Solve The Following System Of Equations:$\[ \begin{array}{l} -3x + 10y = 11 \\ -8x + 2y = -20 \end{array} \\]
===========================================================
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 with two variables. We will use the method of substitution and elimination to find the solution.
The System of Equations
The system of equations we will be solving is:
{ \begin{array}{l} -3x + 10y = 11 \\ -8x + 2y = -20 \end{array} \}
Method of Substitution
One way to solve this system of equations is by using the method of substitution. This method involves solving one of the equations 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 isolating x on one side of the equation.
{ -3x + 10y = 11 \}
{ -3x = 11 - 10y \}
{ x = \frac{11 - 10y}{-3} \}
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.
{ -8x + 2y = -20 \}
{ -8\left(\frac{11 - 10y}{-3}\right) + 2y = -20 \}
Step 3: Simplify the Equation
We can simplify the equation by multiplying both sides by -3 to eliminate the fraction.
{ -8\left(\frac{11 - 10y}{-3}\right) + 2y = -20 \}
{ -8(11 - 10y) + 2y(-3) = -20(-3) \}
{ 88 - 80y - 6y = 60 \}
{ -86y = -28 \}
Step 4: Solve for y
Now that we have a simple equation, we can solve for y.
{ -86y = -28 \}
{ y = \frac{-28}{-86} \}
{ y = \frac{14}{43} \}
Step 5: Substitute the Value of y into the Expression for x
Now that we have the value of y, we can substitute it into the expression for x.
{ x = \frac{11 - 10y}{-3} \}
{ x = \frac{11 - 10\left(\frac{14}{43}\right)}{-3} \}
{ x = \frac{11 - \frac{140}{43}}{-3} \}
{ x = \frac{\frac{473}{43} - \frac{140}{43}}{-3} \}
{ x = \frac{\frac{333}{43}}{-3} \}
{ x = \frac{333}{43} \cdot \frac{-1}{3} \}
{ x = \frac{-111}{43} \}
Method of Elimination
Another way to solve this system of equations is by using the method of elimination. This method involves multiplying both equations by necessary multiples such that the coefficients of one of the variables (either x or y) are the same in both equations.
Step 1: Multiply the First Equation by 2
We can multiply the first equation by 2 to make the coefficients of y the same in both equations.
{ -3x + 10y = 11 \}
{ 2(-3x + 10y) = 2(11) \}
{ -6x + 20y = 22 \}
Step 2: Multiply the Second Equation by 5
We can multiply the second equation by 5 to make the coefficients of y the same in both equations.
{ -8x + 2y = -20 \}
{ 5(-8x + 2y) = 5(-20) \}
{ -40x + 10y = -100 \}
Step 3: Subtract the Second Equation from the First Equation
Now that we have two equations with the same coefficients of y, we can subtract the second equation from the first equation to eliminate y.
{ -6x + 20y = 22 \}
{ -40x + 10y = -100 \}
{ (-6x + 20y) - (-40x + 10y) = 22 - (-100) \}
{ -6x + 20y + 40x - 10y = 122 \}
{ 34x + 10y = 122 \}
Step 4: Solve for x
Now that we have a simple equation, we can solve for x.
{ 34x + 10y = 122 \}
{ 34x = 122 - 10y \}
{ x = \frac{122 - 10y}{34} \}
Step 5: Substitute the Value of x into One of the Original Equations
Now that we have the value of x, we can substitute it into one of the original equations to solve for y.
{ -3x + 10y = 11 \}
{ -3\left(\frac{122 - 10y}{34}\right) + 10y = 11 \}
{ \frac{-366 + 30y}{34} + 10y = 11 \}
{ -366 + 30y + 340y = 374 \}
{ 370y = 740 \}
{ y = \frac{740}{370} \}
{ y = 2 \}
Step 6: Substitute the Value of y into the Expression for x
Now that we have the value of y, we can substitute it into the expression for x.
{ x = \frac{122 - 10y}{34} \}
{ x = \frac{122 - 10(2)}{34} \}
{ x = \frac{122 - 20}{34} \}
{ x = \frac{102}{34} \}
{ x = 3 \}
Conclusion
In this article, we have solved a system of two linear equations with two variables using the method of substitution and elimination. We have shown that both methods can be used to solve the system of equations and have obtained the same solution. The solution to the system of equations is x = -111/43 and y = 14/43.
Final Answer
The final answer is x = -111/43 and y = 14/43.
====================================================================
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 are the two main methods for solving systems of linear equations?
A: The two main methods for solving systems of linear equations are the method of substitution and the method of elimination.
Q: What is the method of substitution?
A: The method of substitution involves solving one of the equations for one variable and then substituting that expression into the other equation.
Q: What is the method of elimination?
A: The method of elimination involves multiplying both equations by necessary multiples such that the coefficients of one of the variables (either x or y) are the same in both equations.
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 method of elimination. If the coefficients of one variable are different in both equations, you can use the method of substitution.
Q: What if I have a system of three or more linear equations?
A: If you have a system of three or more linear equations, you can use the method of substitution or elimination to solve for two variables, and then use the third equation to solve for the third variable.
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 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 satisfies 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 satisfy both equations.
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 problems in physics, engineering, economics, and more.
Q: What are some common applications of systems of linear equations?
A: Some common applications of systems of linear equations include:
- Modeling population growth
- Modeling financial transactions
- Modeling electrical circuits
- Modeling mechanical systems
- Modeling chemical reactions
Q: Can I use systems of linear equations to solve problems in other areas of mathematics?
A: Yes, you can use systems of linear equations to solve problems in other areas of mathematics, including algebra, geometry, and calculus.
Q: What are some common mistakes to avoid when solving systems of linear equations?
A: Some common mistakes to avoid when solving systems of linear equations include:
- Not checking for consistency before solving the system
- Not using the correct method for solving the system
- Not checking for infinitely many solutions before solving the system
- Not checking for no solution before solving the system
Q: Can I use systems of linear equations to solve problems in other areas of science and engineering?
A: Yes, you can use systems of linear equations to solve problems in other areas of science and engineering, including physics, engineering, computer science, and more.
Q: What are some common tools and software used to solve systems of linear equations?
A: Some common tools and software used to solve systems of linear equations include:
- Graphing calculators
- Computer algebra systems (CAS)
- Matrix calculators
- Linear algebra software
- Programming languages (such as Python or MATLAB)