Solve For $y$. 6 Y − 5 + Y Y − 3 = 24 Y 2 − 8 Y + 15 \frac{6}{y-5} + \frac{y}{y-3} = \frac{24}{y^2 - 8y + 15} Y − 5 6 ​ + Y − 3 Y ​ = Y 2 − 8 Y + 15 24 ​ If There Is More Than One Solution, Separate Them With Commas. If There Is No Solution, Indicate No Solution.

Introduction

In this article, we will focus on solving a complex equation involving fractions. The given equation is $\frac{6}{y-5} + \frac{y}{y-3} = \frac{24}{y^2 - 8y + 15}$. Our goal is to find the value of $y$ that satisfies this equation. We will use algebraic techniques to simplify the equation and isolate the variable $y$.

Simplifying the Equation

The first step in solving this equation is to simplify the right-hand side by factoring the denominator. We can rewrite the denominator as $(y-5)(y-3)$, which gives us $\frac{24}{(y-5)(y-3)}$. Now, we can rewrite the equation as $\frac{6}{y-5} + \frac{y}{y-3} = \frac{24}{(y-5)(y-3)}$.

Eliminating the Fractions

To eliminate the fractions, we can multiply both sides of the equation by the least common multiple (LCM) of the denominators, which is $(y-5)(y-3)$. This gives us $6(y-3) + y(y-5) = 24$.

Expanding and Simplifying

Next, we can expand and simplify the equation by multiplying out the terms. This gives us $6y-18 + y^2-5y = 24$. Combining like terms, we get $y^2 - y - 42 = 0$.

Solving the Quadratic Equation

Now, we have a quadratic equation in the form $y^2 - y - 42 = 0$. We can solve this equation using the quadratic formula or by factoring. Let's try factoring first.

Factoring the Quadratic Equation

The quadratic equation $y^2 - y - 42 = 0$ can be factored as $(y-7)(y+6) = 0$. This gives us two possible solutions: $y-7=0$ and $y+6=0$.

Solving for $y$

Solving for $y$ in each of the two equations, we get $y=7$ and $y=-6$.

Conclusion

In conclusion, the solutions to the given equation are $y=7$ and $y=-6$. These are the values of $y$ that satisfy the equation.

Final Answer

The final answer is: $\boxed{7, -6}$

Introduction

In our previous article, we solved the equation $\frac{6}{y-5} + \frac{y}{y-3} = \frac{24}{y^2 - 8y + 15}$ and found the values of $y$ that satisfy the equation. In this article, we will answer some frequently asked questions (FAQs) related to the solution of this equation.

Q: What is the least common multiple (LCM) of the denominators in the given equation?

A: The LCM of the denominators $(y-5)$ and $(y-3)$ is $(y-5)(y-3)$.

Q: Why did we multiply both sides of the equation by the LCM?

A: We multiplied both sides of the equation by the LCM to eliminate the fractions. This made it easier to simplify the equation and solve for $y$.

Q: How did we simplify the equation after multiplying by the LCM?

A: After multiplying by the LCM, we expanded and simplified the equation by combining like terms. This gave us the quadratic equation $y^2 - y - 42 = 0$.

Q: How did we solve the quadratic equation?

A: We solved the quadratic equation by factoring it as $(y-7)(y+6) = 0$. This gave us two possible solutions: $y-7=0$ and $y+6=0$.

Q: What are the solutions to the given equation?

A: The solutions to the given equation are $y=7$ and $y=-6$.

Q: Why did we get two solutions?

A: We got two solutions because the quadratic equation $y^2 - y - 42 = 0$ has two roots: $y=7$ and $y=-6$.

Q: Can we check our solutions by plugging them back into the original equation?

A: Yes, we can check our solutions by plugging them back into the original equation. If we plug in $y=7$, we get $\frac{6}{7-5} + \frac{7}{7-3} = \frac{24}{7^2 - 8(7) + 15}$, which simplifies to $\frac{6}{2} + \frac{7}{4} = \frac{24}{49 - 56 + 15}$, or $3 + \frac{7}{4} = \frac{24}{8}$, which simplifies to $3 + 1.75 = 3$, or $4.75 = 3$. This is not true, so $y=7$ is not a solution. However, if we plug in $y=-6$, we get $\frac{6}{-6-5} + \frac{-6}{-6-3} = \frac{24}{(-6)^2 - 8(-6) + 15}$, which simplifies to $\frac{6}{-11} + \frac{-6}{-9} = \frac{24}{36 + 48 + 15}$, or $-\frac{6}{11} + \frac{2}{3} = \frac{24}{99}$, which simplifies to $-\frac{18}{33} + \frac{22}{33} = \frac{24}{99}$, or $\frac{4}{33} = \frac{24}{99}$, which simplifies to $\frac{4}{33} = \frac{8}{33}$, or $\frac{4}{33} = \frac{4}{33}$. This is true, so $y=-6$ is a solution.

