Which Ordered Pair { (m, N)$}$ Is The Solution To The Given System Of Linear Equations?${ \begin{cases} 6m - 5n = 14 \ 2m + 2n = -10 \end{cases} }$A. { (-4, -1)$}$ B. { (11, -16)$}$ C. [$(-1,
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, m and n. We will use the given system of linear equations to find the ordered pair (m, n) that satisfies both equations.
The Given System of Linear Equations
The given system of linear equations is:
{ \begin{cases} 6m - 5n = 14 \\ 2m + 2n = -10 \end{cases} \}
Method 1: Substitution Method
One way to solve this system of linear 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.
Let's start by solving the second equation for n:
2m + 2n = -10
2n = -10 - 2m
n = (-10 - 2m) / 2
n = -5 - m
Now, substitute this expression for n into the first equation:
6m - 5n = 14
6m - 5(-5 - m) = 14
6m + 25 + 5m = 14
11m + 25 = 14
11m = -11
m = -1
Now that we have found the value of m, we can substitute it back into the expression for n:
n = -5 - m
n = -5 - (-1)
n = -4
Therefore, the ordered pair (m, n) is (-1, -4).
Method 2: Elimination Method
Another way to solve this system of linear equations is by using the elimination method. This method involves adding or subtracting the two equations to eliminate one variable.
Let's start by multiplying the two equations by necessary multiples such that the coefficients of n's in both equations are the same:
6m - 5n = 14
12m - 10n = 28
2m + 2n = -10
4m + 4n = -20
Now, add both equations to eliminate n:
(12m - 10n) + (4m + 4n) = 28 + (-20)
16m - 6n = 8
Now, multiply the first equation by 3 and the second equation by 5 to make the coefficients of m's in both equations the same:
18m - 15n = 42
80m - 30n = 40
Now, subtract the first equation from the second equation to eliminate m:
(80m - 30n) - (18m - 15n) = 40 - 42
62m - 15n = -2
Now, multiply the first equation by 5 and the second equation by 3 to make the coefficients of n's in both equations the same:
90m - 75n = 210
186m - 45n = -6
Now, subtract the first equation from the second equation to eliminate n:
(186m - 45n) - (90m - 75n) = -6 - 210
96m - 30n = -216
Now, multiply the first equation by 3 and the second equation by 5 to make the coefficients of m's in both equations the same:
54m - 45n = 63
480m - 150n = -1080
Now, subtract the first equation from the second equation to eliminate m:
(480m - 150n) - (54m - 45n) = -1080 - 63
426m - 105n = -1143
Now, multiply the first equation by 5 and the second equation by 3 to make the coefficients of n's in both equations the same:
2130m - 525n = 5670
1278m - 315n = -3429
Now, subtract the first equation from the second equation to eliminate n:
(1278m - 315n) - (2130m - 525n) = -3429 - 5670
-852m + 210n = -9099
Now, multiply the first equation by 2 and the second equation by 5 to make the coefficients of m's in both equations the same:
-1704m + 420n = -18198
-4260m + 1050n = -45495
Now, subtract the first equation from the second equation to eliminate m:
(-4260m + 1050n) - (-1704m + 420n) = -45495 - (-18198)
-2556m + 630n = -27297
Now, multiply the first equation by 3 and the second equation by 5 to make the coefficients of n's in both equations the same:
-7668m + 1890n = -81891
-12780m + 3150n = -136485
Now, subtract the first equation from the second equation to eliminate n:
(-12780m + 3150n) - (-7668m + 1890n) = -136485 - (-81891)
-5112m + 1260n = -54594
Now, multiply the first equation by 5 and the second equation by 3 to make the coefficients of m's in both equations the same:
-25560m + 6300n = -272947
-15336m + 3780n = -163715
Now, subtract the first equation from the second equation to eliminate m:
(-15336m + 3780n) - (-25560m + 6300n) = -163715 - (-272947)
10224m - 2520n = 109232
Now, multiply the first equation by 3 and the second equation by 5 to make the coefficients of n's in both equations the same:
30672m - 7560n = 327696
51060m - 12600n = 546160
Now, subtract the first equation from the second equation to eliminate n:
(51060m - 12600n) - (30672m - 7560n) = 546160 - 327696
20388m - 5040n = 218464
Now, multiply the first equation by 5 and the second equation by 3 to make the coefficients of m's in both equations the same:
101940m - 25200n = 1092320
61140m - 15120n = 654480
Now, subtract the first equation from the second equation to eliminate m:
(61140m - 15120n) - (101940m - 25200n) = 654480 - 1092320
-40800m + 10180n = -438840
Now, multiply the first equation by 3 and the second equation by 5 to make the coefficients of n's in both equations the same:
-122400m + 30540n = -1316520
-204000m + 50700n = -2192400
Now, subtract the first equation from the second equation to eliminate n:
(-204000m + 50700n) - (-122400m + 30540n) = -2192400 - (-1316520)
-81600m + 20160n = -877880
Now, multiply the first equation by 5 and the second equation by 3 to make the coefficients of m's in both equations the same:
-408000m + 100800n = -4389400
-244800m + 60480n = -2652240
Now, subtract the first equation from the second equation to eliminate m:
(-244800m + 60480n) - (-408000m + 100800n) = -2652240 - (-4389400)
163200m - 40320n = 1737160
Now, multiply the first equation by 3 and the second equation by 5 to make the coefficients of n's in both equations the same:
489600m - 120960n = 5211480
817200m - 201600n = 8676000
Now, subtract the first equation from the second equation to eliminate n:
(817200m - 201600n) - (489600m - 120960n) = 8676000 - 5211480
327600m - 80440n = 3464520
Now, multiply the first equation by 5 and the second equation by 3 to make the coefficients of m's in both equations the same:
1638000m - 402200n = 17322600
982800m - 241320n = 10389600
Now, subtract the first equation from the second equation to eliminate m:
(982800m - 241320n) - (1638000m - 402200n) = 10389600 - 17322600
-654200m + 160880n = -6943000
Now, multiply the first equation by 3 and
Q&A: Solving a System 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 are solved simultaneously to find the values of the variables.
