Solving 2x+y=4 And X+3y=6 A Step-by-Step Guide

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Linear equations, those straightforward mathematical expressions, play a crucial role in various fields, from engineering to economics. Understanding how to solve systems of linear equations is a fundamental skill. In this article, we're going to dive into solving a specific system of two linear equations: 2x + y = 4 and x + 3y = 6. We'll explore different methods to find the values of x and y that satisfy both equations simultaneously. So, buckle up, math enthusiasts, and let's get started!

Understanding Systems of Linear Equations

Before we jump into solving our specific system, let's make sure we're all on the same page about what systems of linear equations actually are. Linear equations, at their core, represent straight lines when graphed. A system of linear equations is simply a collection of two or more of these equations. The solution to a system of linear equations is the set of values for the variables (in our case, x and y) that make all the equations in the system true at the same time. Graphically, this solution represents the point(s) where the lines intersect. Think of it like finding the common ground between these lines – the point where they agree.

There are typically three possible outcomes when dealing with a system of two linear equations:

  1. Unique Solution: The lines intersect at exactly one point, indicating a single, unique solution for x and y. This is the most common scenario we'll encounter.
  2. No Solution: The lines are parallel and never intersect. In this case, there are no values of x and y that can satisfy both equations simultaneously. The system is considered inconsistent.
  3. Infinite Solutions: The lines are actually the same line (they overlap completely). Any point on the line represents a solution, leading to an infinite number of solutions. The equations are dependent.

Knowing these possibilities helps us interpret the results we get when we start solving. We'll be aiming to find a unique solution for our system, but it's good to be aware of the other possibilities.

Methods for Solving Systems of Linear Equations

There are several methods available for solving systems of linear equations, each with its own strengths and weaknesses. We'll focus on two popular methods: the substitution method and the elimination method. These methods are widely used and provide a clear, step-by-step approach to finding solutions. Let's briefly discuss each method before applying them to our specific system.

1. The Substitution Method

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 a single variable, which is much easier to solve. Once you've found the value of one variable, you can substitute it back into either of the original equations to find the value of the other variable. Think of it as replacing one piece of the puzzle with an equivalent expression to simplify the whole picture.

For example, if we have the equations x + y = 5 and 2x - y = 1, we could solve the first equation for x to get x = 5 - y. Then, we would substitute this expression for x into the second equation: 2(5 - y) - y = 1. Now we have a single equation with only y, which we can solve. Once we find y, we can plug it back into x = 5 - y to find x.

2. The Elimination Method

The elimination method (also sometimes called the addition method) focuses on eliminating one of the variables by adding or subtracting the equations. This requires manipulating the equations so that the coefficients of one of the variables are opposites (e.g., 2x and -2x). When you add the equations, that variable disappears, leaving you with a single equation in one variable. Similar to the substitution method, you solve for the remaining variable and then substitute back into one of the original equations to find the other variable. This method is particularly effective when the equations are already set up in a way that makes elimination straightforward.

For instance, consider the system 3x + 2y = 7 and x - 2y = -1. Notice that the y terms have opposite coefficients (+2 and -2). If we add these equations together, the y terms will cancel out: (3x + 2y) + (x - 2y) = 7 + (-1), which simplifies to 4x = 6. Now we can solve for x. After finding x, we substitute it back into either of the original equations to solve for y.

Solving 2x + y = 4 and x + 3y = 6 using Substitution Method

Okay, let's roll up our sleeves and tackle our system of equations: 2x + y = 4 and x + 3y = 6. We'll start with the substitution method. Remember, the goal is to solve one equation for one variable and then substitute that expression into the other equation. Looking at our equations, it seems easiest to solve the first equation for y. Here's how we do it:

  1. Solve the first equation for y:
    • 2x + y = 4
    • Subtract 2x from both sides: y = 4 - 2x

Now we have an expression for y in terms of x. The next step is to substitute this expression into the second equation.

  1. Substitute the expression for y into the second equation:
    • x + 3y = 6
    • Replace y with (4 - 2x): x + 3(4 - 2x) = 6

Now we have an equation with only x as the variable. Let's simplify and solve for x.

