Solve The System Of Equations:${ \begin{array}{l} 5x - 3y = 10 \ 10x + 6y = 20 \end{array} }$
Introduction
Systems of linear equations are a fundamental concept in mathematics, and solving them is a crucial skill for students and professionals alike. In this article, we will focus on solving a system of two linear equations with two variables. We will use the given system of equations as an example and provide a step-by-step guide on how to solve it.
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 a linear equation, which means it can be written in the form of ax + by = c, where a, b, and c are constants, and x and y are variables. The system of equations is said to be consistent if it has a solution, and inconsistent if it does not have a solution.
The Given System of Equations
The given system of equations is:
{ \begin{array}{l} 5x - 3y = 10 \\ 10x + 6y = 20 \end{array} \}
Method 1: Substitution Method
One way to solve this system of equations is by using the substitution method. This method involves solving one equation for one variable and then substituting that expression into the other equation.
Step 1: Solve the First Equation for x
We can solve the first equation for x by adding 3y to both sides of the equation and then dividing both sides by 5.
{ 5x - 3y = 10 \}
{ 5x = 10 + 3y \}
{ x = \frac{10 + 3y}{5} \}
Step 2: Substitute the Expression for x into the Second Equation
Now that we have an expression for x, we can substitute it into the second equation.
{ 10x + 6y = 20 \}
{ 10\left(\frac{10 + 3y}{5}\right) + 6y = 20 \}
{ 2(10 + 3y) + 6y = 20 \}
{ 20 + 6y + 6y = 20 \}
{ 12y = 0 \}
Step 3: Solve for y
Now that we have an equation with only one variable, we can solve for y.
{ 12y = 0 \}
{ y = \frac{0}{12} \}
{ y = 0 \}
Step 4: Substitute the Value of y into the Expression for x
Now that we have a value for y, we can substitute it into the expression for x.
{ x = \frac{10 + 3y}{5} \}
{ x = \frac{10 + 3(0)}{5} \}
{ x = \frac{10}{5} \}
{ x = 2 \}
Method 2: Elimination Method
Another way to solve this system of equations is by using the elimination method. This method involves adding or subtracting the equations to eliminate one variable.
Step 1: Multiply the First Equation by 2
We can multiply the first equation by 2 to make the coefficients of y in both equations the same.
{ 5x - 3y = 10 \}
{ 2(5x - 3y) = 2(10) \}
{ 10x - 6y = 20 \}
Step 2: Add the Two Equations
Now that we have two equations with the same coefficients of y, we can add them to eliminate y.
{ (10x - 6y) + (10x + 6y) = 20 + 20 \}
{ 20x = 40 \}
Step 3: Solve for x
Now that we have an equation with only one variable, we can solve for x.
{ 20x = 40 \}
{ x = \frac{40}{20} \}
{ x = 2 \}
Step 4: Substitute the Value of x into One of the Original Equations
Now that we have a value for x, we can substitute it into one of the original equations to solve for y.
{ 5x - 3y = 10 \}
{ 5(2) - 3y = 10 \}
{ 10 - 3y = 10 \}
{ -3y = 0 \}
{ y = \frac{0}{-3} \}
{ y = 0 \}
Conclusion
In this article, we have solved a system of two linear equations with two variables using two different methods: the substitution method and the elimination method. We have shown that both methods can be used to solve systems of linear equations, and we have provided a step-by-step guide on how to use each method. We have also emphasized the importance of solving systems of linear equations in mathematics and real-world applications.
Final Answer
The final answer is x = 2 and y = 0.
References
- [1] "Systems of Linear Equations" by Math Open Reference
- [2] "Solving Systems of Linear Equations" by Khan Academy
- [3] "Systems of Linear Equations" by Wolfram MathWorld
Additional Resources
- [1] "Systems of Linear Equations" by MIT OpenCourseWare
- [2] "Solving Systems of Linear Equations" by Coursera
- [3] "Systems of Linear Equations" by edX
Q: What is a system of linear equations?
A: A system of linear equations is a set of two or more linear equations that involve two or more variables. Each equation is a linear equation, which means it can be written 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 if a system of linear equations has a solution?
A: A system of linear equations has a solution if it is consistent, meaning that the equations are not contradictory. If the equations are contradictory, the system is inconsistent and does not have a solution.
Q: What are the two main methods for solving systems of linear equations?
A: The two main methods for solving systems of linear equations are the substitution method and the elimination method.
Q: What is the substitution method?
A: The substitution method involves solving one equation for one variable and then substituting that expression into the other equation.
Q: What is the elimination method?
A: The elimination method involves adding or subtracting the equations to eliminate one variable.
Q: How do I choose which method to use?
A: You can choose which method to use based on the coefficients of the variables in the equations. If the coefficients of one variable are the same in both equations, you can use the elimination method. If the coefficients of one variable are different in both equations, you can use the substitution method.
Q: What if I have a system of linear equations with three or more variables?
A: If you have a system of linear equations with three or more variables, you can use the same methods as before, but you may need to use additional techniques, such as using matrices or graphing.
Q: Can I use a calculator to solve systems of linear equations?
A: Yes, you can use a calculator to solve systems of linear equations. Many calculators have built-in functions for solving systems of linear equations.
Q: What if I get a system of linear equations with no solution?
A: If you get a system of linear equations with no solution, it means that the equations are contradictory and do not have a solution.
Q: What if I get a system of linear equations with infinitely many solutions?
A: If you get a system of linear equations with infinitely many solutions, it means that the equations are dependent and have infinitely many solutions.
Q: Can I use systems of linear equations to model real-world problems?
A: Yes, you can use systems of linear equations to model real-world problems. Systems of linear equations can be used to model a wide range of problems, including business, economics, physics, and engineering.
Q: What are some common applications of systems of linear equations?
A: Some common applications of systems of linear equations include:
- Business: Systems of linear equations can be used to model business problems, such as finding the optimal price and quantity of a product.
- Economics: Systems of linear equations can be used to model economic problems, such as finding the optimal allocation of resources.
- Physics: Systems of linear equations can be used to model physical problems, such as finding the trajectory of a projectile.
- Engineering: Systems of linear equations can be used to model engineering problems, such as finding the optimal design of a bridge.
Conclusion
In this article, we have answered some frequently asked questions about solving systems of linear equations. We have covered topics such as the definition of a system of linear equations, the two main methods for solving systems of linear equations, and some common applications of systems of linear equations. We hope that this article has been helpful in answering your questions and providing you with a better understanding of systems of linear equations.