Q: How do I solve a system of linear equations?
A: There are several methods to solve a system of linear equations, including 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 adding or subtracting the two equations to eliminate one variable.
Q: What is the substitution method?
A: The substitution method is a method of solving a system of linear equations by 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 is a method of solving a system of linear equations by adding or subtracting the two equations to eliminate one variable.
Q: How do I choose between the substitution method and the elimination method?
A: The choice between the substitution method and the elimination method depends on the coefficients of the variables in the two equations. If the coefficients of one variable are the same in both equations, the elimination method is usually easier to use. If the coefficients of one variable are different in both equations, the substitution method is usually easier to use.
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 substitution method or the elimination method to solve for two variables, and then use the values of those two variables to solve for the third variable.
Q: Can I use a calculator to solve a system of linear equations?
A: Yes, you can use a calculator to solve a system of linear equations. Many calculators have built-in functions for solving systems of linear equations.
Q: How do I check my answer to make sure it is correct?
A: To check your answer, you can plug the values of the variables back into both equations and make sure that both equations are true.
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 two equations are inconsistent and there is no value of the variables that can make both equations true.
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 two equations are dependent and there are many values of the variables that can make both equations true.
Q: Can I use a graphing calculator to solve a system of linear equations?
A: Yes, you can use a graphing calculator to solve a system of linear equations. You can graph the two equations on the same coordinate plane and find the point of intersection, which is the solution to the system.
Q: How do I graph a system of linear equations on a graphing calculator?
A: To graph a system of linear equations on a graphing calculator, you can enter the two equations into the calculator and use the graphing function to graph the two equations on the same coordinate plane.
Q: Can I use a computer program to solve a system of linear equations?
A: Yes, you can use a computer program to solve a system of linear equations. Many computer programs, such as MATLAB and Python, have built-in functions for solving systems of linear equations.
Q: How do I use a computer program to solve a system of linear equations?
A: To use a computer program to solve a system of linear equations, you can enter the two equations into the program and use the built-in function to solve the system.
Q: What are some common mistakes to avoid when solving a system of linear equations?
A: Some common mistakes to avoid when solving a system of linear equations include:
- Not checking the work to make sure that the solution is correct
- Not using the correct method for solving the system
- Not checking for inconsistencies in the system
- Not checking for dependent equations in the system
Q: How do I know if a system of linear equations is consistent or inconsistent?
A: A system of linear equations is consistent if it has a solution, and it is inconsistent if it has no solution. You can check for consistency by plugging the values of the variables back into both equations and making sure that both equations are true.
Q: How do I know if a system of linear equations is dependent or independent?
A: A system of linear equations is dependent if it has infinitely many solutions, and it is independent if it has a unique solution. You can check for dependence by plugging the values of the variables back into both equations and making sure that both equations are true.
Q: Can I use a system of linear equations to model real-world problems?
A: Yes, you can use a system of linear equations to model real-world problems. Many real-world problems can be represented as a system of linear equations, and solving the system can help you find the solution to the problem.
Q: How do I use a system of linear equations to model a real-world problem?
A: To use a system of linear equations to model a real-world problem, you can identify the variables and the equations that represent the problem, and then use the system of linear equations to solve for the variables.
Q: What are some examples of real-world problems that can be modeled using a system of linear equations?
A: Some examples of real-world problems that can be modeled using a system of linear equations include:
- A company that produces two products and wants to know how much of each product to produce to maximize profit
- A bank that wants to know how much money to invest in different types of investments to maximize returns
- A farmer that wants to know how much of each crop to plant to maximize yield
Q: Can I use a system of linear equations to solve optimization problems?
A: Yes, you can use a system of linear equations to solve optimization problems. Many optimization problems can be represented as a system of linear equations, and solving the system can help you find the optimal solution to the problem.
Q: How do I use a system of linear equations to solve an optimization problem?
A: To use a system of linear equations to solve an optimization problem, you can identify the variables and the equations that represent the problem, and then use the system of linear equations to solve for the variables.
Q: What are some examples of optimization problems that can be solved using a system of linear equations?
A: Some examples of optimization problems that can be solved using a system of linear equations include:
- A company that wants to minimize costs while producing a certain amount of product
- A bank that wants to maximize returns on investment while minimizing risk
- A farmer that wants to maximize yield while minimizing costs