Problem 33 - Intermediate Algebra

Question

Solve each system using substitution. $$ \left\\{\begin{array}{l} x-8 z=-30 \\ 3 x+y-4 z=5 \\ y+7 z=30 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = 10 \), \( y = -5 \), and \( z = 5 \).

Step 1 - Solve for x in the First Equation

The first equation is \( x - 8z = -30 \). Solve for \( x \) by adding \( 8z \) to both sides to get: \[ x = 8z - 30 \].

Step 2 - Express y in Terms of z from the Third Equation

The third equation is \( y + 7z = 30 \). Solve for \( y \) by subtracting \( 7z \) from both sides to get: \[ y = 30 - 7z \].

Step 3 - Substitute x and y into the Second Equation

Substitute the expressions for \( x \) and \( y \) from Steps 1 and 2 into the second equation, \( 3x + y - 4z = 5 \). This gives: \[ 3(8z - 30) + (30 - 7z) - 4z = 5 \].

Step 4 - Simplify the Equation and Solve for z

Simplify the equation obtained in Step 3: \[ 24z - 90 + 30 - 7z - 4z = 5 \]. Combine like terms: \[ 13z - 60 = 5 \]. Add 60 to both sides: \[ 13z = 65 \]. Finally, divide by 13 to find \( z \): \[ z = 5 \].

Step 5 - Find x Using the Value of z

Substitute \( z = 5 \) back into the expression for \( x \) from Step 1: \[ x = 8(5) - 30 \]. Calculate: \[ x = 40 - 30 \] and \( x = 10 \).

Step 6 - Find y Using the Value of z

Substitute \( z = 5 \) back into the expression for \( y \) from Step 2: \[ y = 30 - 7(5) \]. Calculate: \[ y = 30 - 35 \] and \( y = -5 \).

Key Concepts

Substitution Method

The substitution method is a way to solve systems of equations by replacing one variable with an expression derived from another equation in the system. In simple terms, you solve one of the equations for one of the variables, and then substitute this expression into the other equations. By doing this, you can reduce the number of equations and variables, making it easier to find the solution.
To understand the substitution method better, follow these general steps:

• Identify an equation that can be easily solved for one variable.
• Solve for that variable, expressing it in terms of another variable.
• Substitute your expression into the other equations.
• Solve the resulting equation for another variable.
• Back-substitute to find the remaining variable(s).

In our original exercise, we used the substitution method to express \( x \) and \( y \) in terms of \( z \), and then used these expressions to solve the entire system. This step-wise approach highlights the systematic nature of the substitution method.

Algebra

Algebra is the branch of mathematics that uses symbols and letters to represent numbers and quantities in formulas and equations. It forms the foundation for learning advanced topics in mathematics and is crucial for solving equations and systems.

• In algebra, equations can represent relationships between different variables.
• Solving algebraic equations often involves manipulating these equations by adding, subtracting, multiplying, or dividing both sides by the same amount to isolate the variable representing the unknown quantity.

In the exercise provided, understanding algebra allowed us to solve each equation for different variables, manipulate expressions, and substitute them to find the solution to the system. Algebraic rules and techniques simplify complex problems and enable us to solve equations that model real-world situations.

Linear Equations

Linear equations are mathematical statements that describe a straight line when graphed on a coordinate plane. These equations involve variables raised to the first power and have no variables in the denominator or inside functions like square roots or exponents.
Linear equations can be represented in the standard form \( ax + by + cz = d \), or in forms that make solving and interpreting solutions easier. The primary characteristics of linear equations include:

• They involve two or more variables.
• The graph of these equations is always a straight line.
• Their solutions can be found using methods like substitution, elimination, or graphing.

In the given exercise, we dealt with a system of linear equations. By using the substitution method, we expressed variables in terms of others, which simplified the process of finding an exact solution for all three variables in the system.