Janet Wants To Solve The Equation $y+\frac{y^2-5}{y^2-1}=\frac{y^2+y+2}{y+1}$. What Should She Multiply Both Sides Of The Equation By?A. $y$B. $y^2-1$C. $y+1$D. $y^2+y+2$

Introduction

Rational equations are a type of algebraic equation that involves fractions with polynomials in both the numerator and denominator. Solving these equations can be challenging, but with the right approach, it can be done efficiently. In this article, we will focus on solving a specific rational equation and explore the steps involved in finding the solution.

The Equation

Janet wants to solve the equation:

$$y+\frac{y^2-5}{y^2-1}=\frac{y^2+y+2}{y+1}$$

To solve this equation, we need to eliminate the fractions by multiplying both sides by a suitable expression.

Multiplying Both Sides

When multiplying both sides of an equation by an expression, we need to make sure that the expression cancels out the denominators of the fractions. In this case, we can see that the denominators are $y^2-1$ and $y+1$. To eliminate these denominators, we need to multiply both sides by a factor that contains both $y^2-1$ and $y+1$.

Choosing the Correct Factor

Let's analyze the options given:

A. $y$ B. $y^2-1$ C. $y+1$ D. $y^2+y+2$

We can see that options B and C contain the denominators of the fractions. However, we need to choose a factor that will cancel out both denominators. Let's try multiplying both sides by $y^2-1$.

Multiplying by $y^2-1$

If we multiply both sides of the equation by $y^2-1$, we get:

$$(y^2-1)\left(y+\frac{y^2-5}{y^2-1}\right)=(y^2-1)\left(\frac{y^2+y+2}{y+1}\right)$$

Simplifying the left-hand side, we get:

$$(y^2-1)y+(y^2-1)\frac{y^2-5}{y^2-1}=(y^2-1)\frac{y^2+y+2}{y+1}$$

Cancelling out the common factor of $y^2-1$, we get:

$$y(y^2-1)+(y^2-5)=(y^2-1)\frac{y^2+y+2}{y+1}$$

However, this is not the correct solution. We need to choose a different factor that will cancel out both denominators.

Multiplying by $y+1$

Let's try multiplying both sides of the equation by $y+1$.

$$(y+1)\left(y+\frac{y^2-5}{y^2-1}\right)=(y+1)\left(\frac{y^2+y+2}{y+1}\right)$$

Simplifying the left-hand side, we get:

$$(y+1)y+(y+1)\frac{y^2-5}{y^2-1}=(y+1)\frac{y^2+y+2}{y+1}$$

Cancelling out the common factor of $y+1$, we get:

$$y(y+1)+(y^2-5)=(y^2+y+2)$$

Expanding the left-hand side, we get:

$$y^2+y+(y^2-5)=(y^2+y+2)$$

Simplifying the equation, we get:

$$2y^2-4=y^2+y+2$$

Subtracting $y^2+y+2$ from both sides, we get:

$$y^2-3y-6=0$$

This is a quadratic equation that can be solved using the quadratic formula.

Solving the Quadratic Equation

The quadratic formula is:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

In this case, $a=1$, $b=-3$, and $c=-6$. Plugging these values into the formula, we get:

$$y=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(-6)}}{2(1)}$$

Simplifying the expression, we get:

$$y=\frac{3\pm\sqrt{9+24}}{2}$$

$$y=\frac{3\pm\sqrt{33}}{2}$$

Therefore, the solutions to the equation are:

$$y=\frac{3+\sqrt{33}}{2}$$

$$y=\frac{3-\sqrt{33}}{2}$$

Conclusion

In this article, we solved a rational equation by multiplying both sides by a suitable expression. We analyzed the options given and chose the correct factor to cancel out the denominators. We then simplified the equation and solved the resulting quadratic equation using the quadratic formula. The solutions to the equation are:

$$y=\frac{3+\sqrt{33}}{2}$$

$$y=\frac{3-\sqrt{33}}{2}$$

Final Answer

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Introduction

Rational equations are a type of algebraic equation that involves fractions with polynomials in both the numerator and denominator. Solving these equations can be challenging, but with the right approach, it can be done efficiently. In this article, we will provide a Q&A guide to help you understand the steps involved in solving rational equations.

Q: What is a rational equation?

A: A rational equation is an equation that involves fractions with polynomials in both the numerator and denominator.

Q: How do I solve a rational equation?

A: To solve a rational equation, you need to eliminate the fractions by multiplying both sides of the equation by a suitable expression. This expression should contain both the denominators of the fractions.

Q: What are the steps involved in solving a rational equation?

A: The steps involved in solving a rational equation are:

1. Identify the denominators of the fractions.
2. Choose a suitable expression to multiply both sides of the equation by.
3. Multiply both sides of the equation by the chosen expression.
4. Simplify the equation.
5. Solve the resulting equation.

Q: How do I choose the correct expression to multiply both sides of the equation by?

A: To choose the correct expression, you need to analyze the denominators of the fractions and choose an expression that contains both of them.

Q: What are some common mistakes to avoid when solving rational equations?

A: Some common mistakes to avoid when solving rational equations are:

• Not eliminating the fractions by multiplying both sides of the equation by a suitable expression.
• Choosing an expression that does not contain both denominators.
• Not simplifying the equation after multiplying both sides by the chosen expression.
• Not solving the resulting equation.

Q: Can you provide an example of a rational equation and its solution?

A: Let's consider the equation:

$$y+\frac{y^2-5}{y^2-1}=\frac{y^2+y+2}{y+1}$$

To solve this equation, we need to eliminate the fractions by multiplying both sides by a suitable expression. Let's choose the expression $y+1$.

$$(y+1)\left(y+\frac{y^2-5}{y^2-1}\right)=(y+1)\left(\frac{y^2+y+2}{y+1}\right)$$

Simplifying the left-hand side, we get:

$$(y+1)y+(y+1)\frac{y^2-5}{y^2-1}=(y+1)\frac{y^2+y+2}{y+1}$$

Cancelling out the common factor of $y+1$, we get:

$$y(y+1)+(y^2-5)=(y^2+y+2)$$

Expanding the left-hand side, we get:

$$y^2+y+(y^2-5)=(y^2+y+2)$$

Simplifying the equation, we get:

$$2y^2-4=y^2+y+2$$

Subtracting $y^2+y+2$ from both sides, we get:

$$y^2-3y-6=0$$

This is a quadratic equation that can be solved using the quadratic formula.

Q: What is the quadratic formula?

A: The quadratic formula is:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Q: How do I use the quadratic formula to solve a quadratic equation?

A: To use the quadratic formula, you need to plug in the values of $a$, $b$, and $c$ into the formula.

Q: Can you provide an example of using the quadratic formula to solve a quadratic equation?

A: Let's consider the quadratic equation:

$$y^2-3y-6=0$$

To solve this equation using the quadratic formula, we need to plug in the values of $a$, $b$, and $c$ into the formula.

$$y=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(-6)}}{2(1)}$$

Simplifying the expression, we get:

$$y=\frac{3\pm\sqrt{9+24}}{2}$$

$$y=\frac{3\pm\sqrt{33}}{2}$$

Therefore, the solutions to the equation are:

$$y=\frac{3+\sqrt{33}}{2}$$

$$y=\frac{3-\sqrt{33}}{2}$$

Conclusion

In this article, we provided a Q&A guide to help you understand the steps involved in solving rational equations. We discussed the common mistakes to avoid and provided an example of a rational equation and its solution using the quadratic formula. We hope this guide has been helpful in understanding the concept of rational equations and solving them efficiently.