Problem 28 - Algebra and Trigonometry

Question

The given equation is either linear or equivalent to a linear equation. Solve the equation. \(\frac{2}{3} y+\frac{1}{2}(y-3)=\frac{y+1}{4}\)

Step-by-Step Solution

Answer

The solution is \( y = \frac{21}{11} \).

Step 1: Expand the Equation

To start, let's open up the parentheses by distributing the coefficients across the terms inside. The equation becomes: \( \frac{2}{3}y + \frac{1}{2}y - \frac{3}{2} = \frac{y + 1}{4} \).

Step 2: Find a Common Denominator

In order to simplify the equation, we will find a common denominator for all the terms involved. The denominators are 3, 2, and 4. The least common multiple of these is 12. This gives \( \frac{8}{12}y + \frac{6}{12}y - \frac{18}{12} = \frac{3y + 3}{12} \).

Step 3: Eliminate the Denominators

To eliminate the fractions, multiply every term by 12 (the common denominator). This results in: \( 8y + 6y - 18 = 3y + 3 \).

Step 4: Combine Like Terms

Combine like terms in the equation. \( 14y - 18 = 3y + 3 \) becomes \( 14y = 3y + 21 \).

Step 5: Transpose Terms

Move terms involving \( y \) to one side and constant terms to the other. Subtract \( 3y \) from both sides: \( 11y = 21 \).

Step 6: Solve for y

Finally, solve for \( y \) by dividing both sides by 11: \( y = \frac{21}{11} \).

Key Concepts

Common Denominator

Working with fractions often requires finding a common denominator in order to combine or compare them. A common denominator is a shared multiple of the denominators you have. For instance, in the equation \( \frac{2}{3}y + \frac{1}{2}y - \frac{3}{2} = \frac{y+1}{4} \), the denominators are 3, 2, and 4.
To find a common denominator, it's helpful to find the least common multiple (LCM). In this case, the LCM of 3, 2, and 4 is 12. You can then rewrite each fraction so that they all share this common denominator. This means converting:

• \( \frac{2}{3} \) to \( \frac{8}{12} \)
• \( \frac{1}{2} \) to \( \frac{6}{12} \)
• \( \frac{3}{2} \) to \( -\frac{18}{12} \)
• \( \frac{y+1}{4} \) to \( \frac{3y+3}{12} \)

By having a common denominator, you can more easily eliminate the fractions from the equation and focus on combining terms.

Solving Equations

Solving equations is about finding unknown values that satisfy the equation. In our original problem, after expanding and applying the distributive property, we get:

• \( \frac{8}{12}y + \frac{6}{12}y - \frac{18}{12} = \frac{3y+3}{12} \)

Once all terms have a common denominator, multiply through by this denominator to eliminate it and work with simpler whole numbers: \( 8y + 6y - 18 = 3y + 3 \).
Move all terms involving the variable \( y \) to one side and constant terms to the other side. Simplify by combining like terms, moving all \( y \) terms to one side, and constants to the other. Finally, you divide by the coefficient of the variable to solve for \( y \).

• Subtract \( 3y \) from both sides to get: \( 11y - 18 = 3 \)
• Then add 18 to both sides: \( 11y = 21 \)
• Finally, divide each side by 11 to find \( y = \frac{21}{11} \)

The result is the value for \( y \) that makes the original equation true.

Like Terms

Like terms in algebra are terms that have the same variable raised to the same power, even if their coefficients are different. When solving an equation, combining like terms helps to simplify the equation by reducing the number of terms.
For the equation \( 8y + 6y - 18 = 3y + 3 \), focus on the terms involving \( y \). The terms \( 8y \) and \( 6y \) are like terms because they both involve \( y \). Therefore, you can combine them:

• \( 8y + 6y \rightarrow 14y \)

By combining like terms, the equation becomes less complex and easier to solve. Remember to also look for like constant terms, though in this step, the focus was on \( y \)-terms.
Ultimately, combining like terms lays the groundwork for getting the unknown by itself on one side of the equation.

Distributive Property

The distributive property is a fundamental algebraic principle used to remove parentheses and simplify expressions. It states that \( a(b + c) = ab + ac \).
In the given equation \( \frac{2}{3}y + \frac{1}{2}(y-3) = \frac{y+1}{4} \), the distributive property is applied to the term \( \frac{1}{2}(y-3) \):

• \( \frac{1}{2} \times y \) results in \( \frac{1}{2}y \)
• \( \frac{1}{2} \times -3 \) results in \( -\frac{3}{2} \)

Thus, \( \frac{1}{2}(y-3) \) becomes \( \frac{1}{2}y - \frac{3}{2} \).

Using the distributive property allows you to write expressions without parentheses, making it easier to handle equations and further simplify them by combining like terms. This critical step is part of many algebraic manipulations, ensuring that you correctly expand and manage each component of an equation.