Solving A System Of Equations Find The Value Of Xy

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Introduction

This article delves into solving a system of equations to find the value of $xy$, given that $x$ is negative. We will explore the steps involved in solving the system and highlight key mathematical concepts used in the process. This problem combines algebraic manipulation and substitution to arrive at the solution, making it an excellent example of how different mathematical techniques can be integrated. Understanding such problems enhances problem-solving skills and provides a solid foundation for more advanced mathematical topics. The question requires a methodical approach, starting from simplifying the given equations and then using substitution to find the values of $x$ and $y$. Once these values are determined, we can easily compute the product $xy$. The constraint $x<0$ adds another layer of specificity, ensuring that we select the correct solution from the possible options. This type of question is common in algebra and precalculus courses and is crucial for developing analytical and problem-solving abilities.

Problem Statement

We are given the following system of equations:

$ \begin{cases} x^2 = y \ 6x + 9 = -3(2y - 3) \end{cases} $

We need to find the value of $xy$, given that $x < 0$.

Step-by-Step Solution

1. Simplify the Second Equation

Let's start by simplifying the second equation:

$ 6x + 9 = -3(2y - 3) $

Distribute the -3 on the right side:

$ 6x + 9 = -6y + 9 $

Subtract 9 from both sides:

$ 6x = -6y $

Divide both sides by 6:

$ x = -y $

2. Substitute into the First Equation

Now, we substitute $x = -y$ into the first equation, $x^2 = y$:

$ (-y)^2 = y $

$ y^2 = y $

3. Solve for y

Rearrange the equation to solve for $y$:

$ y^2 - y = 0 $

Factor out $y$:

$ y(y - 1) = 0 $

This gives us two possible solutions for $y$:

$ y = 0 $ or $ y = 1 $

4. Solve for x

Using the relation $x = -y$, we find the corresponding values of $x$:

If $y = 0$, then $x = -0 = 0$. If $y = 1$, then $x = -1$.

5. Apply the Condition $x < 0$

We are given that $x < 0$. From our solutions, we have:

$ x = 0 $ (which does not satisfy $x < 0$) $ x = -1 $ (which satisfies $x < 0$)

Thus, we take $x = -1$ and the corresponding $y = 1$.

6. Calculate $xy$

Now we compute the value of $xy$:

$ xy = (-1)(1) = -1 $

Therefore, the value of $xy$ is -1.

Detailed Explanation of Each Step

Simplifying the Second Equation

In the first step, we simplified the second equation, $6x + 9 = -3(2y - 3)$. The purpose of this simplification is to express one variable in terms of the other, which will be crucial for substitution. We started by distributing the -3 on the right side, resulting in $6x + 9 = -6y + 9$. This expansion allows us to combine like terms and isolate variables. Next, we subtracted 9 from both sides, which eliminated the constant term on both sides, leading to $6x = -6y$. Finally, we divided both sides by 6 to obtain the simplified form $x = -y$. This simple equation provides a direct relationship between $x$ and $y$, making it easier to substitute into the other equation and solve for the variables. This step is a standard algebraic technique, emphasizing the importance of simplification to make equations more manageable.

Substituting into the First Equation

The second step involves substituting the simplified expression $x = -y$ into the first equation, $x^2 = y$. Substitution is a powerful method in solving systems of equations because it allows us to reduce the number of variables in an equation. By replacing $x$ with $-y$ in the first equation, we transformed it into $(-y)^2 = y$. Squaring $-y$ gives us $y^2$, so the equation becomes $y^2 = y$. This new equation involves only one variable, $y$, making it solvable. The substitution technique is a cornerstone of algebra, enabling us to tackle more complex systems by breaking them down into simpler parts. This step effectively sets the stage for finding the possible values of $y$, which will then help us determine the corresponding values of $x$.

Solving for $y$

In the third step, we solved the quadratic equation $y^2 = y$ for $y$. To do this, we first rearranged the equation into the standard quadratic form by subtracting $y$ from both sides, resulting in $y^2 - y = 0$. This form is essential for applying factoring techniques. We then factored out $y$ from the left side, which gave us $y(y - 1) = 0$. This factored form allows us to use the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. Applying this property, we set each factor equal to zero, yielding two possible solutions: $y = 0$ and $y - 1 = 0$. Solving $y - 1 = 0$ gives us $y = 1$. Thus, we have two potential values for $y$, namely 0 and 1. Finding these values is a critical step, as they will be used to find the corresponding values of $x$. This step highlights the importance of understanding quadratic equations and factoring in solving algebraic problems.

Solving for $x$

In the fourth step, we found the corresponding values of $x$ using the relationship $x = -y$, which we derived from simplifying the second original equation. We had two possible values for $y$, which were $y = 0$ and $y = 1$. For each value of $y$, we substituted it into the equation $x = -y$ to find the corresponding $x$ value. When $y = 0$, we have $x = -0 = 0$. When $y = 1$, we have $x = -1$. This process gave us two pairs of solutions: $(0, 0)$ and $(-1, 1)$. Each pair represents a potential solution to the system of equations. However, we must consider the additional condition given in the problem statement to determine which solution is correct. This step demonstrates how a previously established relationship between variables can be used to find specific values, emphasizing the interconnectedness of different parts of a mathematical solution.

Applying the Condition $x < 0$

The fifth step involved applying the condition $x < 0$, which was given in the problem statement. We had two potential solutions for $x$: $x = 0$ and $x = -1$. The condition $x < 0$ means that $x$ must be a negative number. Therefore, we examined our solutions to see which one satisfied this condition. The solution $x = 0$ does not satisfy $x < 0$ because 0 is not a negative number. However, the solution $x = -1$ does satisfy $x < 0$ because -1 is a negative number. Thus, we selected $x = -1$ as the correct value for $x$. The corresponding value of $y$ when $x = -1$ is $y = 1$. This step underscores the importance of considering all given conditions when solving a problem, as these conditions often help narrow down the possible solutions to a single correct answer. This ensures that the final solution not only satisfies the equations but also the specific constraints of the problem.

Calculating $xy$

The final step was to calculate the value of $xy$ using the values of $x$ and $y$ that we determined in the previous steps. We found that $x = -1$ and $y = 1$. To find $xy$, we simply multiplied these values together: $xy = (-1)(1) = -1$. Therefore, the value of $xy$ is -1. This step is a straightforward application of the multiplication operation, but it is a critical step because it provides the final answer to the problem. It demonstrates how all the previous steps come together to solve the original question. Calculating $xy$ is the culmination of the algebraic manipulations and logical deductions made throughout the solution process, highlighting the importance of each step in arriving at the final result.

Final Answer

The value of $xy$ is -1.

Conclusion

In this article, we solved a system of equations to find the value of $xy$, given that $x < 0$. The problem required us to simplify equations, use substitution, solve for variables, and apply a given condition to arrive at the correct solution. The steps included simplifying the second equation, substituting into the first equation, solving for $y$, solving for $x$, applying the condition $x < 0$, and finally calculating $xy$. The detailed explanation of each step highlights the importance of algebraic manipulation and logical reasoning in solving mathematical problems. Understanding and practicing such problems enhances one's problem-solving skills and provides a strong foundation for more advanced mathematical topics. The final answer, $xy = -1$, was obtained by systematically working through each step and applying the given condition. This approach demonstrates the power of a methodical approach in mathematics and the importance of accuracy in each step of the solution process.