College Math: Algebra Linear-Systems - Solving Systems by Elimination Choosing the Best Method

Solving Systems by Elimination – Choosing the Best Method

What Is This?

Solving systems by elimination is a method used to find the solution to a system of linear equations by eliminating one of the variables. This technique involves adding or subtracting equations to eliminate one of the variables, making it easier to solve for the other variable.

Why It Matters

Solving systems by elimination is a fundamental concept in mathematics and has numerous real-world applications, such as: * In economics, it can be used to determine the optimal price and quantity of a product in a market equilibrium. * In engineering, it can be used to design and optimize systems, such as electrical circuits and mechanical systems. * In computer science, it can be used to solve systems of linear equations that arise in machine learning and data analysis.

Core Concepts

• Linear Equations: A linear equation is an equation in which the highest power of the variable(s) is 1. For example, 2x + 3y = 5 is a linear equation.
• System of Linear Equations: A system of linear equations is a set of two or more linear equations that are solved simultaneously. For example, {2x + 3y = 5, x - 2y = -3} is a system of linear equations.
• Elimination Method: The elimination method is a technique used to solve systems of linear equations by eliminating one of the variables.

Step-by-Step: How to Approach Problems

Step 1: Identify the System of Linear Equations

Identify the system of linear equations that needs to be solved.

Step 2: Choose the Method

Choose the elimination method as the method to solve the system of linear equations.

Step 3: Multiply the Equations

Multiply the equations by necessary multiples such that the coefficients of one of the variables (either x or y) are the same in both equations.

Step 4: Add or Subtract the Equations

Add or subtract the equations to eliminate one of the variables.

Step 5: Solve for the Variable

Solve for the variable that is not eliminated.

Step 6: Back-Substitution

Back-substitute the value of the variable into one of the original equations to solve for the other variable.

Solved Examples

Example 1: Solving a System of Linear Equations using Elimination

{2x + 3y = 5, x - 2y = -3}

Problem Statement: Solve the system of linear equations using the elimination method.

Solution:

$$ \begin{align} 2x + 3y &= 5 \ x - 2y &= -3 \end{align} $$

Multiply the second equation by 2:

$$ \begin{align} 2x + 3y &= 5 \ 2x - 4y &= -6 \end{align} $$

Add the two equations:

$$ \begin{align} 7y &= -1 \ y &= -\frac{1}{7} \end{align} $$

Back-substitute the value of y into the first equation:

$$ \begin{align} 2x + 3\left(-\frac{1}{7}\right) &= 5 \ 2x &= \frac{20}{7} \ x &= \frac{10}{7} \end{align} $$

Answer: The solution to the system of linear equations is (x, y) = ($\frac{10}{7}$, -$\frac{1}{7}$).

Example 2: Solving a System of Linear Equations using Elimination with Fractions

{4x + 2y = 6, 2x + y = 2}

Problem Statement: Solve the system of linear equations using the elimination method.

Solution:

$$ \begin{align} 4x + 2y &= 6 \ 2x + y &= 2 \end{align} $$

Multiply the second equation by 2:

$$ \begin{align} 4x + 2y &= 6 \ 4x + 2y &= 4 \end{align} $$

Subtract the two equations:

$$ \begin{align} 0 &= -2 \ y &= 0 \end{align} $$

Back-substitute the value of y into the first equation:

$$ \begin{align} 4x + 2(0) &= 6 \ 4x &= 6 \ x &= \frac{3}{2} \end{align} $$

Answer: The solution to the system of linear equations is (x, y) = ($\frac{3}{2}$, 0).

Example 3: Solving a System of Linear Equations using Elimination with Multiple Variables

{2x + 3y - z = 4, x + 2y + 2z = 5, 3x - y + z = 2}

Problem Statement: Solve the system of linear equations using the elimination method.

Solution:

$$ \begin{align} 2x + 3y - z &= 4 \ x + 2y + 2z &= 5 \ 3x - y + z &= 2 \end{align} $$

Multiply the first equation by 3 and the second equation by 2:

$$ \begin{align} 6x + 9y - 3z &= 12 \ 2x + 4y + 4z &= 10 \end{align} $$

Add the two equations:

$$ \begin{align} 8x + 13y &= 22 \ y &= \frac{22 - 8x}{13} \end{align} $$

Back-substitute the value of y into the second equation:

$$ \begin{align} x + 2\left(\frac{22 - 8x}{13}\right) + 2z &= 5 \ x + \frac{44 - 16x}{13} + 2z &= 5 \ 13x + 44 - 16x + 26z &= 65 \ -3x + 26z &= 21 \end{align} $$

Back-substitute the value of y into the third equation:

$$ \begin{align} 3x - \left(\frac{22 - 8x}{13}\right) + z &= 2 \ 39x - 22 + 8x + 13z &= 26 \ 47x + 13z &= 48 \end{align} $$

Multiply the equation -3x + 26z = 21 by 13:

$$ \begin{align} -39x + 338z &= 273 \end{align} $$

Add the two equations:

$$ \begin{align} -42x + 351z &= 321 \end{align} $$

Back-substitute the value of z into one of the original equations:

$$ \begin{align} 2x + 3y - z &= 4 \ 2x + 3\left(\frac{22 - 8x}{13}\right) - \left(\frac{321 + 42x}{351}\right) &= 4 \end{align} $$

Solve for x:

$$ \begin{align} 2x + \frac{66 - 24x}{13} - \frac{321 + 42x}{351} &= 4 \ 26x + 66 - 24x - 321 - 42x &= 156 \ -80x &= 431 \ x &= -\frac{431}{80} \end{align} $$