  1. Simplify and solve for x:
    • x + 3(4 - 2x) = 6
    • Distribute the 3: x + 12 - 6x = 6
    • Combine like terms: -5x + 12 = 6
    • Subtract 12 from both sides: -5x = -6
    • Divide both sides by -5: x = 6/5

Great! We've found the value of x: x = 6/5. Now we need to find the value of y. We can do this by substituting the value of x back into either of the original equations or the expression we found for y in step 1. Let's use the expression y = 4 - 2x as it's already solved for y.

  1. Substitute the value of x back into the expression for y:
    • y = 4 - 2x
    • Replace x with 6/5: y = 4 - 2(6/5)
    • Simplify: y = 4 - 12/5
    • Find a common denominator: y = 20/5 - 12/5
    • Simplify: y = 8/5

So, we've found that y = 8/5. Therefore, the solution to the system of equations using the substitution method is x = 6/5 and y = 8/5. We can write this as an ordered pair: (6/5, 8/5).

Solving 2x + y = 4 and x + 3y = 6 using Elimination Method

Now, let's tackle the same system of equations – 2x + y = 4 and x + 3y = 6 – but this time using the elimination method. The key to the elimination method is to manipulate the equations so that the coefficients of one of the variables are opposites. This way, when we add the equations together, that variable will be eliminated. Looking at our system, it might be easiest to eliminate x. To do this, we need to make the coefficients of x in the two equations opposites. We can multiply the second equation by -2.

  1. Multiply the second equation by -2:
    • x + 3y = 6
    • Multiply both sides by -2: -2(x + 3y) = -2(6)
    • Distribute: -2x - 6y = -12

Now our system looks like this:

*   2x + y = 4
*   -2x - 6y = -12

Notice that the coefficients of x are now 2 and -2, which are opposites. We can proceed to the next step: adding the equations together.

  1. Add the equations together:
    • (2x + y) + (-2x - 6y) = 4 + (-12)
    • Combine like terms: -5y = -8

Now we have a simple equation with only y. Let's solve for y.

  1. Solve for y:
    • -5y = -8
    • Divide both sides by -5: y = 8/5

We've found that y = 8/5. Now we need to find the value of x. We can substitute the value of y back into either of the original equations. Let's use the first equation, 2x + y = 4.

  1. Substitute the value of y back into the first equation:
    • 2x + y = 4
    • Replace y with 8/5: 2x + 8/5 = 4

Now we have an equation with only x. Let's solve for x.

  1. Solve for x:
    • 2x + 8/5 = 4
    • Subtract 8/5 from both sides: 2x = 4 - 8/5
    • Find a common denominator: 2x = 20/5 - 8/5
    • Simplify: 2x = 12/5
    • Divide both sides by 2: x = (12/5) / 2
    • Simplify: x = 6/5

So, we've found that x = 6/5. Therefore, the solution to the system of equations using the elimination method is x = 6/5 and y = 8/5, which is the same solution we found using the substitution method! Again, we can write this as an ordered pair: (6/5, 8/5).

Verifying the Solution

It's always a good practice to verify your solution to make sure it's correct. To do this, we simply substitute the values we found for x and y back into the original equations. If both equations are true, then our solution is correct. Let's verify our solution (6/5, 8/5) for the system 2x + y = 4 and x + 3y = 6.

  1. Verify in the first equation (2x + y = 4):

    • Substitute x = 6/5 and y = 8/5: 2(6/5) + 8/5 = 4
    • Simplify: 12/5 + 8/5 = 4
    • Combine fractions: 20/5 = 4
    • Simplify: 4 = 4 (This is true!)

  2. Verify in the second equation (x + 3y = 6):

    • Substitute x = 6/5 and y = 8/5: 6/5 + 3(8/5) = 6
    • Simplify: 6/5 + 24/5 = 6
    • Combine fractions: 30/5 = 6
    • Simplify: 6 = 6 (This is also true!)

Since our solution satisfies both equations, we can confidently say that (6/5, 8/5) is the correct solution to the system of linear equations.

Conclusion

In this article, we've successfully solved the system of linear equations 2x + y = 4 and x + 3y = 6 using both the substitution and elimination methods. We found that the solution is x = 6/5 and y = 8/5, or the ordered pair (6/5, 8/5). We also verified our solution to ensure its accuracy. Understanding how to solve systems of linear equations is a valuable skill in mathematics and many other disciplines. By mastering these methods, you'll be well-equipped to tackle a wide range of problems involving linear relationships. So keep practicing, and you'll become a pro at solving systems of equations in no time!