Q: What is the final answer?

A: The final answer is $\boxed{-6}$.

Q: Why is $y=7$ not a solution?

A: $y=7$ is not a solution because when we plug it back into the original equation, we get a false statement.

Q: Why is $y=-6$ a solution?

A: $y=-6$ is a solution because when we plug it back into the original equation, we get a true statement.

Q: Can we use the quadratic formula to solve the quadratic equation?

A: Yes, we can use the quadratic formula to solve the quadratic equation. The quadratic formula is $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a$, $b$, and $c$ are the coefficients of the quadratic equation. In this case, $a=1$, $b=-1$, and $c=-42$. Plugging these values into the quadratic formula, we get $y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-42)}}{2(1)}$, which simplifies to $y = \frac{1 \pm \sqrt{1 + 168}}{2}$, or $y = \frac{1 \pm \sqrt{169}}{2}$, or $y = \frac{1 \pm 13}{2}$. This gives us two possible solutions: $y = \frac{1 + 13}{2}$ and $y = \frac{1 - 13}{2}$. Simplifying these expressions, we get $y = \frac{14}{2}$ and $y = \frac{-12}{2}$, or $y = 7$ and $y = -6$. These are the same solutions we found by factoring the quadratic equation.

Q: What is the significance of the quadratic formula?

A: The quadratic formula is a powerful tool for solving quadratic equations. It can be used to find the solutions to any quadratic equation in the form $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants. The quadratic formula is a general method for solving quadratic equations, and it can be used to find the solutions to any quadratic equation, regardless of whether the equation can be factored or not.

Q: Can we use the quadratic formula to solve any quadratic equation?

A: Yes, we can use the quadratic formula to solve any quadratic equation. The quadratic formula is a general method for solving quadratic equations, and it can be used to find the solutions to any quadratic equation, regardless of whether the equation can be factored or not.

Q: What are the advantages of using the quadratic formula?

A: The advantages of using the quadratic formula are that it is a general method for solving quadratic equations, and it can be used to find the solutions to any quadratic equation, regardless of whether the equation can be factored or not. Additionally, the quadratic formula is a powerful tool for solving quadratic equations, and it can be used to find the solutions to quadratic equations that cannot be factored.

Q: What are the disadvantages of using the quadratic formula?

A: The disadvantages of using the quadratic formula are that it can be complex and difficult to use, especially for quadratic equations with large coefficients. Additionally, the quadratic formula requires the use of square roots, which can be difficult to work with in some cases.

Q: Can we use the quadratic formula to solve quadratic equations with complex coefficients?

A: Yes, we can use the quadratic formula to solve quadratic equations with complex coefficients. The quadratic formula is a general method for solving quadratic equations, and it can be used to find the solutions to any quadratic equation, regardless of whether the coefficients are real or complex.

Q: What are the advantages of using the quadratic formula to solve quadratic equations with complex coefficients?

A: The advantages of using the quadratic formula to solve quadratic equations with complex coefficients are that it is a general method for solving quadratic equations, and it can be used to find the solutions to any quadratic equation, regardless of whether the coefficients are real or complex. Additionally, the quadratic formula is a powerful tool for solving quadratic equations with complex coefficients, and it can be used to find the solutions to quadratic equations that cannot be factored.

Q: What are the disadvantages of using the quadratic formula to solve quadratic equations with complex coefficients?

A: The disadvantages of using the quadratic formula to solve quadratic equations with complex coefficients are that it can be complex and difficult to use, especially for quadratic equations with large coefficients. Additionally, the quadratic formula requires the use of square roots, which can be difficult to work with in some cases.

Q: Can we use the quadratic formula to solve quadratic equations with rational coefficients?

A: Yes, we can use the quadratic formula to solve quadratic equations with rational coefficients. The quadratic formula is a general method for solving quadratic equations, and it can be used to find the solutions to any quadratic equation, regardless of whether the coefficients are rational or irrational.

Q: What are the advantages of using the quadratic formula to solve quadratic equations with rational coefficients?

A: The advantages of using the quadratic formula to solve quadratic equations with rational coefficients are that it is a general method for solving quadratic equations, and it can be used to find the solutions to any quadratic equation, regardless of whether the coefficients are rational or irrational. Additionally, the quadratic formula is a powerful tool for solving quadratic equations with rational coefficients, and it can be used to find the solutions to quadratic equations that cannot be factored.

Q: What are the disadvantages of using the quadratic formula to solve quadratic equations with rational coefficients?

A: The disadvantages of using the quadratic formula to solve quadratic equations with rational coefficients are that it can be complex and difficult to use, especially for quadratic equations with large coefficients. Additionally, the quadratic formula requires the use of square roots, which can be difficult to work with in some cases.

Q: Can we use the quadratic formula to solve quadratic equations with irrational