5/6y-x-2=0 12x-5y=-9
Introduction
In mathematics, a system of linear equations is a set of two or more linear equations that involve the same set of variables. Solving a system of linear equations is a fundamental concept in algebra and is used to find the values of the variables that satisfy all the equations in the system. In this article, we will discuss how to solve a system of linear equations using the method of substitution and elimination.
The Problem
The problem we will be solving is a system of two linear equations:
5/6y - x - 2 = 0 12x - 5y = -9
Understanding the Equations
The first equation is:
5/6y - x - 2 = 0
This equation can be rewritten as:
5y/6 - x = 2
The second equation is:
12x - 5y = -9
Solving the System of Linear Equations
To solve this system of linear equations, we can use the method of substitution or elimination. In this case, we will use the method of elimination.
Step 1: Multiply the Equations by Necessary Multiples
To eliminate one of the variables, we need to multiply the equations by necessary multiples such that the coefficients of the variable to be eliminated are the same.
Let's multiply the first equation by 6 to eliminate the fraction:
30y - 6x - 12 = 0
Now, let's multiply the second equation by 1 (since it's already in the simplest form):
12x - 5y = -9
Step 2: Add the Equations
Now that we have the equations in the simplest form, we can add them to eliminate one of the variables.
Let's add the two equations:
(30y - 6x - 12) + (12x - 5y) = 0 + (-9)
This simplifies to:
25y - 6x - 12 = -9
Step 3: Simplify the Equation
Now, let's simplify the equation by adding 12 to both sides:
25y - 6x = -9 + 12
This simplifies to:
25y - 6x = 3
Step 4: Solve for One Variable
Now that we have the equation in the simplest form, we can solve for one variable.
Let's solve for y:
25y = 3 + 6x
This simplifies to:
25y = 6x + 3
Now, let's divide both sides by 25:
y = (6x + 3)/25
Step 5: Substitute the Value of y into One of the Original Equations
Now that we have the value of y, we can substitute it into one of the original equations to solve for x.
Let's substitute the value of y into the first original equation:
5/6y - x - 2 = 0
This simplifies to:
5/6((6x + 3)/25) - x - 2 = 0
Step 6: Simplify the Equation
Now, let's simplify the equation:
(5/6)(6x + 3)/25 - x - 2 = 0
This simplifies to:
(5x + 5)/30 - x - 2 = 0
Step 7: Multiply the Equation by 30
Now, let's multiply the equation by 30 to eliminate the fraction:
(5x + 5) - 30x - 60 = 0
This simplifies to:
-25x - 55 = 0
Step 8: Add 55 to Both Sides
Now, let's add 55 to both sides:
-25x = 55
Step 9: Divide Both Sides by -25
Now, let's divide both sides by -25:
x = -55/25
This simplifies to:
x = -11/5
Step 10: 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.
Let's substitute the value of x into the first original equation:
5/6y - x - 2 = 0
This simplifies to:
5/6y - (-11/5) - 2 = 0
Step 11: Simplify the Equation
Now, let's simplify the equation:
5/6y + 11/5 + 2 = 0
This simplifies to:
5/6y + 11/5 + 10/5 = 0
Step 12: Combine the Fractions
Now, let's combine the fractions:
5/6y + 21/5 = 0
Step 13: Subtract 21/5 from Both Sides
Now, let's subtract 21/5 from both sides:
5/6y = -21/5
Step 14: Multiply Both Sides by 6/5
Now, let's multiply both sides by 6/5:
y = (-21/5)(6/5)
This simplifies to:
y = -126/25
Conclusion
In this article, we discussed how to solve a system of linear equations using the method of substitution and elimination. We used the method of elimination to solve the system of linear equations:
5/6y - x - 2 = 0 12x - 5y = -9
We found that the values of x and y that satisfy both equations are:
x = -11/5 y = -126/25
Final Answer
The final answer is:
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 the same set of variables. Solving a system of linear equations is a fundamental concept in algebra and is used to find the values of the variables that satisfy all the equations in the system.
Q: What are the different methods for solving systems of linear equations?
A: There are two main methods for solving systems of linear equations: the method of substitution and the method of elimination. The method of substitution involves solving one equation for one variable and then substituting that expression into the other equation. The method of elimination involves adding or subtracting the equations to eliminate one of the variables.
Q: What is the method of substitution?
A: The method of substitution involves solving one equation for one variable and then substituting that expression into the other equation. This method is useful when one of the equations is already solved for one variable.
Q: What is the method of elimination?
A: The method of elimination involves adding or subtracting the equations to eliminate one of the variables. This method is useful when the coefficients of the variables are the same in both equations.
Q: How do I choose which method to use?
A: To choose which method to use, you need to look at the coefficients of the variables in both equations. If the coefficients of one variable are the same in both equations, you can use the method of elimination. If one of the equations is already solved for one variable, you can use the method of substitution.
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 following the order of operations (PEMDAS)
- Not simplifying the equations before solving
- Not checking the solutions to make sure they satisfy both equations
- Not using the correct method for the given equations
Q: How do I check my solutions to make sure they satisfy both equations?
A: To check your solutions, you need to substitute the values of the variables into both equations and make sure they are true. If the values satisfy both equations, then they are the correct solutions.
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:
- Finding the intersection point of two lines
- Determining the cost of producing a product
- Calculating the amount of money in a bank account
- Solving problems in physics and engineering
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. However, it's always a good idea to check your solutions by hand to make sure they are correct.
Q: How do I graph a system of linear equations?
A: To graph a system of linear equations, you need to graph both equations on the same coordinate plane. The point of intersection of the two lines is the solution to the system.
Q: What are some common types of systems of linear equations?
A: Some common types of systems of linear equations include:
- Consistent systems: These are systems that have a unique solution.
- Inconsistent systems: These are systems that have no solution.
- Dependent systems: These are systems that have an infinite number of solutions.
Q: How do I determine the type of system of linear equations?
A: To determine the type of system of linear equations, you need to look at the coefficients of the variables in both equations. If the coefficients are the same, the system is consistent. If the coefficients are different, the system is inconsistent. If the coefficients are the same and the constant terms are different, the system is dependent.