What Is The Solution To The System Of Equations Below?${ \begin{align*} 7x - 2y &= -23 \ 4x - 5y &= -17 \ \end{align*} }$A) { (-3, 6)$}$ B) { (-3, 1)$}$ C) { (0, 1)$}$ D) { (-3, 7)$}$ E)

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Introduction

Solving a system of linear equations is a fundamental concept in mathematics, particularly in algebra and geometry. It involves finding the values of variables that satisfy multiple equations simultaneously. In this article, we will explore a system of two linear equations and provide a step-by-step solution to find the values of the variables.

The System of Equations

The given system of equations is:

7xβˆ’2y=βˆ’234xβˆ’5y=βˆ’17\begin{align*} 7x - 2y &= -23 \\ 4x - 5y &= -17 \\ \end{align*}

Method 1: Substitution Method

One way to solve this system of 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.

Step 1: Solve the First Equation for x

We can start by solving the first equation for x:

7xβˆ’2y=βˆ’237x - 2y = -23

Add 2y to both sides:

7x=βˆ’23+2y7x = -23 + 2y

Divide both sides by 7:

x=βˆ’23+2y7x = \frac{-23 + 2y}{7}

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:

4xβˆ’5y=βˆ’174x - 5y = -17

Substitute the expression for x:

4(βˆ’23+2y7)βˆ’5y=βˆ’174\left(\frac{-23 + 2y}{7}\right) - 5y = -17

Step 3: Simplify the Equation

Simplify the equation by multiplying the numerator and denominator of the fraction:

4(βˆ’23+2y)7βˆ’5y=βˆ’17\frac{4(-23 + 2y)}{7} - 5y = -17

Multiply the numerator and denominator of the fraction:

βˆ’92+8y7βˆ’5y=βˆ’17\frac{-92 + 8y}{7} - 5y = -17

Multiply both sides by 7 to eliminate the fraction:

βˆ’92+8yβˆ’35y=βˆ’119-92 + 8y - 35y = -119

Combine like terms:

βˆ’92βˆ’27y=βˆ’119-92 - 27y = -119

Add 92 to both sides:

βˆ’27y=βˆ’27-27y = -27

Divide both sides by -27:

y=1y = 1

Step 4: Find the Value of x

Now that we have the value of y, we can substitute it into the expression for x:

x=βˆ’23+2y7x = \frac{-23 + 2y}{7}

Substitute the value of y:

x=βˆ’23+2(1)7x = \frac{-23 + 2(1)}{7}

Simplify the expression:

x=βˆ’23+27x = \frac{-23 + 2}{7}

x=βˆ’217x = \frac{-21}{7}

Divide both sides by 7:

x=βˆ’3x = -3

Method 2: Elimination Method

Another way to solve this system of equations is by using the elimination method. This method involves adding or subtracting the equations to eliminate one variable.

Step 1: Multiply the Equations by Necessary Multiples

To eliminate one variable, we need to multiply the equations by necessary multiples such that the coefficients of y's in both equations are the same.

Multiply the first equation by 5 and the second equation by 2:

35xβˆ’10y=βˆ’11535x - 10y = -115

8xβˆ’10y=βˆ’348x - 10y = -34

Step 2: Subtract the Second Equation from the First Equation

Subtract the second equation from the first equation to eliminate the y variable:

(35xβˆ’10y)βˆ’(8xβˆ’10y)=βˆ’115βˆ’(βˆ’34)(35x - 10y) - (8x - 10y) = -115 - (-34)

Simplify the equation:

27x=βˆ’8127x = -81

Divide both sides by 27:

x=βˆ’3x = -3

Step 3: Find the Value of y

Now that we have the value of x, we can substitute it into one of the original equations to find the value of y.

Substitute the value of x into the first equation:

7xβˆ’2y=βˆ’237x - 2y = -23

Substitute the value of x:

7(βˆ’3)βˆ’2y=βˆ’237(-3) - 2y = -23

Simplify the equation:

βˆ’21βˆ’2y=βˆ’23-21 - 2y = -23

Add 21 to both sides:

βˆ’2y=βˆ’2-2y = -2

Divide both sides by -2:

y=1y = 1

Conclusion

In this article, we have solved a system of two linear equations using the substitution method and the elimination method. Both methods have led to the same solution: x = -3 and y = 1. This solution satisfies both equations, and it is the only solution that satisfies both equations.

Final Answer

The final answer is:

(-3, 1)$<br/> # **Frequently Asked Questions (FAQs) About Solving Systems of Linear Equations**

Introduction

Solving systems of linear equations is a fundamental concept in mathematics, particularly in algebra and geometry. In this article, we will address some frequently asked questions (FAQs) about 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 variables. Each equation is a linear equation, which means it is an equation in which the highest power of the variable(s) is 1.

Q: How do I know if a system of linear equations has a solution?

A: To determine if a system of linear equations has a solution, you can use the following methods:

  • Check if the equations are consistent (i.e., they have the same solution).
  • Check if the equations are inconsistent (i.e., they have no solution).
  • Check if the equations are dependent (i.e., they have infinitely many solutions).

Q: What is the difference between a consistent and an inconsistent system of linear equations?

A: A consistent system of linear equations has a solution, while an inconsistent system of linear equations has no solution.

Q: What is the difference between a dependent and an independent system of linear equations?

A: A dependent system of linear equations has infinitely many solutions, while an independent system of linear equations has a unique solution.

Q: How do I solve a system of linear equations using the substitution method?

A: To solve a system of linear equations using the substitution method, follow these steps:

  1. Solve one equation for one variable.
  2. Substitute the expression for the variable into the other equation.
  3. Solve for the other variable.
  4. Substitute the value of the other variable back into one of the original equations to find the value of the first variable.

Q: How do I solve a system of linear equations using the elimination method?

A: To solve a system of linear equations using the elimination method, follow these steps:

  1. Multiply the equations by necessary multiples such that the coefficients of the variables are the same.
  2. Add or subtract the equations to eliminate one variable.
  3. Solve for the remaining variable.
  4. Substitute the value of the remaining variable back into one of the original equations to find the value of the other variable.

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 if the equations are consistent or inconsistent.
  • Not checking if the equations are dependent or independent.
  • Not following the correct steps when using the substitution or elimination method.
  • Not checking if the solution satisfies both equations.

Q: How do I check if a solution satisfies both equations?

A: To check if a solution satisfies both equations, substitute the values of the variables into both equations and check if the equations are true.

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:

  • Physics: Solving systems of linear equations is used to describe the motion of objects in physics.
  • Engineering: Solving systems of linear equations is used to design and optimize systems in engineering.
  • Economics: Solving systems of linear equations is used to model economic systems and make predictions about economic trends.
  • Computer Science: Solving systems of linear equations is used in computer science to solve problems in computer graphics, game development, and machine learning.

Conclusion

In this article, we have addressed some frequently asked questions (FAQs) about solving systems of linear equations. We have covered topics such as the definition of a system of linear equations, how to determine if a system has a solution, and how to solve systems using the substitution and elimination methods. We have also discussed some common mistakes to avoid and real-world applications of solving systems of linear equations.