Solving systems of linear equations is a fundamental concept in algebra with broad applications in various fields, from physics and engineering to economics and computer science. One of the most common methods for solving these systems is the substitution method. This guide will walk you through the process, providing clear explanations, examples, and, of course, an answer key to practice problems.
Understanding Linear Systems and the Substitution Method
A system of linear equations consists of two or more linear equations with the same variables. A solution to the system is a set of values for the variables that satisfy all equations simultaneously. The substitution method involves solving one equation for one variable and then substituting that expression into the other equation to solve for the remaining variable.
Steps for Solving Linear Systems by Substitution:
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Solve one equation for one variable: Choose one equation and solve it for one of its variables. Select the equation and variable that will make the process easiest (often, this involves selecting a variable with a coefficient of 1).
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Substitute: Substitute the expression you found in step 1 into the other equation. This will create a new equation with only one variable.
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Solve for the remaining variable: Solve the new equation from step 2 for the remaining variable.
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Substitute back: Substitute the value you found in step 3 back into either of the original equations to solve for the other variable.
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Check your solution: Substitute both values back into both original equations to verify they satisfy both.
Examples with Detailed Explanations
Let's work through a few examples to solidify your understanding.
Example 1:
Solve the system:
x + y = 5 x - y = 1
Solution:
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Solve for one variable: Let's solve the first equation for x: x = 5 - y
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Substitute: Substitute this expression for x into the second equation: (5 - y) - y = 1
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Solve for the remaining variable: Simplify and solve for y: 5 - 2y = 1 => -2y = -4 => y = 2
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Substitute back: Substitute y = 2 into either original equation. Let's use the first: x + 2 = 5 => x = 3
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Check: x + y = 3 + 2 = 5 (Correct) ; x - y = 3 - 2 = 1 (Correct)
Example 2:
Solve the system:
2x + y = 7 x - 3y = -10
Solution:
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Solve for one variable: Let's solve the second equation for x: x = 3y - 10
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Substitute: Substitute this into the first equation: 2(3y - 10) + y = 7
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Solve for the remaining variable: Simplify and solve for y: 6y - 20 + y = 7 => 7y = 27 => y = 27/7
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Substitute back: Substitute y = 27/7 into x = 3y - 10: x = 3(27/7) - 10 = 81/7 - 70/7 = 11/7
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Check: (Substitute these fractional values back into both original equations to verify—this step is crucial to ensure accuracy).
Practice Problems with Answer Key
Now it's your turn! Try solving these systems using the substitution method.
Problem 1:
x + 2y = 4 x - y = 1
Problem 2:
3x + y = 10 x - 2y = -5
Problem 3:
2x + 3y = 1 x - y = 2
Answer Key:
Problem 1: x = 2, y = 1
Problem 2: x = 5/7, y = 55/7
Problem 3: x = 7/5, y = -3/5
Conclusion
Mastering the substitution method for solving linear systems is a vital skill in algebra. By carefully following the steps outlined above and practicing with various examples, you can confidently solve a wide range of linear systems. Remember to always check your solutions to ensure accuracy. This method provides a systematic approach that will be invaluable in more advanced mathematical applications.
