2(-3c - 4) + (2c + 1) + 3c - 2d = -4 \).- Simplify: \( -6c - 8 + 2c + 1 + 3c - 2d = -4 \).- Further simplify to \( -1c - 7 - 2d = -4 \).- Rearranging gives \( -c - 2d = 3 \).

Step 6: Solve for c in terms of d

From \( -c - 2d = 3 \), express \( c \) in terms of \( d \):\( c = -2d - 3 \).

Step 7: Substitute c back into Equations (1) and (2)

Substitute \( c = -2d - 3 \) back into: - Equation (1) \( a + c + 2d = -4 \) becomes \( a + (-2d - 3) + 2d = -4 \), simplifying to \( a - 3 = -4 \), so \( a = -1 \).- Check consistency in other equations using these expressions to verify

Step 8: Final Step: Substitute Expressions and Solve for Variables

Substitute \( c = -2d - 3 \), \( a = -1 \), and \( b = 2c + 1 \) to find final values for \( b \) and \( d \):- Using \( c = -2d - 3 \) in \( b = 2c + 1 \): \( b = 2(-2d - 3) + 1 = -4d - 6 + 1 \), resulting in \( b = -4d - 5 \).- Since you have \( c = -2d - 3 \), \( a = -1 \), and \( b = -4d - 5 \), pick any consistent \( d \) to satisfy the system, for example, if \( d = 0 \), then \( c = -3 \), \( a = -1 \), and \( b = -5 \).

Key Concepts

Linear Equations

Linear equations form the backbone of solving systems of equations. These are equations of the first order, meaning they involve variables raised to the power of one. A linear equation generally takes the form \( ax + by + cz + ... = d \), where \( a, b, c,...\) are constants and \( x, y, z,...\) are variables.
Solving a system means finding values for these variables that make all the equations true.
Understanding linear equations helps decipher complex algebraic systems by revealing relationships between different quantities.
When addressing a system of linear equations like the one given in the exercise, it is essential to manipulate these equations to isolate and solve for the variables inside them. This usually involves using other related techniques such as substitution or elimination to systematically reduce the equations to a form where the solution is apparent.

Variables

Variables are symbols that represent numbers or values in mathematical expressions and equations. In the system of equations we are examining, the variables are \( a, b, c, \) and \( d \).
Each of these symbols represents an unknown value that we aim to find.
The beauty of variables is their flexibility; they allow us to express general relationships without specifying exact numbers until the final equation is solved.
Understanding how to manipulate these symbols is crucial for solving complex algebraic expressions. The goal in manipulating a set of linear equations is to isolate each variable, often chosen systematically to simplify the process of finding a unique solution or set of solutions that satisfy all equations.

Substitution Method

The substitution method is a technique used to solve a system of equations by expressing a variable from one equation in terms of another variable. This method starts by solving one of the equations for one variable and then substituting this expression in the other equations.
This process simplifies the system, reducing the number of variables step by step. For example, in the given system, the substitution method helped express \( b \) in terms of \( c \) using equation (2). - Solve for \( b = 2c + 1 \).- Substitute \( b \) into equation (3) to find relationships with other variables.
This method is particularly effective when one of the equations is easily solvable for one of the variables, as seen in the solution process. It allows you to reduce a system of multiple equations into a simpler form eventually leading to a solution.
The substitution method is systematic and detailed, ensuring that each step gets you closer to solving the equation.

Algebraic Expressions

Algebraic expressions appear throughout the process of solving a system of equations. An algebraic expression is a combination of variables, numbers, and operations (such as addition and multiplication). In this exercise, expressions like \( b = 2c + 1 \) or \( a = -3c - 4 \) simplify the parts of the system into forms that are easier to work with.
These expressions are critical in reducing and solving the system, facilitating the smooth transition from one step to the next.

• They allow for simplification, breaking down complex equations into simpler functions of a single variable.
• Enable comparison, where simplified expressions can be compared or checked against each other to ensure consistency and correctness.

Understanding how to manipulate and simplify algebraic expressions is fundamental in solving linear systems efficiently. Once simplified, these expressions lead directly to the determination of the variable values.