Problem 14 - Understanding Elementary Algebra with Geometry

Question

$$\text { In Exercises } 1-14, \text { solve the system of equations using the elimination method.}$$ $$\left\\{\begin{array}{l} 14 w+9 z=6 \\ 21 w+15 z=11 \end{array}\right.$$

Step-by-Step Solution

Answer

w = \(\frac{-3}{7}\), z = \(\frac{4}{3}\)

Step 1: Multiply both equations

To use the elimination method, align the coefficients of either variable. Start by multiplying the first equation by 3 and the second equation by 2 to make the coefficients of \(w\) the same. Equation 1: \(3(14w + 9z) = 3(6)\) Equation 2: \(2(21w + 15z) = 2(11)\) This yields: \(42w + 27z = 18\) and \(42w + 30z = 22\)

Step 2: Subtract the second equation from the first

Subtract the second altered equation from the first one to eliminate \(w\): \( (42w + 27z) - (42w + 30z) = 18 - 22\) This simplifies to: \(-3z = -4\)

Step 3: Solve for \(z\)

Divide both sides of the equation by -3 to find \(z\): \(z = \frac{-4}{-3} = \frac{4}{3}\)

Step 4: Substitute \(z\) back into one of the original equations

Substitute \(z = \frac{4}{3}\) back into one of the original equations to find \(w\). Using the first original equation: \(14w + 9 \left(\frac{4}{3}\right) = 6\) Simplify this to: \(14w + 12 = 6\)

Step 5: Solve for \(w\)

Isolate \(w\) by subtracting 12 from both sides: \(14w = 6 - 12\) This simplifies to: \(14w = -6\) Divide both sides by 14: \(w = \frac{-6}{14} = \frac{-3}{7}\)

Key Concepts

System of Equations

A system of equations consists of two or more equations that share common variables. The goal is to find values for the variables that satisfy all the equations simultaneously.
Each equation in the system represents a relationship between the variables.
For example, in the provided exercise:
\[ \begin{cases} 14w + 9z = 6 \ 21w + 15z = 11 \end{cases} \] You have two equations involving the variables \(w\) and \(z\).
The solutions to the system will be the \(w\) and \(z\) values that make both equations true at the same time.
Different methods can be used to solve such systems, including the elimination method and the substitution method.

Algebraic Solutions

Finding solutions algebraically involves manipulating the equations to isolate one of the variables.
This often requires different steps including:

• Aligning the coefficients of one of the variables.
• Performing basic arithmetic operations like addition, subtraction, multiplication, or division.
• Substituting found values back into the equations to find the remaining variables.

In our exercise, we used the elimination method:

• First, we multiplied each equation to make the coefficients of \(w\) the same.
• Then, we subtracted one equation from the other to eliminate \(w\).
• This helped us find the value of \(z\).
• Finally, we substituted \(z\) back into one of the original equations to solve for \(w\).

Algebraic solutions require careful steps to ensure that you're performing the right operations to simplify the system correctly.

Substitution Method

The substitution method is another technique to solve a system of equations. This method involves:

• Isolating one of the variables in one of the equations.
• Substituting that expression into the other equation(s).
• Solving for the other variable(s).

Let's apply the substitution method to our exercise:
1. From the first equation: \(14w + 9z = 6\), let's solve for \(w\):
\[ w = \frac{6 - 9z}{14} \]
2. Substitute this expression into the second equation: \(21w + 15z = 11\)
\[ 21\frac{6 - 9z}{14} + 15z = 11 \]
3. Simplify and solve for \(z\):
\[ \frac{126 - 189z}{14} + 15z = 11 \]
\[ 126 - 189z + 210z = 154 \]
\[ 21z = 28 \]
\[ z = \frac{28}{21} = \frac{4}{3} \]
4. Substitute \(z = \frac{4}{3}\) back into \(w = \frac{6 - 9z}{14}\):
\[ w = \frac{6 - 9(\frac{4}{3})}{14} = \frac{6 - 12}{14} = \frac{-6}{14} = \frac{-3}{7} \]
So we find the same values for \(w\) and \(z\) as with the elimination method. Each method has its own situations where it is more convenient to use.