Problem 70 - Algebra 2

Question

Solve each equation. Check your solutions. $$ \frac{1}{y+1}-\frac{3}{y-3}=2 $$

Step-by-Step Solution

Answer

The solutions are \(y = 0\) and \(y = 1\).

Step 1: Write the Equation

We are given the equation: \( \frac{1}{y+1} - \frac{3}{y-3} = 2 \).

Step 2: Find a Common Denominator

To eliminate the fractions, find a common denominator, which is \((y+1)(y-3)\).

Step 3: Multiply Through by the Common Denominator

Multiply every term in the equation by \((y+1)(y-3)\) to clear the fractions:\[(y+1)(y-3) \cdot \frac{1}{y+1} - (y+1)(y-3) \cdot \frac{3}{y-3} = 2(y+1)(y-3)\].

Step 4: Simplify the Equation

Simplify by canceling terms:\[(y-3) - 3(y+1) = 2(y+1)(y-3)\]Expanding gives:\[y - 3 - (3y + 3) = 2(y^2 - 3y + y - 3)\].

Step 5: Simplify Further

Continue simplifying:\[ y - 3 - 3y - 3 = 2(y^2 - 2y - 3) \].Combine like terms:\[ -2y - 6 = 2(y^2 - 2y - 3)\].

Step 6: Expand the Right Side

Distribute the 2 on the right side:\[ -2y - 6 = 2y^2 - 4y - 6 \].

Step 7: Arrange in Polynomial Form

Move all terms to one side to set the equation to 0:\[ 0 = 2y^2 - 2y \].

Step 8: Factor the Quadratic

Factor out the common term among the remaining terms:\[ 0 = 2y(y - 1) \].

Step 9: Solve for y

Set each factor equal to zero: 1. \(2y = 0\) implies \(y = 0\). 2. \(y - 1 = 0\) implies \(y = 1\).

Step 10: Check Solutions

Check both solutions in the original equation:- For \(y = 0\): \( \frac{1}{0+1} - \frac{3}{0-3} = 1 - (-1) = 2\), so \(y = 0\) is valid.- For \(y = 1\): \( \frac{1}{1+1} - \frac{3}{1-3} = \frac{1}{2} - (-\frac{3}{2}) = 2\), so \(y = 1\) is also valid.

Key Concepts

Rational Equations

Rational equations are equations that involve at least one fraction whose numerator and/or denominator is a polynomial. These equations can seem intimidating, but they're quite manageable once you know the right steps.

• Recognize that a rational equation involves fractions with variables in the denominator.
• Remember that the key objective is to remove the fractions to simplify the equation-solving process.
• This is usually achieved by finding a common denominator for all the fractions involved.

In our example, \[\frac{1}{y+1} - \frac{3}{y-3} = 2\], the left side consists of two rational expressions. The equation can be solved by eliminating these fractions through a common denominator.

Polynomial Factoring

Polynomial factoring is an essential skill in algebra, particularly when solving equations with polynomials. It involves breaking down a complex expression into simpler components, known as factors.

• Factoring transforms equations into products of terms that allow us to easily find the roots.
• Start by identifying common factors, such as a constant or a variable, shared by all terms.
• In our problem, the factored form of a particular polynomial might be like \[x^2 - 3x = x(x - 3)\].

In context, \[0 = 2y(y - 1)\], shows the polynomial is factored, enabling us to solve for the variable by setting each factor equal to zero.

Solving Equations

Solving algebraic equations involves isolating the variable to determine its value. This process requires careful manipulation of the equation, ensuring all operations are performed correctly.

• Begin by getting rid of any fractions, often by multiplying through by a common denominator.
• Rearrange the equation to consolidate like terms on one side while keeping the constants on the opposite side.
• Simplify the expression until you can clearly identify the value of the variable.

For instance, in our solution, after simplification we have \[-2y - 6 = 2y^2 - 4y - 6\], then arranged it to set the equation to zero.

Common Denominator

The common denominator is crucial in solving rational equations. It refers to a common multiple of the denominators of various fractions, enabling you to combine or eliminate fractions.

• To find it, determine the least common multiple (LCM) of all the denominators in the equation.
• Multiplying every term by this common denominator eliminates the fractions.
• This step simplifies the equation significantly, allowing you to solve for the unknown variable as if it were a simple polynomial expression.

In our case, \[(y+1)(y-3)\] was chosen as the common denominator, effectively turning our rational equation into a polynomial for simpler resolution.