Equation 1: 6 X + 4 Y = 34 6x + 4y = 34 6 X + 4 Y = 34 Equation 2: 5 X − 2 Y = 15 5x - 2y = 15 5 X − 2 Y = 15 Decide Whether Each ( X , Y (x, Y ( X , Y ] Pair Is A Solution To One Equation, Both Equations, Or Neither Of The Equations:1. ( 3 , 4 (3, 4 ( 3 , 4 ] 2. ( 4 , 2.5 (4, 2.5 ( 4 , 2.5 ] 3. $(5,

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Introduction

In mathematics, a system of linear equations is a set of two or more linear equations that are solved simultaneously to find the values of the variables. In this article, we will focus on solving a system of two linear equations using the substitution method. We will also discuss how to determine whether a given pair of values is a solution to one equation, both equations, or neither of the equations.

The Equations

We are given two linear equations:

  • Equation 1: 6x+4y=346x + 4y = 34
  • Equation 2: 5x−2y=155x - 2y = 15

Our goal is to find the values of xx and yy that satisfy both equations.

Method 1: Substitution Method

To solve the system of equations using the substitution method, we can solve one of the equations for one variable and then substitute that expression into the other equation.

Step 1: Solve Equation 1 for y

We can solve Equation 1 for yy by isolating yy on one side of the equation:

6x+4y=346x + 4y = 34

Subtract 6x6x from both sides:

4y=34−6x4y = 34 - 6x

Divide both sides by 4:

y=34−6x4y = \frac{34 - 6x}{4}

Step 2: Substitute the Expression for y into Equation 2

Now that we have an expression for yy in terms of xx, we can substitute this expression into Equation 2:

5x−2y=155x - 2y = 15

Substitute y=34−6x4y = \frac{34 - 6x}{4} into Equation 2:

5x−2(34−6x4)=155x - 2\left(\frac{34 - 6x}{4}\right) = 15

Simplify the equation:

5x−34−6x2=155x - \frac{34 - 6x}{2} = 15

Multiply both sides by 2 to eliminate the fraction:

10x−34+6x=3010x - 34 + 6x = 30

Combine like terms:

16x−34=3016x - 34 = 30

Add 34 to both sides:

16x=6416x = 64

Divide both sides by 16:

x=4x = 4

Step 3: Find the Value of y

Now that we have the value of xx, we can substitute it into the expression for yy:

y=34−6x4y = \frac{34 - 6x}{4}

Substitute x=4x = 4 into the expression:

y=34−6(4)4y = \frac{34 - 6(4)}{4}

Simplify the expression:

y=34−244y = \frac{34 - 24}{4}

y=104y = \frac{10}{4}

y=2.5y = 2.5

Conclusion

We have found the values of xx and yy that satisfy both equations:

  • x=4x = 4
  • y=2.5y = 2.5

These values satisfy both Equation 1 and Equation 2.

Example 1: (3, 4)

To determine whether the pair (3,4)(3, 4) is a solution to one equation, both equations, or neither of the equations, we can substitute these values into both equations:

  • Equation 1: 6x+4y=346x + 4y = 34 Substitute x=3x = 3 and y=4y = 4 into Equation 1:

    6(3)+4(4)=346(3) + 4(4) = 34

    Simplify the equation:

    18+16=3418 + 16 = 34

    34=3434 = 34

    The pair (3,4)(3, 4) satisfies Equation 1.

  • Equation 2: 5x−2y=155x - 2y = 15 Substitute x=3x = 3 and y=4y = 4 into Equation 2:

    5(3)−2(4)=155(3) - 2(4) = 15

    Simplify the equation:

    15−8=1515 - 8 = 15

    7=157 = 15

    The pair (3,4)(3, 4) does not satisfy Equation 2.

Conclusion

The pair (3,4)(3, 4) is a solution to Equation 1 but not a solution to Equation 2.

Example 2: (4, 2.5)

To determine whether the pair (4,2.5)(4, 2.5) is a solution to one equation, both equations, or neither of the equations, we can substitute these values into both equations:

  • Equation 1: 6x+4y=346x + 4y = 34 Substitute x=4x = 4 and y=2.5y = 2.5 into Equation 1:

    6(4)+4(2.5)=346(4) + 4(2.5) = 34

    Simplify the equation:

    24+10=3424 + 10 = 34

    34=3434 = 34

    The pair (4,2.5)(4, 2.5) satisfies Equation 1.

