Solving System Of Equations X + 2y = 8 And 2xy = 7 By Substitution Method

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Hey guys! Let's dive into solving systems of equations using the substitution method. We'll tackle a specific problem and break down each step, so you can ace these types of questions. We will discuss in detail how to solve the system of equations x + 2y = 8 and 2xy = 7, employing the substitution method to find the solution. We will provide a step-by-step explanation to ensure clarity and understanding of the process.

Problem Overview

The problem we're going to solve is this: Find the solution to the system of equations x + 2y = 8 and 2xy = 7 using the substitution method. We have four options to choose from:

A) (2, 3) B) (3, 2) C) (1, 3.5) D) (4, 0)

So, let's get started!

Step-by-Step Solution Using the Substitution Method

Step 1 Isolating a Variable in One Equation

The first thing we need to do in the substitution method is to isolate one variable in one of the equations. Looking at our two equations:

  1. x + 2y = 8
  2. 2xy = 7

The first equation, x + 2y = 8, looks simpler to manipulate. Let’s isolate x in this equation. To do this, we subtract 2y from both sides:

x = 8 - 2y

Now we have x isolated, which is a crucial step for the substitution method. Isolating a variable makes it easier to substitute its value into the other equation, simplifying the process of finding the solution. This initial step sets the stage for the subsequent steps, where we will substitute this expression into the second equation and solve for y.

Step 2 Substituting the Isolated Variable into the Other Equation

Now that we have x = 8 - 2y, we can substitute this expression for x in the second equation, which is 2xy = 7. By replacing x with (8 - 2y), we will have an equation in terms of y only. This substitution is a key step in solving the system of equations, as it allows us to reduce the system to a single equation with a single variable. This makes the equation solvable using algebraic techniques. Substituting x, we get:

2(8 - 2y)y = 7

This new equation now only involves the variable y, which we can solve in the next steps. The substitution process is a powerful technique for solving systems of equations because it simplifies the problem by reducing the number of variables in the equation. This makes it easier to find the values of the variables that satisfy both equations simultaneously.

Step 3: Expanding and Simplifying the Equation

Let's expand and simplify the equation we got after the substitution: 2(8 - 2y)y = 7. First, distribute the 2y across the terms inside the parentheses:

16y - 4y² = 7

Now, to solve this quadratic equation, let’s rearrange it into the standard form (ay² + by + c = 0). Subtract 7 from both sides to set the equation to zero:

-4y² + 16y - 7 = 0

To make it easier to work with, we can multiply the entire equation by -1:

4y² - 16y + 7 = 0

Now we have a standard quadratic equation. This simplification is essential because it prepares the equation for solving using methods like factoring, completing the square, or the quadratic formula. By rearranging the terms and setting the equation to zero, we create a form that is conducive to these solution techniques. This step is a bridge between the substitution and the actual solving of the variable's value.

Step 4 Solving the Quadratic Equation

We've got the quadratic equation 4y² - 16y + 7 = 0. We can solve this using the quadratic formula, which is given by:

y = (-b ± √(b² - 4ac)) / (2a)

In our equation:

a = 4 b = -16 c = 7

Plug these values into the formula:

y = (16 ± √((-16)² - 4 * 4 * 7)) / (2 * 4) y = (16 ± √(256 - 112)) / 8 y = (16 ± √144) / 8 y = (16 ± 12) / 8

So we have two possible solutions for y:

y₁ = (16 + 12) / 8 = 28 / 8 = 3.5 y₂ = (16 - 12) / 8 = 4 / 8 = 0.5

Thus, the solutions for y are 3.5 and 0.5. Solving the quadratic equation is a critical step as it provides the possible values for the variable y. These values are essential for finding the corresponding values of x, which will eventually give us the solution pairs for the system of equations. The quadratic formula is a reliable method for solving quadratic equations, especially when factoring is not straightforward.

Step 5 Finding the Corresponding Values of x

Now that we have the values for y, y₁ = 3.5 and y₂ = 0.5, we need to find the corresponding values of x. We can use the equation we derived earlier, x = 8 - 2y:

For y₁ = 3.5: x₁ = 8 - 2 * 3.5 x₁ = 8 - 7 x₁ = 1

For y₂ = 0.5: x₂ = 8 - 2 * 0.5 x₂ = 8 - 1 x₂ = 7

So, we have two pairs of solutions: (1, 3.5) and (7, 0.5). This step is crucial because it completes the process of finding the solutions to the system of equations. By substituting each value of y back into the equation for x, we obtain the corresponding x values. This gives us the ordered pairs (x, y) that satisfy both equations in the system. This process ensures that we find all possible solutions to the system.

Step 6 Verifying the Solutions

It’s always a good idea to verify our solutions by plugging them back into the original equations to make sure they hold true.

For the pair (1, 3.5):

  1. x + 2y = 8 1 + 2 * 3.5 = 1 + 7 = 8 (True)
  2. 2xy = 7 2 * 1 * 3.5 = 7 (True)

For the pair (7, 0.5):

  1. x + 2y = 8 7 + 2 * 0.5 = 7 + 1 = 8 (True)
  2. 2xy = 7 2 * 7 * 0.5 = 7 (True)

Both pairs of solutions satisfy the original equations. This verification step is vital to ensure the accuracy of our solutions. By plugging the values back into the original equations, we confirm that the solutions satisfy both equations simultaneously. This step helps to catch any errors that might have occurred during the solving process and provides confidence in the final answer. If a solution does not satisfy both equations, it is not a valid solution to the system.

Final Answer and Justification

From the given options, we found that (1, 3.5) is one of the solutions. The other solution we found was (7, 0.5), but this was not among the options. So, the correct answer is:

C) (1, 3.5)

We arrived at this answer by using the substitution method, which involves isolating one variable in one equation, substituting that expression into the other equation, solving for the remaining variable, and then finding the corresponding value of the first variable. We verified our solution by plugging the values back into the original equations.

Conclusion

Solving systems of equations using the substitution method might seem tricky at first, but with practice, you'll get the hang of it. Remember to isolate, substitute, solve, and verify! This step-by-step approach can be applied to various systems of equations, making it a valuable tool in algebra. Always remember to check your solutions to ensure accuracy. Keep practicing, and you’ll become a pro at solving these problems. If you guys have any questions, feel free to ask! Understanding these methods opens up a wide range of problem-solving capabilities in mathematics and related fields.