Solving 3w - 4v = 62 And 2w - V = 8 A Step-by-Step Guide

by Scholario Team 57 views

This article provides a comprehensive guide to solving the system of linear equations:

\begin{align*} 3w - 4v &= 62 \ 2w - v &= 8 \end{align*}

We will explore various methods, including substitution, elimination, and graphical approaches, to find the values of w and v that satisfy both equations. Understanding these methods is crucial for various applications in mathematics, science, and engineering. Let's dive in!

Introduction to Systems of Linear Equations

Before we tackle the specific system, let's briefly discuss what systems of linear equations are and why they are important. A system of linear equations is a set of two or more linear equations containing the same variables. The solution to a system of linear equations is the set of values for the variables that make all the equations true simultaneously. In simpler terms, it's the point where the lines represented by the equations intersect on a graph.

Systems of linear equations pop up everywhere in real-world applications. Think about mixing solutions in chemistry, balancing budgets in finance, or even optimizing routes in logistics. The ability to solve these systems efficiently is a valuable skill. Now, let's get back to our specific example. We have two equations:

\begin{align*} 3w - 4v &= 62 \ 2w - v &= 8 \end{align*}

Our goal is to find the values of w and v that satisfy both of these equations. We'll start with the method of substitution, a powerful technique for solving such systems.

Method 1: Solving by Substitution

The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This effectively reduces the system to a single equation with one variable, which we can then solve. Let's apply this to our system.

  1. Solve one equation for one variable: Looking at our equations, the second equation, 2w - v = 8, seems easier to manipulate. Let's solve it for v:

    \begin{align*} 2w - v &= 8 \ -v &= 8 - 2w \ v &= 2w - 8 \end{align*}

    We now have an expression for v in terms of w: v = 2w - 8. This is a crucial step in the substitution process.

  2. Substitute into the other equation: Now, we'll substitute this expression for v into the first equation, 3w - 4v = 62:

    \begin{align*} 3w - 4(2w - 8) &= 62 \end{align*}

    Notice that we've replaced v with (2w - 8). This gives us a single equation with only w as the variable.

  3. Solve for the remaining variable: Let's simplify and solve for w:

    \begin{align*} 3w - 8w + 32 &= 62 \ -5w &= 30 \ w &= -6 \end{align*}

    So, we've found that w = -6. This is one piece of our solution. Now we need to find v.

  4. Substitute back to find the other variable: We can substitute the value of w back into either of our original equations or the expression we found for v. Let's use v = 2w - 8:

    \begin{align*} v &= 2(-6) - 8 \ v &= -12 - 8 \ v &= -20 \end{align*}

    Therefore, v = -20. We now have both values: w = -6 and v = -20.

  5. Check the solution: It's always a good idea to check our solution by plugging the values of w and v back into the original equations:

    • Equation 1: 3w - 4v = 62 becomes 3(-6) - 4(-20) = -18 + 80 = 62 (Correct!)
    • Equation 2: 2w - v = 8 becomes 2(-6) - (-20) = -12 + 20 = 8 (Correct!)

    Since our solution satisfies both equations, we've successfully solved the system using the substitution method. This meticulous approach ensures accuracy and builds confidence in our answer. Now, let's explore another method: elimination.

Method 2: Solving by Elimination

The elimination method involves manipulating the equations so that when they are added together, one of the variables cancels out. This leaves us with a single equation in one variable, which we can solve. Then, we substitute back to find the other variable. Let's apply this to our system:

\begin{align*} 3w - 4v &= 62 \ 2w - v &= 8 \end{align*}

  1. Multiply equations to match coefficients: Our goal is to make either the w coefficients or the v coefficients the same (but with opposite signs). Let's focus on the v coefficients. We can multiply the second equation by -4:

    \begin{align*} -4(2w - v) &= -4(8) \ -8w + 4v &= -32 \end{align*}

    Now our system looks like this:

    \begin{align*} 3w - 4v &= 62 \ -8w + 4v &= -32 \end{align*}

    Notice that the v coefficients are now -4 and 4, which are opposites.

  2. Add the equations: Now, we add the two equations together:

    \begin{align*} (3w - 4v) + (-8w + 4v) &= 62 + (-32) \ -5w &= 30 \end{align*}

    The v terms have canceled out, leaving us with an equation in just w.

  3. Solve for the remaining variable: We can now solve for w:

    \begin{align*} -5w &= 30 \ w &= -6 \end{align*}

    We've found w = -6, which agrees with our result from the substitution method. This consistency is reassuring! Now, let's find v.

  4. Substitute back to find the other variable: We can substitute w = -6 into either of the original equations. Let's use the second equation, 2w - v = 8:

    \begin{align*} 2(-6) - v &= 8 \ -12 - v &= 8 \ -v &= 20 \ v &= -20 \end{align*}

    So, v = -20, which again matches our previous result.

  5. Check the solution: Just as before, let's verify our solution by plugging w = -6 and v = -20 back into the original equations:

    • Equation 1: 3w - 4v = 62 becomes 3(-6) - 4(-20) = -18 + 80 = 62 (Correct!)
    • Equation 2: 2w - v = 8 becomes 2(-6) - (-20) = -12 + 20 = 8 (Correct!)

    Our solution w = -6 and v = -20 satisfies both equations, confirming that we've correctly solved the system using the elimination method. This method provides an alternative approach, showcasing the flexibility in solving linear systems. Let's briefly touch on a third method: graphical solutions.

Method 3: Graphical Solution (Brief Overview)

While we've focused on algebraic methods, it's worth mentioning the graphical method for solving systems of linear equations. Each equation represents a line on a coordinate plane. The solution to the system is the point where the lines intersect. To solve graphically:

  1. Graph both equations on the same coordinate plane.
  2. Identify the point of intersection.
  3. The coordinates of the intersection point represent the solution (w, v).

In our case, graphing 3w - 4v = 62 and 2w - v = 8 would show that they intersect at the point (-6, -20), visually confirming our algebraic solutions. While graphical methods are useful for visualization, algebraic methods tend to be more precise for complex systems.

Conclusion

We have thoroughly explored how to solve the system of linear equations:

\begin{align*} 3w - 4v &= 62 \ 2w - v &= 8 \end{align*}

We successfully found the solution w = -6 and v = -20 using both the substitution and elimination methods. We also briefly discussed the graphical method. Understanding these different techniques provides a robust toolkit for tackling systems of linear equations in various contexts. Remember to always check your solution to ensure accuracy. Mastering these methods is a valuable asset in mathematics and its applications.

This detailed walkthrough aims to provide a clear and comprehensive understanding of solving systems of linear equations. By understanding the nuances of each method, you'll be well-equipped to tackle a wide range of problems in mathematics and beyond. Keep practicing, and you'll become a confident problem-solver!