Solving 2x + 5y = 12 And 15y = -6x By Substitution

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When faced with a system of equations, the goal is to find the values of the variables that satisfy all equations simultaneously. There are several methods to achieve this, and one of the most powerful and versatile is the substitution method. In this article, we will delve deep into the substitution method, breaking it down into manageable steps and illustrating its application with a detailed example. We aim to provide a comprehensive understanding of how to use substitution to solve systems of equations, enhancing your problem-solving skills in mathematics and related fields. This article provides a detailed solution to the system of equations 2x + 5y = 12 and 15y = -6x using the substitution method. Understanding how to solve systems of equations is crucial in various fields, including mathematics, physics, engineering, and economics. Mastering this method will equip you with a powerful tool for tackling more complex problems.

The substitution method involves solving one equation for one variable and substituting that expression into the other equation. This process transforms the system into a single equation with a single 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. This method is particularly useful when one of the equations is already solved for one variable or can be easily manipulated to do so. The key to the substitution method is to isolate one variable in one of the equations. This isolated variable's expression is then substituted into the other equation, eliminating one variable and creating a solvable equation in a single variable. This process simplifies the system, making it easier to find the solution. Let's explore the general steps involved in solving systems of equations using the substitution method:

  1. Isolate a Variable: Choose one equation and solve it for one of the variables. This means expressing one variable in terms of the other. For example, if you have the equation x + y = 5, you might solve for x to get x = 5 - y.
  2. Substitute: Substitute the expression obtained in step 1 into the other equation. This will result in a new equation with only one variable. For example, if you have another equation 2x - y = 4, you would substitute 5 - y for x to get 2(5 - y) - y = 4.
  3. Solve: Solve the new equation for the remaining variable. This will give you the numerical value of one of the variables. Continuing the example, 2(5 - y) - y = 4 simplifies to 10 - 2y - y = 4, then -3y = -6, and finally y = 2.
  4. Back-Substitute: Substitute the value found in step 3 back into either of the original equations (or the expression from step 1) to find the value of the other variable. In our example, substituting y = 2 into x = 5 - y gives x = 5 - 2, so x = 3.
  5. Check: Check your solution by substituting both values into both original equations to ensure they are satisfied. This step verifies the accuracy of your solution. For our example, substituting x = 3 and y = 2 into x + y = 5 gives 3 + 2 = 5, which is true. Substituting into 2x - y = 4 gives 2(3) - 2 = 4, which is also true. Thus, our solution is correct.

By following these steps, you can systematically solve any system of equations using the substitution method. This technique is a fundamental tool in algebra and is essential for solving a wide range of mathematical problems.

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

Step 1: Isolate a Variable

We have two equations:

  1. 2x + 5y = 12
  2. 15y = -6x

The second equation, 15y = -6x, looks easier to manipulate to isolate a variable. Let's solve this equation for y:

  • Divide both sides by 15: y = (-6x) / 15
  • Simplify the fraction: y = -2x / 5

Now we have expressed y in terms of x. This is a crucial step in the substitution method, as it allows us to replace y in the other equation with an expression involving only x. This simplification is key to solving the system.

Step 2: Substitute

Substitute the expression for y (y = -2x / 5) into the first equation:

  • Original first equation: 2x + 5y = 12
  • Substitute y: 2x + 5(-2x / 5) = 12

This substitution replaces y in the first equation, resulting in a new equation that contains only x. This is the essence of the substitution method: reducing the system to a single equation with one variable.

Step 3: Solve

Now, solve the resulting equation for x:

  • Simplify: 2x - 2x = 12
  • Combine like terms: 0 = 12

This result, 0 = 12, is a contradiction. It indicates that there is no value of x that can satisfy this equation. This outcome is significant because it tells us something important about the system of equations we are trying to solve.

Step 4: Interpret the Result

The contradiction 0 = 12 implies that the system of equations has no solution. This means that the two equations represent lines that do not intersect in the Cartesian plane. They are parallel lines.

In the context of solving systems of equations, encountering a contradiction like this is a definitive indicator that the system is inconsistent. Inconsistent systems have no solutions, meaning there are no values for the variables that will satisfy all equations simultaneously. This is a common outcome when dealing with systems of linear equations, and it's important to recognize and interpret such results correctly.

To further solidify our understanding that the system has no solution, let's consider alternative approaches to verify this result. These methods provide additional perspectives and can be useful in different scenarios.

Method 1: Convert Both Equations to Slope-Intercept Form

Converting equations to slope-intercept form (y = mx + b) allows us to easily compare their slopes and y-intercepts. If the slopes are the same and the y-intercepts are different, the lines are parallel and there is no solution.

  1. Equation 1: 2x + 5y = 12

    • Subtract 2x from both sides: 5y = -2x + 12
    • Divide by 5: y = (-2/5)x + 12/5

  2. Equation 2: 15y = -6x

    • Divide by 15: y = (-6/15)x
    • Simplify: y = (-2/5)x

Comparing the two equations in slope-intercept form:

  • Equation 1: y = (-2/5)x + 12/5 (slope = -2/5, y-intercept = 12/5)
  • Equation 2: y = (-2/5)x (slope = -2/5, y-intercept = 0)

Both equations have the same slope (-2/5) but different y-intercepts (12/5 and 0). This confirms that the lines are parallel and do not intersect, hence there is no solution.

Method 2: Check for Proportionality

Another way to check for no solution is to see if the coefficients of x and y are proportional, but the constants are not. This also indicates parallel lines.

  1. Equation 1: 2x + 5y = 12
  2. Equation 2: 15y = -6x, rewrite as 6x + 15y = 0

Check the ratios of the coefficients:

  • Ratio of x-coefficients: 6/2 = 3
  • Ratio of y-coefficients: 15/5 = 3

The ratios of the x and y coefficients are the same (both are 3), but the ratio of the constants is different (0/12 = 0). This indicates that the lines are parallel and there is no solution.

Importance of Verification

Verifying the result using alternative methods is a crucial step in problem-solving. It not only confirms the correctness of the solution but also enhances understanding of the underlying concepts. In this case, by converting the equations to slope-intercept form and checking for proportionality, we gain additional insights into why the system has no solution.

In summary, when we applied the substitution method to the system of equations 2x + 5y = 12 and 15y = -6x, we arrived at a contradiction (0 = 12). This contradiction definitively indicates that the system has no solution. Further verification by converting the equations to slope-intercept form and checking for proportionality confirmed that the lines are parallel and do not intersect.

Understanding how to recognize and interpret contradictions is crucial in solving systems of equations. It allows us to identify inconsistent systems and avoid wasting time trying to find a solution that does not exist. The substitution method, along with alternative verification techniques, provides a comprehensive approach to solving and analyzing systems of equations. Mastering these skills will be invaluable in various mathematical and real-world applications.