Solve The System Of Equations:${ \begin{array}{l} y = -\frac{4}{5}x \ 4x + 5y = 0 \end{array} }$

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Introduction

Solving systems of linear equations is a fundamental concept in mathematics, particularly in algebra and geometry. It involves finding the solution to a set of two or more linear equations with two or more variables. In this article, we will focus on solving a system of two linear equations with two variables using the substitution method.

What are Systems of Linear Equations?

A system of linear equations is a set of two or more linear equations that involve two or more variables. Each equation is in the form of ax + by = c, where a, b, and c are constants, and x and y are variables. The goal is to find the values of x and y that satisfy all the equations in the system.

The System of Equations

The system of equations we will be solving is:

y=−45x4x+5y=0\begin{array}{l} y = -\frac{4}{5}x \\ 4x + 5y = 0 \end{array}

Step 1: Understand the Equations

The first equation is a simple linear equation in the form of y = mx + b, where m is the slope and b is the y-intercept. In this case, the slope is -4/5 and the y-intercept is 0.

The second equation is a linear equation in the form of ax + by = c, where a, b, and c are constants. In this case, a = 4, b = 5, and c = 0.

Step 2: Solve the First Equation for y

We can solve the first equation for y by isolating y on one side of the equation.

y=−45xy = -\frac{4}{5}x

Step 3: Substitute the Expression for y into the Second Equation

Now that we have an expression for y, we can substitute it into the second equation.

4x+5(−45x)=04x + 5(-\frac{4}{5}x) = 0

Step 4: Simplify the Equation

We can simplify the equation by combining like terms.

4x−4x=04x - 4x = 0

Step 5: Solve for x

The equation is now a simple equation in the form of ax = 0, where a is a constant. We can solve for x by dividing both sides of the equation by a.

x=0x = 0

Step 6: Find the Value of y

Now that we have the value of x, we can find the value of y by substituting x into the expression for y.

y=−45(0)y = -\frac{4}{5}(0)

y=0y = 0

Conclusion

We have solved the system of equations using the substitution method. The solution is x = 0 and y = 0.

Why is this Method Important?

The substitution method is an important technique in solving systems of linear equations. It allows us to solve systems of equations with two or more variables by substituting one equation into another. This method is particularly useful when one of the equations is already solved for one of the variables.

Real-World Applications

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

  • Physics and Engineering: Solving systems of linear equations is used to model real-world problems, such as the motion of objects and the behavior of electrical circuits.
  • Computer Science: Solving systems of linear equations is used in computer graphics, game development, and machine learning.
  • Economics: Solving systems of linear equations is used to model economic systems, including supply and demand curves.

Common Mistakes to Avoid

When solving systems of linear equations, there are several common mistakes to avoid, including:

  • Not checking the solution: Make sure to check the solution to ensure that it satisfies all the equations in the system.
  • Not using the correct method: Choose the correct method for solving the system of equations, such as the substitution method or the elimination method.
  • Not simplifying the equation: Simplify the equation as much as possible to make it easier to solve.

Conclusion

Introduction

Solving systems of linear equations is a fundamental concept in mathematics, particularly in algebra and geometry. In our previous article, we discussed the substitution method for solving systems of linear equations. In this article, we will answer some frequently asked questions about solving systems of linear equations.

Q: What is a system of linear equations?

A system of linear equations is a set of two or more linear equations that involve two or more variables. Each equation is in the form of ax + by = c, where a, b, and c are constants, and x and y are variables.

Q: How do I know which method to use to solve a system of linear equations?

There are several methods for solving systems of linear equations, including the substitution method and the elimination method. The choice of method depends on the specific system of equations and the variables involved. If one of the equations is already solved for one of the variables, the substitution method may be the best choice. If the coefficients of the variables are the same, the elimination method may be the best choice.

Q: What is the difference between the substitution method and the elimination method?

The substitution method involves substituting one equation into another to solve for one of the variables. The elimination method involves adding or subtracting the equations to eliminate one of the variables.

Q: How do I check if the solution is correct?

To check if the solution is correct, substitute the values of x and y into each equation and make sure that the equation is true. If the equation is not true, the solution is incorrect.

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

Some common mistakes to avoid when solving systems of linear equations include:

  • Not checking the solution
  • Not using the correct method
  • Not simplifying the equation
  • Not considering the possibility of no solution or infinite solutions

Q: How do I handle systems of linear equations with no solution or infinite solutions?

If a system of linear equations has no solution, it means that the equations are inconsistent and there is no value of x and y that satisfies both equations. If a system of linear equations has infinite solutions, it means that the equations are dependent and there are many values of x and y that satisfy both equations.

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

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

  • Physics and Engineering: Solving systems of linear equations is used to model real-world problems, such as the motion of objects and the behavior of electrical circuits.
  • Computer Science: Solving systems of linear equations is used in computer graphics, game development, and machine learning.
  • Economics: Solving systems of linear equations is used to model economic systems, including supply and demand curves.

Q: How do I use technology to solve systems of linear equations?

There are many software programs and online tools that can be used to solve systems of linear equations, including:

  • Graphing calculators: Graphing calculators can be used to graph the equations and find the solution.
  • Computer algebra systems: Computer algebra systems, such as Mathematica and Maple, can be used to solve systems of linear equations.
  • Online tools: Online tools, such as Wolfram Alpha and Symbolab, can be used to solve systems of linear equations.

Conclusion

Solving systems of linear equations is a fundamental concept in mathematics, particularly in algebra and geometry. By understanding the system of equations and using the correct method, we can solve systems of linear equations with ease. This Q&A guide provides answers to some frequently asked questions about solving systems of linear equations and provides a comprehensive overview of the topic.

Additional Resources

For more information on solving systems of linear equations, check out the following resources:

  • Textbooks: "Algebra and Trigonometry" by Michael Sullivan and "Linear Algebra and Its Applications" by Gilbert Strang.
  • Online tutorials: Khan Academy, MIT OpenCourseWare, and Coursera.
  • Software programs: Mathematica, Maple, and Wolfram Alpha.

Practice Problems

Try solving the following systems of linear equations:

  • \begin{array}{l}

x + y = 2 \ 2x - 3y = -1 \end{array}$

  • \begin{array}{l}

x - 2y = -3 \ 3x + 4y = 5 \end{array}$

Answer Key

  • \begin{array}{l}

x + y = 2 \ 2x - 3y = -1 \end{array}$ + Solution: x = 1, y = 1

  • \begin{array}{l}

x - 2y = -3 \ 3x + 4y = 5 \end{array}$ + Solution: x = 1, y = 1