Solving The System Of Equations 4x + 7y = 3 And 4x - Y = -5

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When confronted with a system of equations, the challenge lies in identifying the values of the variables that simultaneously satisfy all the equations. In this article, we will delve into a step-by-step solution for the given system of equations:

4x+7y=34x−y=−5\begin{array}{l} 4x + 7y = 3 \\ 4x - y = -5 \end{array}

We will explore the methods of substitution and elimination, providing a comprehensive understanding of how to tackle such problems. This detailed explanation aims to equip you with the skills to confidently solve similar systems of equations.

Understanding Systems of Equations

Before diving into the solution, let's clarify what a system of equations is. A system of equations is a set of two or more equations containing the same variables. The solution to a system of equations is the set of values for the variables that make all the equations true simultaneously. Graphically, the solution represents the point(s) where the lines or curves representing the equations intersect. Solving systems of equations is a fundamental concept in algebra and has wide applications in various fields, including engineering, economics, and computer science.

Systems of equations can be categorized based on the number of solutions they have:

  • Consistent Systems: These systems have at least one solution. Consistent systems can be further classified into:
    • Independent Systems: Have exactly one solution, meaning the lines intersect at a single point.
    • Dependent Systems: Have infinitely many solutions, meaning the equations represent the same line.
  • Inconsistent Systems: These systems have no solution, meaning the lines are parallel and never intersect.

Methods for Solving Systems of Equations

There are several methods for solving systems of equations, each with its own advantages and suitability depending on the specific system. The most common methods include:

  1. Substitution Method: This method involves solving one equation for one variable and substituting that expression into the other equation. This reduces the system to a single equation with one variable, which can then be easily solved. Once the value of one variable is found, it can be substituted back into either of the original equations to find the value of the other variable.

  2. Elimination Method (or Addition Method): This method involves manipulating the equations so that the coefficients of one of the variables are opposites. Adding the equations then eliminates that variable, resulting in a single equation with one variable. This equation can be solved, and the value can be substituted back into one of the original equations to find the value of the other variable.

  3. Graphing Method: This method involves graphing both equations on the same coordinate plane. The solution to the system is the point(s) where the lines intersect. This method is particularly useful for visualizing the solutions and understanding the nature of the system (consistent, inconsistent, independent, or dependent). However, it may not be the most accurate method for finding exact solutions, especially if the solutions are not integers.

  4. Matrix Methods: For systems with more than two variables, matrix methods such as Gaussian elimination and matrix inversion can be used. These methods provide a systematic way to solve complex systems of equations.

In the following sections, we will demonstrate the substitution and elimination methods to solve the given system of equations.

Solving the System Using the Elimination Method

The elimination method, also known as the addition method, is a powerful technique for solving systems of equations. It involves manipulating the equations so that the coefficients of one of the variables are opposites. By adding the equations together, one variable is eliminated, leaving a single equation with one variable that can be easily solved.

Let's apply the elimination method to the given system:

4x+7y=34x−y=−5\begin{array}{l} 4x + 7y = 3 \\ 4x - y = -5 \end{array}

Notice that the coefficients of x in both equations are already the same (4). To eliminate x, we can multiply the second equation by -1 and then add the two equations together.

Multiply the second equation by -1:

−1(4x−y)=−1(−5)⇒−4x+y=5-1(4x - y) = -1(-5) \Rightarrow -4x + y = 5

Now we have the following system:

4x+7y=3−4x+y=5\begin{array}{l} 4x + 7y = 3 \\ -4x + y = 5 \end{array}

Add the two equations together:

(4x+7y)+(−4x+y)=3+5(4x + 7y) + (-4x + y) = 3 + 5

8y=88y = 8

Divide both sides by 8 to solve for y:

y=88=1y = \frac{8}{8} = 1

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

4x+7y=34x + 7y = 3

Substitute y = 1:

4x+7(1)=34x + 7(1) = 3

4x+7=34x + 7 = 3

Subtract 7 from both sides:

4x=3−74x = 3 - 7

4x=−44x = -4

Divide both sides by 4 to solve for x:

x=−44=−1x = \frac{-4}{4} = -1

Therefore, the solution to the system of equations using the elimination method is x = -1 and y = 1.

Solving the System Using the Substitution Method

The substitution method is another effective technique for solving systems of equations. It involves solving one equation for one variable and then substituting that expression into the other equation. This process eliminates one variable and leaves a single equation with one unknown, which can be easily solved.

Let's apply the substitution method to the given system:

4x+7y=34x−y=−5\begin{array}{l} 4x + 7y = 3 \\ 4x - y = -5 \end{array}

First, we need to solve one of the equations for one variable. Let's solve the second equation for y:

4x−y=−54x - y = -5

Subtract 4x from both sides:

−y=−5−4x-y = -5 - 4x

Multiply both sides by -1:

y=4x+5y = 4x + 5

Now we have an expression for y in terms of x. Substitute this expression into the first equation:

4x+7y=34x + 7y = 3

4x+7(4x+5)=34x + 7(4x + 5) = 3

Distribute the 7:

4x+28x+35=34x + 28x + 35 = 3

Combine like terms:

32x+35=332x + 35 = 3

Subtract 35 from both sides:

32x=3−3532x = 3 - 35

32x=−3232x = -32

Divide both sides by 32 to solve for x:

x=−3232=−1x = \frac{-32}{32} = -1

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

y=4x+5y = 4x + 5

Substitute x = -1:

y=4(−1)+5y = 4(-1) + 5

y=−4+5y = -4 + 5

y=1y = 1

Therefore, the solution to the system of equations using the substitution method is x = -1 and y = 1.

Verification of the Solution

To ensure the accuracy of our solution, it's crucial to verify that the values we found for x and y satisfy both original equations. Let's substitute x = -1 and y = 1 into the original equations:

Equation 1:

4x+7y=34x + 7y = 3

4(−1)+7(1)=34(-1) + 7(1) = 3

−4+7=3-4 + 7 = 3

3=33 = 3

The first equation is satisfied.

Equation 2:

4x−y=−54x - y = -5

4(−1)−1=−54(-1) - 1 = -5

−4−1=−5-4 - 1 = -5

−5=−5-5 = -5

The second equation is also satisfied.

Since the values x = -1 and y = 1 satisfy both equations, we can confidently conclude that this is the correct solution to the system of equations.

Conclusion

In this article, we have demonstrated two methods, elimination and substitution, for solving the system of equations:

4x+7y=34x−y=−5\begin{array}{l} 4x + 7y = 3 \\ 4x - y = -5 \end{array}

Both methods led us to the same solution: x = -1 and y = 1. We also verified the solution by substituting the values back into the original equations. Mastering these techniques will empower you to solve a wide range of systems of equations, a valuable skill in various mathematical and real-world contexts. Understanding the properties of systems of equations and choosing the most efficient method for solving them are key to success in algebra and beyond. Whether you prefer the structured approach of elimination or the direct substitution method, practice and familiarity will make you proficient in solving these problems. Remember to always verify your solutions to ensure accuracy and build confidence in your problem-solving abilities.