Solve The System Of Equations:$\[ \begin{array}{l} 5u - 4v = 3 \\ 2u - 4v = 6 \end{array} \\]
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 involves finding the values of the variables that satisfy all the equations in the system. In this article, we will focus on solving a system of two linear equations with two variables.
What are Linear Equations?
A linear equation is an equation in which the highest power of the variable(s) is 1. For example, the equation 2x + 3y = 5 is a linear equation because the highest power of x and y is 1. Linear equations can be written in the form ax + by = c, where a, b, and c are constants.
The System of Equations
The system of equations we will be solving is:
{ \begin{array}{l} 5u - 4v = 3 \\ 2u - 4v = 6 \end{array} \}
This system consists of two linear equations with two variables, u and v.
Method 1: Substitution Method
One way to solve this system 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 u
We can solve the first equation for u by adding 4v to both sides of the equation:
5u - 4v + 4v = 3 + 4v
This simplifies to:
5u = 3 + 4v
Next, we can divide both sides of the equation by 5 to solve for u:
u = (3 + 4v) / 5
Step 2: Substitute the Expression for u into the Second Equation
Now that we have an expression for u, we can substitute it into the second equation:
2u - 4v = 6
Substituting u = (3 + 4v) / 5 into this equation gives:
2((3 + 4v) / 5) - 4v = 6
Step 3: Simplify the Equation
To simplify the equation, we can multiply both sides of the equation by 5 to eliminate the fraction:
2(3 + 4v) - 20v = 30
Expanding the left-hand side of the equation gives:
6 + 8v - 20v = 30
Combine like terms:
-12v + 6 = 30
Subtract 6 from both sides:
-12v = 24
Step 4: Solve for v
Finally, we can solve for v by dividing both sides of the equation by -12:
v = -24 / 12
v = -2
Step 5: Find the Value of u
Now that we have the value of v, we can substitute it into the expression for u:
u = (3 + 4v) / 5
Substituting v = -2 into this equation gives:
u = (3 + 4(-2)) / 5
u = (3 - 8) / 5
u = -5 / 5
u = -1
Method 2: Elimination Method
Another way to solve this system is by using the elimination method. This method involves multiplying both equations by necessary multiples such that the coefficients of one of the variables (in this case, v) are the same in both equations.
Step 1: Multiply the First Equation by 4 and the Second Equation by 1
Multiplying the first equation by 4 gives:
20u - 16v = 12
Multiplying the second equation by 1 gives:
2u - 4v = 6
Step 2: Subtract the Second Equation from the First Equation
Subtracting the second equation from the first equation gives:
(20u - 16v) - (2u - 4v) = 12 - 6
This simplifies to:
18u - 12v = 6
Step 3: Solve for u
We can solve for u by dividing both sides of the equation by 18:
u = (6 + 12v) / 18
Step 4: Substitute the Expression for u into the Second Equation
Now that we have an expression for u, we can substitute it into the second equation:
2u - 4v = 6
Substituting u = (6 + 12v) / 18 into this equation gives:
2((6 + 12v) / 18) - 4v = 6
Step 5: Simplify the Equation
To simplify the equation, we can multiply both sides of the equation by 18 to eliminate the fraction:
2(6 + 12v) - 72v = 108
Expanding the left-hand side of the equation gives:
12 + 24v - 72v = 108
Combine like terms:
-48v + 12 = 108
Subtract 12 from both sides:
-48v = 96
Step 6: Solve for v
Finally, we can solve for v by dividing both sides of the equation by -48:
v = -96 / 48
v = -2
Step 7: Find the Value of u
Now that we have the value of v, we can substitute it into the expression for u:
u = (6 + 12v) / 18
Substituting v = -2 into this equation gives:
u = (6 + 12(-2)) / 18
u = (6 - 24) / 18
u = -18 / 18
u = -1
Conclusion
In this article, we have solved a system of two linear equations with two variables using the substitution method and the elimination method. Both methods have given us the same solution: u = -1 and v = -2. This solution satisfies both equations in the system, and it is the only solution that does so.
Final Answer
The final answer is:
Introduction
In our previous article, we solved a system of two linear equations with two variables using the substitution method and the elimination method. In this article, we will answer some common questions that students often have when solving systems of linear equations.
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. For example:
{ \begin{array}{l} 2x + 3y = 5 \\ x - 2y = -3 \end{array} \}
This system consists of two linear equations with two variables, x and y.
Q: How do I know which method to use to solve a system of linear equations?
A: There are two common methods to solve a system of linear equations: the substitution method and the elimination method. The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. The elimination method involves multiplying both equations by necessary multiples such that the coefficients of one of the variables are the same in both equations.
Q: What is the difference between the substitution method and the elimination method?
A: The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. The elimination method involves multiplying both equations by necessary multiples such that the coefficients of one of the variables are the same in both equations.
Q: How do I know if a system of linear equations has a solution?
A: A system of linear equations has a solution if and only if the two equations are consistent. In other words, if the two equations have the same solution, then the system has a solution.
Q: What is the solution to a system of linear equations?
A: The solution to a system of linear equations is the set of values that satisfy both equations in the system. For example, if we have the system:
{ \begin{array}{l} 2x + 3y = 5 \\ x - 2y = -3 \end{array} \}
Then the solution to this system is x = 2 and y = 1.
Q: How do I check if my solution is correct?
A: To check if your solution is correct, you can plug the values of x and y into both equations in the system and see if they are true. For example, if we have the solution x = 2 and y = 1, then we can plug these values into both equations in the system:
2(2) + 3(1) = 5 (2) - 2(1) = -3
Both equations are true, so we know that the solution x = 2 and y = 1 is correct.
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, then the two equations are inconsistent. In other words, the two equations have no common solution.
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, then the two equations are dependent. In other words, the two equations are equivalent and have the same solution.
Conclusion
In this article, we have answered some common questions that students often have when solving systems of linear equations. We have discussed the substitution method and the elimination method, and we have explained how to check if a solution is correct. We have also discussed what to do if a system of linear equations has no solution or infinitely many solutions.
Final Answer
The final answer is:
- A system of linear equations is a set of two or more linear equations that involve the same set of variables.
- The substitution method involves solving one equation for one variable and then substituting that expression into the other equation.
- The elimination method involves multiplying both equations by necessary multiples such that the coefficients of one of the variables are the same in both equations.
- A system of linear equations has a solution if and only if the two equations are consistent.
- The solution to a system of linear equations is the set of values that satisfy both equations in the system.
- To check if a solution is correct, you can plug the values of x and y into both equations in the system and see if they are true.