Back-substitute the value of x into one of the original equations:

$$ \begin{align} 2x + 3y - z &= 4 \ 2\left(-\frac{431}{80}\right) + 3y - z &= 4 \ -\frac{862}{40} + 3y - z &= 4 \ 3y - z &= \frac{862}{40} + 4 \ 3y - z &= \frac{862 + 160}{40} \ 3y - z &= \frac{1022}{40} \ z &= 3y - \frac{1022}{40} \end{align} $$

Back-substitute the value of x into one of the original equations:

$$ \begin{align} x + 2y + 2z &= 5 \ -\frac{431}{80} + 2y + 2\left(3y - \frac{1022}{40}\right) &= 5 \ -\frac{431}{80} + 2y + 6y - \frac{2044}{40} &= 5 \ -8y &= \frac{431}{80} + \frac{2044}{40} - 5 \ -8y &= \frac{431 + 4080 - 400}{80} \ -8y &= \frac{3711}{80} \ y &= -\frac{3711}{640} \end{align} $$

Back-substitute the value of y into one of the original equations:

$$ \begin{align} 3x - y + z &= 2 \ 3\left(-\frac{431}{80}\right) - \left(-\frac{3711}{640}\right) + \left(3\left(-\frac{3711}{640}\right) - \frac{1022}{40}\right) &= 2 \ -\frac{1293}{80} + \frac{3711}{640} - \frac{11133}{640} + \frac{1022}{40} &= 2 \ -\frac{1293}{80} + \frac{3711 - 11133 + 40880}{640} &= 2 \ -\frac{1293}{80} + \frac{30758}{640} &= 2 \ -\frac{1293}{80} + \frac{30758}{640} &= \frac{1280}{640} \ \frac{30758 - 51840}{640} &= \frac{1280}{640} \ \frac{-21082}{640} &= \frac{1280}{640} \ -21082 &= 1280 \ z &= \frac{1022}{40} - 3\left(-\frac{3711}{640}\right) \ z &= \frac{1022}{40} + \frac{11133}{640} \ z &= \frac{1022 \cdot 16}{40 \cdot 16} + \frac{11133}{640} \ z &= \frac{16352 + 11133}{640} \ z &= \frac{27485}{640} \end{align} $$

Answer: The solution to the system of linear equations is (x, y, z) = ($-\frac{431}{80}$, $-\frac{3711}{640}$, $\frac{27485}{640}$).

Common Pitfalls & Mistakes

• Not multiplying the equations by necessary multiples to eliminate one of the variables.
• Not adding or subtracting the equations correctly to eliminate one of the variables.
• Not solving for the variable that is not eliminated.
• Not back-substituting the value of the eliminated variable into one of the original equations to solve for the other variable.

Best Practices & Study Tips

• Check your work by plugging the solution back into the original equations.
• Use a table to keep track of the coefficients and constants.
• Use a calculator to check your work.
• Practice, practice, practice!

Tools & Software

• Graphing calculators (TI-84, Desmos)
• Statistical software (R, Python libraries like NumPy/SciPy, Excel)
• Symbolic math tools (Wolfram Alpha, Symbolab)

Real-World Use Cases

• In economics, it can be used to determine the optimal price and quantity of a product in a market equilibrium.
• In engineering, it can be used to design and optimize systems, such as electrical circuits and mechanical systems.
• In computer science, it can be used to solve systems of linear equations that arise in machine learning and data analysis.

Check Your Understanding (MCQs)

Question 1

What is the elimination method used to solve systems of linear equations?

A) By adding or subtracting the equations to eliminate one of the variables. B) By multiplying the equations by necessary multiples to eliminate one of the variables. C) By solving for one of the variables and substituting it into the other equation. D) By using a calculator to solve the system of linear equations.

Correct Answer: A) By adding or subtracting the equations to eliminate one of the variables.

Explanation: The elimination method is used to solve systems of linear equations by adding or subtracting the equations to eliminate one of the variables.

Why the Distractors Are Tempting:

• B) Multiplying the equations by necessary multiples is a step in the elimination method, but it is not the entire method.
• C) Solving for one of the variables and substituting it into the other equation is a step in the substitution method, not the elimination method.
• D) Using a calculator to solve the system of linear equations is not a method of solving systems of linear equations.

Question 2

What is the purpose of back-substitution in the elimination method?

A) To solve for the variable that is not eliminated. B) To check the work by plugging the solution back into the original equations. C) To eliminate one of the variables. D) To multiply the equations by necessary multiples.

Correct Answer: A) To solve for the variable that is not eliminated.

Explanation: Back-substitution is used to solve for the variable that is not eliminated in the elimination method.

Why the Distractors Are Tempting:

• B) Checking the work by plugging the solution back into the original equations is a good practice, but it is not the purpose of back-substitution.
• C) Eliminating one of the variables is a step in the elimination method, but it is not the purpose of back-substitution.
• D) Multiplying the equations by necessary multiples is a step in the elimination method, but it is not the purpose of back-substitution.

Question 3

What is the benefit of using the elimination method to solve systems of linear equations?

A) It is faster than the substitution method. B) It is easier than the substitution method. C) It is more accurate than the substitution method. D) It is more efficient than the substitution method.

Correct Answer: D) It is more efficient than the substitution method.

Explanation: The elimination method is more efficient than the substitution method because it eliminates one of the variables, making it easier to solve for the other variable.

Why the Distractors Are Tempting:

• A) The elimination method may be faster than the substitution method, but it is not necessarily faster.
• B) The elimination method may be easier than the substitution method, but it is not necessarily easier.
• C) The elimination method may be more accurate than the substitution method, but it is not necessarily more accurate.