  • Equation 2: 5x−2y=155x - 2y = 15 Substitute x=4x = 4 and y=2.5y = 2.5 into Equation 2:

    5(4)−2(2.5)=155(4) - 2(2.5) = 15

    Simplify the equation:

    20−5=1520 - 5 = 15

    15=1515 = 15

    The pair (4,2.5)(4, 2.5) satisfies Equation 2.

Conclusion

The pair (4,2.5)(4, 2.5) is a solution to both Equation 1 and Equation 2.

Example 3: (5, 3)

To determine whether the pair (5,3)(5, 3) is a solution to one equation, both equations, or neither of the equations, we can substitute these values into both equations:

  • Equation 1: 6x+4y=346x + 4y = 34 Substitute x=5x = 5 and y=3y = 3 into Equation 1:

    6(5)+4(3)=346(5) + 4(3) = 34

    Simplify the equation:

    30+12=3430 + 12 = 34

    42=3442 = 34

    The pair (5,3)(5, 3) does not satisfy Equation 1.

  • Equation 2: 5x−2y=155x - 2y = 15 Substitute x=5x = 5 and y=3y = 3 into Equation 2:

    5(5)−2(3)=155(5) - 2(3) = 15

    Simplify the equation:

    25−6=1525 - 6 = 15

    19=1519 = 15

    The pair (5,3)(5, 3) does not satisfy Equation 2.

Conclusion

The pair (5,3)(5, 3) is not a solution to either Equation 1 or Equation 2.

Conclusion

Q: What is a system of linear equations?

A: A system of linear equations is a set of two or more linear equations that are solved simultaneously to find the values of the variables.

Q: What is the substitution method?

A: The substitution method is a technique used to solve a system of linear equations by solving one of the equations for one variable and then substituting that expression into the other equation.

Q: How do I determine whether a pair of values is a solution to one equation, both equations, or neither of the equations?

A: To determine whether a pair of values is a solution to one equation, both equations, or neither of the equations, you can substitute the values into both equations and check if the resulting equations are true.

Q: What if the pair of values satisfies one equation but not the other?

A: If the pair of values satisfies one equation but not the other, then it is a solution to the first equation but not a solution to the second equation.

Q: What if the pair of values does not satisfy either equation?

A: If the pair of values does not satisfy either equation, then it is not a solution to either equation.

Q: Can I use the substitution method to solve a system of three or more linear equations?

A: Yes, you can use the substitution method to solve a system of three or more linear equations. However, it may be more complicated and require more steps.

Q: What if I have a system of linear equations with no solution?

A: If you have a system of linear equations with no solution, it means that the equations are inconsistent and there is no value of the variables that can satisfy both equations.

Q: What if I have a system of linear equations with infinitely many solutions?

A: If you have a system of linear equations with infinitely many solutions, it means that the equations are dependent and there are many values of the variables that can satisfy both equations.

Q: Can I use a graphing calculator to solve a system of linear equations?

A: Yes, you can use a graphing calculator to solve a system of linear equations. You can graph the equations on the calculator and find the point of intersection, which represents the solution to the system.

Q: What are some common mistakes to avoid when solving a system of linear equations?

A: Some common mistakes to avoid when solving a system of linear equations include:

  • Not checking if the equations are linear
  • Not solving for one variable in terms of the other
  • Not substituting the expression into the other equation
  • Not checking if the resulting equation is true
  • Not considering the possibility of no solution or infinitely many solutions

Q: How can I practice solving systems of linear equations?

A: You can practice solving systems of linear equations by working through examples and exercises in a textbook or online resource. You can also try solving systems of linear equations on your own and then check your answers with a calculator or a teacher.

Q: What are some real-world applications of solving systems of linear equations?

A: Solving systems of linear equations has many real-world applications, including:

  • Physics: Solving systems of linear equations can be used to model the motion of objects and solve problems involving forces and velocities.
  • Engineering: Solving systems of linear equations can be used to design and optimize systems, such as electrical circuits and mechanical systems.
  • Economics: Solving systems of linear equations can be used to model economic systems and solve problems involving supply and demand.
  • Computer Science: Solving systems of linear equations can be used to solve problems involving computer graphics and game development.

Conclusion

Solving systems of linear equations is an important skill in mathematics and has many real-world applications. By understanding the substitution method and how to determine whether a pair of values is a solution to one equation, both equations, or neither of the equations, you can solve systems of linear equations with confidence.