Type The Correct Answer In Each Box. Use Numerals Instead Of Words.Consider This System Of Linear Equations.${ \begin{align*} 4x - 3y &= -8 \ 3x + 2y &= -6 \end{align*} }$The Solution To This System Is ( $\square$ ,

Introduction

When dealing with a system of linear equations, we are given multiple equations that involve variables, and our goal is to find the values of these variables that satisfy all the equations simultaneously. In this article, we will focus on solving a system of two linear equations using the method of substitution or elimination.

The System of Linear Equations

We are given the following system of linear equations:

{ \begin{align*} 4x - 3y &= -8 \\ 3x + 2y &= -6 \end{align*} \}

Our task is to find the values of $x$ and $y$ that satisfy both equations.

Method of Substitution

One way to solve this system is by using the method of substitution. We can solve one of the equations for one variable and then substitute that expression into the other equation.

Let's solve the first equation for $x$:

$4x - 3y = -8$

$4x = -8 + 3y$

$x = \frac{-8 + 3y}{4}$

Now, substitute this expression for $x$ into the second equation:

$3\left(\frac{-8 + 3y}{4}\right) + 2y = -6$

$-6 + \frac{9y}{4} + 2y = -6$

$\frac{9y}{4} + 2y = 0$

$\frac{9y + 8y}{4} = 0$

$17y = 0$

$y = 0$

Now that we have found the value of $y$, we can substitute it back into one of the original equations to find the value of $x$. Let's use the first equation:

$4x - 3(0) = -8$

$4x = -8$

$x = -2$

Method of Elimination

Another way to solve this system is by using the method of elimination. We can multiply both equations by necessary multiples such that the coefficients of $y$'s in both equations are the same:

$4x - 3y = -8$

$3x + 2y = -6$

Multiply the first equation by 2 and the second equation by 3:

$8x - 6y = -16$

$9x + 6y = -18$

Now, add both equations to eliminate the variable $y$:

$(8x + 9x) + (-6y + 6y) = -16 - 18$

$17x = -34$

$x = -2$

Now that we have found the value of $x$, we can substitute it back into one of the original equations to find the value of $y$. Let's use the first equation:

$4(-2) - 3y = -8$

$-8 - 3y = -8$

$-3y = 0$

$y = 0$

Conclusion

In this article, we have solved a system of two linear equations using the method of substitution and the method of elimination. We have found that the solution to the system is $x = -2$ and $y = 0$. This means that the values of $x$ and $y$ that satisfy both equations are $x = -2$ and $y = 0$.

Discussion

Solving a system of linear equations is an important concept in mathematics, and it has many real-world applications. For example, in physics, we can use systems of linear equations to model the motion of objects. In economics, we can use systems of linear equations to model the behavior of markets.

Final Answer

The final answer is: $\boxed{-2, 0}$

Introduction

In our previous article, we solved a system of two linear equations using the method of substitution and the method of elimination. 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 variables. The goal is to find the values of the variables that satisfy all the equations simultaneously.

Q: How do I know which method to use, substitution or elimination?

A: Both methods can be used to solve a system of linear equations. The choice of method depends on the coefficients of the variables in the equations. If the coefficients of one variable are the same in both equations, it is easier to use the elimination method. If the coefficients of one variable are different in both equations, it is easier to use the substitution method.

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 same methods as before, but you may need to use a combination of both methods. For example, you can use the elimination method to reduce the system to two equations, and then use the substitution method to solve the resulting system.

Q: How do I know if the system has a unique solution, no solution, or infinitely many solutions?

A: To determine the number of solutions, you can use the following criteria:

• If the system has a unique solution, the equations are consistent and the coefficients of the variables are not proportional.
• If the system has no solution, the equations are inconsistent and the coefficients of the variables are proportional.
• If the system has infinitely many solutions, the equations are consistent and the coefficients of the variables are proportional.

Q: What if I have a system of linear equations with fractions?

A: If you have a system of linear equations with fractions, you can multiply both sides of each equation by the least common multiple (LCM) of the denominators to eliminate the fractions.

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 represents the solution to the system.

Q: How do I check my solution to a system of linear equations?

A: To check your solution, you can substitute the values of the variables back into both original equations and verify that they are true.

Conclusion

In this article, we have answered some common questions that students often have when solving systems of linear equations. We hope that this Q&A article has been helpful in clarifying any doubts you may have had.

Final Answer

The final answer is: $\boxed{0}$