1-6 Additional Practice Linear Systems Answer Key

1-6 Additional Practice Linear Systems: Answer Key and Comprehensive Guide

Solving systems of linear equations is a fundamental concept in algebra, with applications spanning various fields like engineering, economics, and computer science. This comprehensive guide provides detailed solutions and explanations for six additional practice problems, focusing on different methods and highlighting common pitfalls. Understanding these examples will solidify your grasp of linear systems and enhance your problem-solving skills.

Problem 1: Solving Using Elimination

Problem: Solve the following system of equations using the elimination method:

3x + 2y = 7 x - 2y = 1

Solution:

The elimination method, also known as the addition method, involves manipulating the equations to eliminate one variable. In this case, notice that the '2y' terms have opposite signs. Adding the two equations directly eliminates 'y':

(3x + 2y) + (x - 2y) = 7 + 1 4x = 8 x = 2

Now, substitute the value of x (x=2) back into either of the original equations to solve for y. Let's use the second equation:

2 - 2y = 1 -2y = -1 y = 1/2

Therefore, the solution to the system is x = 2 and y = 1/2. You can verify this solution by substituting these values back into both original equations.

Key takeaway: The elimination method is particularly efficient when the coefficients of one variable are opposites or can be easily made opposites by multiplying one or both equations by a constant.

Problem 2: Solving Using Substitution

Problem: Solve the following system using the substitution method:

y = 2x - 1 x + y = 5

Solution:

The substitution method involves solving one equation for one variable and substituting that expression into the other equation. The first equation is already solved for y: y = 2x - 1. Substitute this expression for 'y' into the second equation:

x + (2x - 1) = 5 3x - 1 = 5 3x = 6 x = 2

Now, substitute the value of x (x=2) back into either original equation to find y. Using the first equation:

y = 2(2) - 1 y = 3

Therefore, the solution is x = 2 and y = 3. Always check your solution by plugging the values into both original equations.

Problem 3: A System with Infinite Solutions

Problem: Solve the following system:

2x + 4y = 6 x + 2y = 3

Solution:

Let's use the elimination method. Multiply the second equation by -2:

-2(x + 2y) = -2(3) -2x - 4y = -6

Now, add this modified equation to the first equation:

(2x + 4y) + (-2x - 4y) = 6 + (-6) 0 = 0

This result (0 = 0) indicates that the two equations are dependent; they represent the same line. Therefore, there are infinitely many solutions. Any point on the line x + 2y = 3 satisfies both equations.

Problem 4: A System with No Solution

Problem: Solve the following system:

x + y = 4 x + y = 6

Solution:

Let's use the elimination method. Subtract the first equation from the second equation:

(x + y) - (x + y) = 6 - 4 0 = 2

This result (0 = 2) is a contradiction. It means the two equations represent parallel lines that never intersect. Therefore, there is no solution to this system.

Problem 5: Solving a System with Three Variables

Problem: Solve the following system:

x + y + z = 6 2x - y + z = 3 x + 2y - z = 3

Solution:

Solving systems with three variables often requires a systematic approach. We can use elimination or substitution, but elimination is generally preferred.

Step 1: Eliminate one variable from two pairs of equations. Let's eliminate 'z' from the first and second equations, and then from the first and third equations.

Adding the first and second equations gives: 3x + z = 9 (Equation 4) Adding the first and third equations gives: 2x + 3y = 9 (Equation 5)

Step 2: Now we have a system of two equations with two variables (x and y). Solve this system. Let's use elimination again. Multiply Equation 5 by 3 and Equation 4 by -3 to eliminate x:

-9x - 3z = -27 6x + 9y = 27

Adding these gives: 9y - 3z = 0 This simplifies to: 3y - z = 0 (Equation 6)

Now solve Equation 6 and Equation 5. Substitute 'z = 3y' from Equation 6 into equation 5: 2x + 3y = 9

Step 3: Solve for x and y.

2x + 3y = 9 z = 3y

Let's assume y = 1, then z = 3 and 2x + 3(1) = 9, which gives x = 3.

Therefore, one solution is x = 3, y = 1, z = 3. You should check this solution by substituting the values into all three original equations.

Problem 6: A Real-World Application

Problem: A farmer has 100 acres of land to plant corn and soybeans. Corn requires 2 hours of labor per acre and soybeans require 1 hour of labor per acre. The farmer has 150 hours of labor available. If the profit from corn is $50 per acre and the profit from soybeans is $40 per acre, how many acres of each crop should the farmer plant to maximize profit?

Solution:

Let's define:

• x = acres of corn
• y = acres of soybeans

We can set up a system of equations:

• x + y = 100 (Total acres)
• 2x + y = 150 (Total labor hours)

Let's use elimination to solve this system. Subtract the first equation from the second:

(2x + y) - (x + y) = 150 - 100 x = 50

Substitute x = 50 into the first equation:

50 + y = 100 y = 50

So the farmer should plant 50 acres of corn and 50 acres of soybeans.

To calculate the maximum profit:

Profit = 50(50) + 40(50) = $4500

Therefore, the farmer should plant 50 acres of corn and 50 acres of soybeans to maximize profit at $4500.

This example showcases how systems of linear equations can be applied to real-world optimization problems.

Conclusion: Mastering Linear Systems

These six examples demonstrate various techniques for solving systems of linear equations. Remember to always check your solutions by substituting them back into the original equations. Consistent practice and understanding the underlying concepts will build your confidence and proficiency in solving even more complex linear systems. Don't be afraid to experiment with different methods to find the most efficient approach for each problem. Mastering linear systems is a crucial stepping stone in your mathematical journey, opening doors to more advanced topics and applications. Remember to practice regularly and seek help when needed to solidify your understanding.